Genética 1
Guía de Problemas Segundo Cuatrimestre de 2007
Profesores responsables: Dres. Esteban Hopp, Beatriz Saidman y Liliana Mola
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2
1. CONCEPTOS Y TÉCNICAS BÁSICOS DE GENÉTICA MOLECULAR Bibliografía: Repaso IBMyC, Alberts, Griffiths Cap 1 y Recombinant DNA (Watson et al.).
Guía teórica:
1) ¿Qué son, de dónde se obtienen y para qué sirven las enzimas (mejor llamadas endonucleasas) de restricción? ¿Cómo hacen las bacterias que producen ER para evitar que su genoma se vea afectado por las mismas? ¿Qué tipos de especificidad presentan? ¿Qué tipos de extremos pueden generar en el ADN sustrato? R: Las endonucleasas de restricción son producidas por bacterias como mecanismo de defensa contra los fagos. El genoma se encuentra protegido por metilación específica del sitio de restricción mediante una metilasa con igual especificidad que la endonucleasa. Cortan el ADN en sitios específicos (secuencias definidas de ADN de 4 ó 6 bases, generalmente). Algunos dejan extremos cohesivos, otros dejan extremos romos. Siempre el sitio blanco (de reconocimiento) es un palíndrome, aunque el sitio de corte puede no coincidir con el sitio de reconocimiento. Los fragmentos generados se pueden separar fácilmente mediante electroforesis en geles de agarosa o poliacrilamida. La velocidad de los fragmentos depende de su longitud: los más pequeños migran más rápido. 2) ¿En qué consiste la técnica de secuenciación de Sanger? ¿Qué trascendencia tuvo y tiene en Genética y Genómica? ¿Porqué desplazó a la técnica de Maxam & Gilbert? ¿Cuál es la ventaja de la secuenciación automática y en qué se diferencia de la manual? R: En la secuenciación manual, se suele marcar con radioisótopos y se corren 4 calles por secuencia (una para cada nucleótido). Se revela por autoradiografía. Secuenciación de Sanger (ver esquema de libro): secuenciación por dideoxinucleótidos - ADN simple cadena - Oligonucleótido iniciador ( primer ) - Incubación c/ 4 diferentes dideoxinucleótidos (uno para cada calle) - (Mono)deoxinucleótidos (uno de ellos marcado, usualmente α[32P]dCTP) - Electroforesis en gel de poliacrilamida Existen métodos no radiactivos (como tinción con nitrato de plata), pero son menos utilizados. En la secuenciación automática se marca con fluorocromos
secuenciación masiva de genes y genomas posibilitando el advenimiento de la genómica y proyectos como el genoma humano. La técnica de M&G es un método degradativo y tiene 2 reacciones “de terminaciones de nucleótidos” ines pecíficas (revela 2 nucleótidos y el verdadero se saca por comparación con otra calle del gel). Esta razón dificulta, además, la automatización. Por otro lado resultan más difíciles de controlar las condiciones. 3) ¿En qué consiste la PCR de punto final? ¿Es una técnica cualitativa o cuantitativa? ¿Qué diferencias existen entre PCR de punto final y PCR de tiempo real? ¿Es posible hacer una PCR a partir de RNA? ¿Cómo? ¿Puede utilizarse para diagnóstico (la detección de HIV, por ejemplo)? ¿Qué ventajas o desventa jas tendría respecto a otras técnicas diagnósticas como el ELISA? R: PCR: polymerase chain reaction (esquema del Recombinant DNA de Watson et al.) - Utiliza la Taq polimerasa: Thermus aquaticus - Permite amplificar segmentos específicos de ADN. - No requiere digestión del ADN. - No requiere clonar el segmento - Se requiere poca cantidad de ADN. Es una técnica fundamentalmente cualitativa, debido a que llega a una meseta de amplificación por limitantes que no suelen ser la concentración del templado. La PCR en tiempo real va siguiendo la cinética de amplificación permitiendo comparar las fases logarítmicas de amplificación entre sí y así cuantificar con precisión los templados frente a una referencia o estándar. Para realizar una PCR a partir de RNA se debe hacer RT-PCR (se llama igual que la otra –una por RetroTranscriptasa y la otra por Real Time). Se realiza la primera reacción con retrotranscriptasa y después se continua con Taq pol. Es posible utilizar PCR para diagnóstico y tiene como ventaja que se trata de un método directo (detecta al virus y no los anticuerpos con los que reacciona el paciente). Por eso sirve para la detección en recién nacidos o personal médico de alto riesgo que trabaja con pacientes, antes de los 6 meses que tarda en detectarse una respuesta inmunológica. No da falsos positivos (salvo por contaminación de amplicones: un pro blema general de las PCRs). Desventajas: más com plejo, caro y menos automatizable que un ELISA. Hay que purificar RNA e implica mayor exposición al virus por parte del personal que hace el diagnóstico. 4) ¿En qué consiste un Southern blot? ¿y el northern? ¿y el western? ¿Para qué se usan y qué tipo de molé-
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3
- Se parte de ADN (genómico de alto peso molecular en el caso de los RFLP) y se lo digiere con enzimas de restricción. - Los fragmentos de ADN se separan por electroforesis en gel de agarosa. - Se transfiere a un filtro de nylon o nitrocelulosa y se desnaturaliza el ADN. - Se incuba con una sonda marcada específica de la secuencia en estudio. - Lavado y exposición del filtro. - Detección autoradiográfica o fluorográfica. Esta técnica permite determinar: a) Presencia y número de sitios de corte con una determinada enzima de restricción en un fragmento de ADN dado y establecer la existencia de polimorfismos para fines diagnósticos o de estudio genético (RFLP). b) Número de copias (aproximado) del gen en el genoma, por ejemplo cuando se hacen transgénicos. Los RFLP surgen de comparar los patrones de restricción de Southern blot entre distintos individuos, especies, etc. observando diferencias (polimorfismos) en sitios de corte. Aplicaciones de los RFLP: - mapeo genético - diagnóstico prenatal - diferenciación de individuos ( fingerprinting ) - estudios evolutivos Northern blotting: - El ARN celular total o mRNA se separa por tamaño utilizando electroforesis en gel de agarosa (en presencia de un agente desnaturalizante como el formaldeído para evitar la formación de estructuras secundarias en el ARN durante la corrida). - Se transfiere a un filtro de nitrocelulosa o a un papel tratado químicamente. - Hibridación con una sonda apropiada marcada seguida por autoradiografía. Se revelan bandas que indican el número y tamaño de los tipos de ARN com plementarios a la sonda. Permite cuantificar aproximadamente y evaluar niveles de expresión de genes (para mejor precisión se usa RT-PCR en tiempo real). También es útil, entre otras cosas, como auxiliar del clonado de cDNA porque el tamaño de un mRNA específico puede compararse con el tamaño de los cDNA copiados de ese mRNA, y revela si este cDNA está completo. En el Western blotting se transfieren proteínas corridas en un gel de poliacrilamida a una membrana de nitrocelulosa o PVDF y luego se las visualiza por su reacción con anticuerpos específicos. La PCR ha desplazado, paulatinamente, al Southern
gnósticos. En algunos casos, se puede tener una sonda pero no conocer la secuencia nucleotídica para diseñar los iniciadores, por lo que se puede hacer un Southern si no se tiene la información como para hacer PCR. Northern y RT-PCR, ídem que con Southern. La PCR “común” (denominada de punto final) no es cuantitativa, para ello se debe hacer PCR de tiempo real cuantitativa (qRT-PCR). El western es más específico que el ELISA (se identifica la banda). No da falsos positivos. Es más comple jo y caro. Se utiliza para confirmación del HIV, por ej. 5) Indique si las siguientes afirmaciones son verdaderas o falsas y justifique su respuesta: a) Un experimento de northern requiere desnaturalización de la doble cadena de ADN. b) En una reacción de PCR estándar se utilizan dos primers complementarios a la cadena codificante. c) La secuenciación por el método de Sanger no requiere, en ningún paso la desnaturalización de la do ble cadena del ADN. d) Al leer una autoradiografía en una secuenciación de Sanger el extremo 5´ del fragmento secuenciado se encuentra más cerca del polo positivo y el extremo 3´ más cerca del polo negativo. R: a) FALSO. Se trata de ARN de simple cadena el que se somete a electroforesis. b) FALSO. Se utilizan dos primers complementarios a las dos cadenas. c) FALSO. En esta secuenciación se requiere la desnaturalización del ADN. d) VERDADERO: Dado que es un método de secuenciación por síntesis, los fragmentos más pequeños corresponderán a la región 5´ del fragmento a secuenciar. Estos fragmentos pequeños migran más y lo hacen hacia el polo positivo (ya que el ADN tiene carga negativa).
Problemas:
1) Suponiendo un alineamiento al azar de nucleótidos en un fragmento de ADN, ¿cuál es la probabilidad de que una endonucleasa de restricción cuyo sitio de reconocimiento consta de cuatro bases pueda cortar el ADN? ¿y para una enzima de seis bases? ¿y para una de ocho? R: A T C G 4 bases ¼ X ¼ X ¼ X ¼* ( ¼ )4 = 3.9 10-3 *(probabilidad de que c/base se encuentre y que al lado se encuentre una base cualquiera) 6 bases ( ¼ ) 6 = 2.4 10-4 8 bases ( ¼ ) 8 = 1.5 10-5 Fórmula general: (1/4) n, donde n = longitud de la secuencia que se busca.
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4
que tienen las defensas inmunológicas disminuidas. a) ¿Cuál será el patrón de bandas del RFLP para un Ese fragmento de ADN, que tiene 8 kpb de longitud, individuo A1A1 utilizando la sonda 1? ¿Y para uno se inserta en el plásmido bacteriano pBR322 en su A2A2? ¿Y para el heterocigota? único sitio para EcoRI. El plásmido recombinante es b) ¿Y con la sonda 2? clivado por dos endonucleasas de restricción (por se parado o juntas) y luego es sometido a electroforesis R: Sonda 1 Sonda 2 en un gel de agarosa y teñido con bromuro de etidio. En la figura se muestra el patrón de bandas (visualizaA1A1 A2A2 A1A2 A1A1 A2A2 A1A2 das por medio de luz UV) originado por cada uno de los tratamientos: 6 kpb 6 kpb A) EcoRI
B) BamHI
C) E + B
4 kpb 2 kpb
8 kpb 6 kpb
4 kpb
5,5 kpb 4,5 kpb 4 kpb
4 kpb 3,5 kpb 3 kpb 2,5 kpb 1 kpb
4) Un investigador está estudiando gen bacteriano que presenta tres dominios y desea introducirle mutaciones a dicho gen con el fin de estudiar su funcionalidad. Para ello parte del gen, del cual conoce su seAl realizar Southern blots de los geles "B" y "C", ¿qué cuencia completa, clonado en un plásmido para su fragmentos hibridarán con una sonda del gen de la expresión en bacterias. cápside del virus? a) ¿Cómo haría para delecionar el dominio intermedio R: del gen? EcoRI BamHI b) ¿Cómo haría para verificar que el vector generado 3 porta la deleción? c) ¿Qué elementos de secuencia debe tener el plásmi2,5 5,5 do para poder mantenerse, amplificarse y expresarse 8 BamHI en bacterias? 4 4 6 R: a) Un método posible es cortando con una misma 4,5 enzima de restricción (o dos enzimas distintas que 3,5 dejen extremos compatibles) flanqueando el dominio 1 BamHI intermedio. Luego se religa el vector al cual le faltará el domino intermedio. En caso de que no se encuenEcoRI tren dichos sitios de restricción se pueden generar me Aclaración: Las distancias que se encuentran dentro diante mutagénesis sitio dirigida. del círculo son las correspondientes a los sitios de cor- b) Para verificar que el vector porta la deleción se tes BamHI. Las subrayadas, más externas, marcan la puede hacer una PCR con primers flanqueado la sedistancia entre los dos sitios EcoR I y las que están en cuencia del dominio intermedio. Si está deletado dará negrita corresponden a las distancias entre cada sitio un producto de PCR de menor tamaño. de corte (representado por las flechas). Los fragmen- c) El vector deberá tener un origen de replicación para tos de 3, 4 y 1 kpb del gel C están dentro del segmento poder duplicarse en la bacteria y un marcador de sede 8 kpb, y estos hibridan con la sonda del gen de la lección (generalmente un gen de resistencia a antibicápside del virus. Dado que todos los fragmentos del ótico) de forma de poder seleccionar a las bacterias gel B contienen una porción del gen viral todos los transformadas. En el caso de un vector cuya finalidad fragmentos de este gel hibridarán con la sonda. sea la expresión en bacterias, deberá tener además una 3) Un investigador está estudiando un polimorfismo secuencia promotora río arriba del sitio múltiple de en el largo de los fragmentos de restricción (RFLP) clonado (región con sitios únicos de corte para muubicado en el cromosoma 7, que presenta dos alelos. chas enzimas de restricción). El sitio de corte para EcoRI indicado con la flecha 5) El interferón α está codificado por un gen que no está presente en uno de los alelos (A1) y ausente en el contiene intrones. Por clivaje con BamHI, se genera otro (A2). un único fragmento de restricción conteniendo al gen completo, que puede ser identificado en un Southern Sonda 1
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5
western blot representa ivo utilizando anticuerpos anti-interferó α. Los ind viduos 1 y 2 no son afe ctados (sanos). En base a los resultados btenidos en Southern, north rn y wester : Southern
1 2
3 4
Nor hern
5
1 2
3 4
estern
5
1 2
3
4
5
a) ¿Cuál sería la posible causa de a inmunodeficiencia en los p acientes 3, 4 , y 5? b) ¿Qué característica particular tiene el individuo 2 para el gen del interferón α? c) ¿Import a partir de qué tejido se realizó el Southern? ¿Y el Northern y el Western? R: a) Paciente 3: En b se al northern notamos que el paciente 3 no presenta mensajer o funcional. Esto podría deberse a una m tación puntual en el pr motor del gen, que inhibe su expresión. Dado que el fragmento de r estricción es de mayor amaño (cor e menos en el outhern) podría tratars de una mu tación puntual en el promoto perdiéndo e un sitio lanco para la en ima de re tricción utilizada en para el Southern. ambién pu de ser inse ción en la región promotora por lo tanto no dará mRNA. El hecho de no present r mensajer trae com consecuencia la
ausencia de pr oteína lo cual puede observarse a p rtir del western. Paciente 4: pr senta el ale lo de tamaño normal e el Southern, mensajero de ta año salvaj , pero una roteína defectuosa de meno tamaño. P dría deberse a una mutación puntual que genera un codón Stop rem turo y una p roteína de enor tamañ . Paciente 5: el tamaño del f ragmento de restricción, del m nsajero y l proteína parecieran ser normales, sin e bargo, preenta inmu odeficienci . Esto po ría ex plicarse por una mutación puntual ue provoca un ca bio amino cídico y la proteína pie de su activi dad (c mbio conformacional) o bien que el origen d la enfermedad n tenga rela ión con esta proteína. Por ej mplo podrí tratarse de una mutación en el rece tor de interferón lo cual dar ía un fenoti o similar al de lo otros paci ntes. Habría que demostrar, en un siguiente paso, i la proteín es o no fu cional. ¿Se les ocurre como h cerlo? b) El individuo 2 es hete ocigota par a dos varia tes al licas del ge del interfe ón α, mientras que los esta tes son ho ocigotas pa a uno (indi iduos 1, 4 5) u tro alelo (individuo 3). c) Dado que el Southern se hace a partir de D A genómico y t das las cél las tienen l mismo D A entonces no i porta a pa tir de qué ejido se ex rae mismo. En ca bio en el orthern y en el wester sí im porta dado que se está evaluando la expresión de lo genes, la c al puede se tejido específica (como en es e caso).
1. DIVI IÓN CELULAR b) mitosis y meiosis II c) meiosis I y meiosi s II C nsidere las iferencias tanto en el mecanismo como en los product s finales. 3) El gráfico r epresenta la variación del contenido de A N durante l ciclo vital de una célula:
Bibliogra ía: ía: •
•
Alberts y col. Cap. 17. Secciones: La estrategia general del ciclo celular, Cap. 20 Células germinales y fertilización: Beneficios del sexo, M eiosis Griffiths y col Cap. 3: Mitosis y meiosis, Compor-
¿Qué ocurr e en el inter alo de tiem o de 2 a 3?
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6
cromosomas y cromátidas en los siguientes períodos del ciclo celular en la especie humana? Marque con H ó D si la célula es haploide o diploide.
8) Otra cigota posee un complemento de dos pares de cromosomas homólogos A y a, B y b: a) ¿cuáles de los siguientes complementos cabría es perar en las células somáticas durante el crecimiento: valor Nºcromo- Nº cromá- H/D AaBB, AABb, AABB o aabb? “C” somas tidas b) ¿Podría encontrarse más de una combinación? G1 premitótica Cuando este individuo llegue a adulto: c) ¿cuáles de las siguientes combinaciones de cromoG2 premitótica somas cabría esperar en las gametas?: Profase mitótica Aa, AA, aa, Bb, BB, bb; Metafase mitótica Aa, Bb; A, a, B, b; AB, Ab, aB, ab; Telofase mitótica (en Aa, Ab, aB, Bb. c/núcleo) 9) Dibuje la configuración de los bivalentes en PaquiEspermatozoide tene, *Diplotene, Diacinesis y Metafase I de una espeProfase meiótica I cie con 2n= 4, donde uno de los pares es metacéntrico leptoteno y muestra regularmente durante toda la meiosis un Metafase meiótica I quiasma intersticial a cada lado del centrómero; el Anafase meiótica I otro es un acrocéntrico con un único quiasma intersticial en el brazo largo. Use lápices de colores para diMetafase meiótica II ferenciar a los dos homólogos. Telofase meiótica II 10) Los conceptos de división reduccional o división (cada polo) ecuacional se refieren a: Interfase prea) fenómenos citológicos o genéticos? meiótica (antes S) Interfase preb) ¿Hay paralelismo entre dichos aspectos citológicos meiótica (después S) y genéticos? Si se tiene un par de cromosomas homólogos acrocén#5) Albert y Swift (1953) encontraron las siguientes tricos donde donde se ubican los genes A y a: cantidades relativas de ADN (en unidades arbitrarias) c) ¿En qué estadio efectúa reducción el segmento que por núcleo en varios tipos de células del anélido Sabe- contiene el gen A si ocurre un intercambio en la rellaria: gión I? a) Primer cuerpo polar: 127 + 3 d) Realice el mismo razonamiento si el intercambio b) Esperma: 61 + 1 ocurre en II. c) Profase mitótica: 263 + 10 I II d) Telofase mitótica: 124 + 3 __________ A ____________ Explique estas diferencias. _______________________ 6) Una especie animal tiene 2n cromosomas: ¿Qué A proporción de las gametas formadas tendrán los __________ a ____________ centrómeros: _______________________ i) de origen paterno únicamente; I a II ii) de origen materno únicamente; #10) En qué organismos esperaría la mayor variabiliiii) de origen materno y paterno simultáneamente? 7) Una cigota posee un 2n = 8; en el primer par de dad: cromosomas homólogos uno de los miembros presen- a) en los que se reproducen asexualmente ó en los que ta un abultamiento ("knob") intersticial; en el segundo lo hacen sexualmente? par uno de los miembros posee un satélite en posición b) Suponga que el proceso meiótico nunca hubiera terminal, en el tercero uno de los cromosomas presen- aparecido en la evolución biológica: ¿cuál piensa Ud. ta un segmento heterocromático terminal y el cuarto que habría sido la consecuencia? par es acrocéntrico. Realice un esquema de las si- c) En las especies animales que poseen una meiosis masculina aquiasmática, ¿hay recombinación genética guientes etapas de la mitosis y meiosis. en esas especies? a) metafase mitótica 11) IMPORTANTE La dirección b) metafase de meiosis I www.biologia.arizona.edu/cell/tutor/mitosis/01q.html c) metafase de meiosis II
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7
2. LEYES DE MEN EL Y EXTENSIONES DEL ANÁLISIS
Bibliografí a: Griffiths col. Cap. 2 y 4 Guía teóri a 1) Señale la importancia del concepto de segregación, por u lado, e in ependencia, por el otro , de la herencia de caracteres erivada de las leyes de Mendel. ¿Cuál es la impor ancia de la diploidía e todo esto? ¿Serí lo mismo i las arveja fueran mo oploides en vez e diploides R: Además de repasar as leyes de Mendel, se resalta que la segr egación e independenci de la herencia de los caracter es es, valga la redunda cia, indepe diente de si los or ganismos so n haploides o diploides y permite la rec mbinación, entendida omo nueva com binaciones de alelos e distintos genes. Se puede hacer, com ejercicio, ué hubiese pasado con as generaciones paternales, “F1” y “F2” si las arvej s fueran haploides. Ahora bien, la segre ación de al los se ve enmasc rada en los diploides por las interacciones de dominancia-recesivi ad. Por eso , antes de endel, no se le en ontraba lógica a que 2 i dividuos d fenotipo dominante, tuvier n descendi ntes con fenotipo recesivo, por ejemplo. Destacar la importancia del análisis nu érico-estadístico para stablecer estas leyes y que pocos aplicar n previame te. 2) ¿Por q é procesos recombina los genes en la meiosis? R: Después de Mend l se pudie on incorpo ar los conceptos erivados d los avances en citog nética (teoría cro osómica de la herenci ) y reformu lar, en su acepció moderna ue “los ge es situados sobre diferentes romosomas recombina mediante la segregación l azar de los cromosomas en la an fase y los genes situados en el mismo cro osoma se r ecom binan por edio del ent recruzamiento”. 3) a) ¿Qué es un mono íbrido? ¿Q é es un híbr ido en sentido agr onómico o comercial ( or ej. un íbrido superrendi or para Clarín Rural)? b) ¿Qué si nifica hacer un cr zamiento dihíbrido o tri híbrido? R :a) :a) Monohíbrido es s inónimo de heterocigota para un único g en. Un híb ido comercial es también un heterocigota (pero con obredomin ncia para el carác-
ENDEL ANO I
La retrocruza s un cruzam iento entre l heterocigota (F1) o cualqui ra de las sig uientes gen raciones y cualquiera de los padres o riginales. 5) En qué con iste la prue a de compl mentación? R: Cruzamien o entre dos mutantes h mocigotas con fe otipo simil r (por ejem plo albino) ara determinar si la F1 tiene fenotipo sal vaje (indicativa de que las m taciones están localizadas en distintos genes). E s el caso típico de una vía etabólica d varios pa os. Tanto si hay na mutació para el primer paso d la ví , como par el segundo, la consecu ncia es que no se formará el producto fi al (por eje plo antocianas de color rojo, dando albinismo). Si se cruza una uta te para el g n codificante para la pri era enzim de la vía metaból ca con otro para la seg nda, la F1 será heterocigota para ambos enes. Com generalmente lo alelos salvajes (enzim normal ac iva) son do minantes, las en zimas sinte izadas se complementarán
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8
Benzer para establecer las “distancias” mínimas o problemas que se refieren a estos ejemplos dado que, “unidades” de recombinación en sus experimentos con experimentalmente, cuando se obtiene una progenie fagos. Hoy sabemos que esta distancia o unidad es el segregante, normalmente no se sabe de antemano si el nucleótido. comportamiento es de tipo epistático y menos aún a 6) ¿Qué ventajas tiene el uso de caracteres, genes o qué variante de epístasis se corresponde. En este conmarcadores codominantes (respecto a los dominantes) texto, resulta útil disponer del cuadrito de abajo que en un análisis genético de segregación? muestra algunos ejemplos del tipo de proporciones R : Se puede distinguir los heterocigotas de los homo- fenotípicas posibles. cigotas, por lo que se dispone de mayor información Problemas: (estadística) en una Tipo de interacción génica 9 3 3 1 Razón F2, por ejemplo. ReA-B- A-bb aaB- aabb fenotípica trocruzar F1 con el Ninguna (cuatro fenotipos distintos) 9 3 3 1 9:3:3:1 doble recesivo (cruAcción génica complementaria 9 7 9:7 zamiento prueba), en Supr Supres esió iónn domi domina nant ntee de A sobr sobree el alel aleloo domi domina nant ntee B 12 3 1 13:3 13:3 vez de autofecundar. Epístasis recesiva de aa sobre los alelos B y b 9 3 4 9:3:4 7) ¿Cuál es la base Epístasis dominante de A sobre los alelos B y b 12 3 1 12:3:1 molecular de las inGenes duplicados 15 1 15:1 teracciones génicas, dominancia, semidominancia, codominancia, sobre- 1) Si dispone de las siguientes cepas de Drosophila iii) bw_vg ; dominancia (=heterosis o vigor híbrido), y dominancia i) bw e ; ii) salvaje y bw e bw vg incompleta? R: Repasar concepto un gen-un polipéptido y qué im- y desea comprobar las leyes de Mendel, indique: plica a nivel de fenotipo (ej. grupos sanguíneos). Es a) Todos los cruzamientos que le permitirían hacerlo y la segregación de qué carácter (o caracteres) estudiaría decir, en la codominancia se expresan los dos alelos. La sobredominancia implica que la F1 supera en el en cada caso. carácter estudiado (usualmente rendimiento agronó- b) ¿Es equivalente el cruzamiento directo al recíproco? mico) a ambos padres. Dominancia incompleta: el heterocigota presenta ras- c) ¿Deben usarse hembras vírgenes en todos los cagos intermedios entre cada tipo de homocigota, no es sos? 2) Si dos pares de genes A:a y B:b se transmiten inigual a ninguno de ellos. dependientemente y se sabe que A es dominante sobre 8) Normalmente, de un carácter autosómico doa y B dominante sobre b, ¿cuál es la probabilidad de minante relacionado con alguna enfermedad se obtener: espera que cada individuo afectado tenga al meAaBb? nos un progenitor también afectado. Pero, ¿puede a) una gameta AB a partir de un individuo AaBb? Bb? b) una gameta Ab a partir de un individuo AA Bb? suceder que ambos padres no manifiesten manifiesten la enc) una cigota AABB a partir de aabb X AABB? AABB? fermedad? d) un fenotipo AB a partir de AaBb X AaBb? AaBb? R: Penetrancia (incompleta) e) un fenotipo AB a partir de AABB X aabb? aabb? f) un fenotipo aB a partir de AaBb X AaBB? AaBB? 3) Considere el cruzamiento AaBbCcDdEe x aa aaBbccDdee. aaBbccDdee. (a) ¿Qué proporción de la descendencia será fenotípicamente como: 1) el primer parental, 2) el segundo parental, 3) cualquiera de los dos parentales y
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9
ficados de acuerdo con la disminución de intensidad fueron los siguientes: 92 rojas, 30 cremas, 41 blancas. del color de pelo de los homocigotas) son: Explique. C = color completo (tipo salvaje) 10) En el ratón hay un alelo mutante que causa encorvamiento de la cola. A partir de los resultados de los cch= chinchilla cruzamientos que se indican a continuación deduzca cd = dilución extrema el tipo de herencia de este carácter: c = albino a) ¿Es recesivo o dominante? Las relaciones de dominancia son C> cch> cd> c. b) ¿Es autosómico o ligado al sexo? a) Haga un esquema de un cruzamiento entre un ratón c) ¿Cuáles son los genotipos de los progenitores y de salvaje heterocigota para dilución extrema y un ratón las progenies en cada uno de los cruzamientos? chinchilla heterocigota para el albinismo. Cruza Progenitores Progenie b) Otro gen que afecta el color del pelo en el ratón Madr Madree Padr Padree Hija Hijass Hijo Hijoss tiene su locus en otro cromosoma, y presenta los aleNormall Curva Curva Todas Todas curvas curvas Todas Todas normal normales es 1 Norma los B para pelo negro (tipo salvaje) y b para castaño. 0,5 curvas 0,5 curvas Curva Normal 2 Ampliar el esquema anterior incluyendo el dato de 0,5 normales 0,5 normales que los dos ratones que se cruzan son heterocigotas Curva Curva Normal Normal Todas Todas curvas curvas Todas Todas curvas curvas 3 para ese gen. Norma Normal Normal Todas Todas normale normaless Todas Todas normal normales es 4 Norma 7) Un fitopatólogo llamado Flor vio que algunas vaCurva Curva Todas curvas Todas curvas 5 riedades de lino presentan diferente resistencia a dis0,5 curvas tintas razas de un hongo, Melampsora lini. La varieCurva Curva Todas curvas 6 0,5 normales dad de lino 770B es resistente a la raza 24 y suscepti ble a la raza 22, mientras que la raza Bombay es resis- 11) En Drosophila se aparearon 2 moscas de fenotipo tente a la 22 y susceptible a la 24. Si se cruzan las va- "alas curvadas" ( Curly) y "setas anormales" ( Stubble). riedades 770B y Bombay, el híbrido obtenido es resis- Se analizó un gran número de descendientes adultos y CySb: 2 CySb+: 2 tente a ambas razas 22 y 24. Cuando autofecundó la la proporción fue la siguiente: 4 CySb: F1 se produjo una F2 con las siguientes proporciones Cy+Sb Sb:: 1 Cy+Sb+. Explique estos resultados. fenotípicas: 12) El color del pelaje en los perros depende de la acRAZA 22 ción de por lo menos 2 genes. En un locus, un inhibiRAZA 24 Resistente Susceptible dor epistático dominante (I ( I) de la pigmentación evita Resistente 110 43 la expresión de los alelos del color que se ubican en otro locus de segregación independiente y en conseSusceptible 32 9 a) Proponga una hipótesis para explicar la base gené- cuencia produce pelaje blanco. Cuando se da la conditica de la resistencia del lino a estas razas particulares ción recesiva ii , los alelos del locus hipostástico pueden expresarse: iiN_ produce iiN_ produce color negro, iinn produde hongos. b) En base a su hipótesis, ¿qué número de cada una ce color café. En un cruzamiento dihíbrido para am bos loci, determine: de las cuatro categorías aparece en F 2? 2 a) proporciones fenotípicas esperadas en la progenie. c) Pruebe su hipótesis usando χ . LEYES DE MENDEL Y EXTENSIONES DEL b) probabilidad de hallar, en la progenie de color blanco, un individuo que sea homocigota para ambos loci. ANÁLISIS MENDELIANO II 8) Si el carácter antenas largas (a ( a) está ligado al sexo 13) En una especie cultivada se realizó un cruzamien(determinación XX-XY) en el mosquito Sinospica to entre plantas de flores rojas y plantas de flores rascaremus, qué probabilidad hay de obtener un mos- blancas. La totalidad de la F 1 presentó flores rojas. En la F los resultados fueron los siguientes: 92 rojas, 30
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10
emplumadas y sólo 1 sin plumas. Diagrame los cru- si los dos genes fueran en realidad uno sólo. Sin emzamientos y explique los resultados. bargo, durante la primera fase de la meiosis, se produ16) En la rata, dos parejas alélicas A:a y R :r :r interact- cen entrecruzamientos entre cromosomas homólogos úan de la siguiente forma: que resulta en un intercambio físico como lo demosA_R_ : pelaje gris. aaR_ : pelaje negro. traron Creighton y Mc Clintock en 1931 utilizando A_rr: A_rr: pelaje amarillo. aarr: aarr: pelaje color crema. marcadores citológicos combinados con marcadores Estos genotipos sólo pueden expresarse en presencia genéticos. Estos entrecruzamientos se visualizan como de un alelo dominante de un tercer gen, D, mientras quiasmas a nivel citológico. Si imaginamos una distrique su alelo recesivo d causa albinismo. Cuatro líneas bución aleatoria de los entrecruzamientos a lo largo albinas homocigotas diferentes fueron cruzadas con del cromosoma, la probabilidad de que ocurra un enuna línea pura de color gris, y estos cruzamientos pro- trecruzamiento es aproximadamente proporcional a la dujeron las correspondientes F 1 grises. Estas F 1 produ- distancia que separa a los genes en el cromosoma: jeron las siguientes F 2: cuanto más separados se encuentren, más probable es Líneas Cantidad de individuos en las distintas F 2 que ocurra un entrecruzamiento entre ellos. En consealbinas Gris Gris Am Amar aril illo lo Negr Negroo Crema Crema Albi Albino no cuencia, determinando la frecuencia de recombinantes, podemos obtener una medida de la distancia entre 1 174 0 65 0 80 los genes ligados. Esta es la base para la construcción 2 48 0 0 0 16 de mapas genéticos. 104 33 0 0 44 3 El mapeo genético se realiza mediante análisis de cru4 292 87 88 32 171 zamientos planificados, pruebas de progenie y análisis a) ¿Cuál es el genotipo de cada línea albina? b) ¿Qué tipo de interacción -si la hay- está implicada de pedigrí. Es decir, mediante el análisis comparativo de fenotipos recombinantes. El mapeo físico se realien estos cruzamientos? za mediante el análisis de clones cl ones genómicos (mediante mapeo de contigs de los cuales se conocen, por ejemIMPORTANTE: Una vez resueltos los problemas de la guía, las siguientes direcciones de Internet contie- plo, sus mapas de restricción), combinado con senen problemas que, a manera de repaso, permiten in- cuenciación nucleotídica, por hibridación in situ, u cursionar en las técnicas de autoaprendizaje con auto- otros métodos que no involucran cruzamientos y análisis de recombinantes. recombinantes. En este caso, las distancias se evaluación. http://www.biologia.arizona.edu/mendel/sets/mono/m calculan en nucleótidos o mediante medición de cromosomas. ono.html (cruza monohíbrida) http://www.biología.arizona.edu/mendel/sets/di/02Q.h 2) ¿Cuándo se dice que 2 genes están en acoplamiento o en repulsión? tml (cruza dihíbrida) R: Alelos mutantes ubicados en el mismo cromosoma o separados en los 2 cromosomas homólogos. 3. MAPEO CROMOSÓMICO Acoplamiento: AB ; Repulsión: Ab Bibliografía: Bibliografía: Griffiths o cualquier otro libro de genéab aB tica. 3) ¿Qué relación existe entre ligamiento y mapeo? Guía Teórica: R : Establecer el ligamiento y calcular la distancia 1) ¿Cómo y cuando se encontró el fenómeno de liga- permite en base a las distancias relativas de muchos miento entre genes? ¿Por qué contradice las leyes de genes establecer un mapa genético. Mendel? 4) ¿Cuál es la base física del mapeo genético? ¿Qué R: A principios del siglo XX se encontraron genes relación tienen los estudios de mapeo genético con los que no respondían a la segunda ley de Mendel. En ese entrecruzamientos y quiasmas que se observan en las
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11
plo, sus mapas de restricción), combinado con se- lejos como para que siempre exista un entrecruzacuenciación nucleotídica, por hibridación in situ, u miento entre ellos, la máxima proporción de recombiotros métodos que no involucran cruzamientos y aná- nantes será del 50%. lisis de recombinantes. Por lo tanto, 2 genes pueden estar ligados a más de 50 Las distancias se calculan en nucleótidos o mediante u.m., pero no lo puedo determinar mediante una simmedición de cromosomas. ple prueba de 2 puntos entre ambos genes analizados, El orden físico de los genes en el cromosoma coincide porque la máxima distancia determinable por este con el orden que surge de su mapeo genético. Existe, método es de 50 u.m. Necesito un gen “puente” (o además, una correspondencia proporcional relativa varios) para poder integrar a los dos genes en el misentre distancia genética y distancia calculada en nu- mo grupo de ligamient o. Por lo tanto, un grupo de cleótidos, aunque la correspondencia no es absoluta y ligamiento está formado por el conjunto de genes (o varía en distintas partes del cromosoma. Es decir, una marcadores genéticos) que se comportan como ligados unidad de mapa significa un número distinto de nu- entre sí en forma directa o a través de otros genes cleótidos cerca del centrómero que cerca del telómero, “puente”. El grupo de ligamiento se corresponde a porque las frecuencias de entrecruzamiento (forma- todos los genes o marcadores genéticos localizados en ción de quiasmas) son distintas. La vecindad de regio- el mismo cromosoma o molécula de ADN. Se llama nes heterocromáticas también afecta la frecuencia de desequilibrio de ligamiento al fenómeno que se da aparición de quiasmas y la relación absoluta entre dis- entre dos alelos de dos genes distintos que se heredan tancia genética/distancia física. juntos con una probabilidad mayor que la dada por el 5) ¿Qué es y cómo se hace una prueba de 2 puntos? azar. Esto se debe en general a que los genes están ¿Qué significa la palabra punto en prueba de 2 pun- ligados. tos? ¿Implica ligamiento? 8) ¿Qué es una prueba de 3 puntos? En una prueba R: Se estudia el ligamiento en cruzamientos de dobles de 2 puntos, ¿se puede establecer un orden de los geheterocigotas por dobles recesivos (Ver libro para una nes? ¿y en una de 3? ¿hasta qué punto? explicación más completa) R: Cuando se estudia el ligamiento simultáneo entre 3 6) ¿Qué es una unidad de mapa? ¿Es sinónimo de genes (por ej. A, B y C), se puede establecer un orden centiMorgan, frecuencia de recombinación o de uni- ya que (suponiendo que el orden sea ABC), la distandad de recombinación? ¿A cuántos nucleótidos equi- cia (prueba de 2 puntos) entre A y C es aproximadavale una unidad de mapa? mente igual a la suma de la distancia entre A-B + B-C. R: Se define como unidad de mapa a la distancia entre En una prueba de 2 ptos. no se puede establecer un genes (o marcadores) para los cuales se observa un orden, sólo distancias relativas. En la de 3 sí, pero no recombinante con una probabilidad de uno en 100 se sabe si los 2 puntos extremos (A y C) están “a la productos de la meiosis. Una unidad de mapa sería, derecha” o “a la izquierda”. entonces, equivalente a una frecuencia de recombina- 9) ¿Porqué no hacer 2 o 3 pruebas de 2 puntos? ¿Dan ción del 1%. Es decir, que se calcula mediante la distintos resultados? ¿Por qué? ¿Qué es interferencia y fórmula del % de recombinantes sobre totales y es cómo se calcula el coeficiente de coincidencia? sinónimo de unidad de recombinación. Los cM tienen R: Pueden dar resultados distintos por el fenómeno de otra fórmula de cálculo que toma en cuenta la posible subestimación de dobles entrecruzamientos y/o de existencia de dobles entrecruzamientos y la interfe- interferencia. En el ejemplo de arriba, la distancia enrencia. No siempre coinciden. Como se mencionó, la tre A y C (medida como prueba de dos puntos) suele unidad de mapa genético varía (en número de nucleó- ser menor a la de A-B + B-C. Esto se debe a la pretidos) según la especie, el cromosoma y localización sencia de de entrecruzamientos dobles entre Ay C cudentro del mismo cromosoma. Groseramente, 1 cM yo resultado final es la formación de gametas parenta-
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12
que se produzcan quiasmas muy cercanos entre sí (se “interfieren”) por lo que el número de recombinantes entre genes muy cercanos entre sí es menor al proba bilísticamente esperado. Algunas funciones de mapeo más refinadas, toman en cuenta este fenómeno. Estas funciones se derivan de la función de Poisson y son, fundamentalmente, dos: la de Haldane (que considera los dobles entrecruzamientos pero no considera la interferencia) y la de Kosambi (que considera ambas). Por lo general, todos los mapas que se publican se basan en la función de Kosambi. 10) ¿Por qué, en una prueba de 3 puntos, realizar un cruzamiento prueba facilita el análisis? R: Porque el cruzamiento con el homocigota recesivo permite concentrarse directamente en las gametas de la F1 permitiendo distinguir las gametas que portan alelos recesivos de dominantes mediante una sencilla relación de 1 a 1. Esto es particularmente útil para genes en los que no se puede distinguir el fenotipo heterocigota del homocigota dominante. Toda la progenie segregante será homocigota recesiva o heterocigota (no habrán homocigotas dominantes) 11) ¿Es sencillo aplicar los sistemas arriba mencionados para mapeo en humanos? ¿Por qué? ¿Es posible utilizar información de pedigrí para hacer estudios de ligamiento? ¿Existen otros métodos alternativos de mapeo? ¿En qué consiste el método de mapeo con híbridos somáticos (fusión interespecífica de células)? ¿En qué consiste la hibridación in situ? R: No R: No se pueden “planificar” cruzamientos en humanos, obtener líneas puras (depresión por endocría) y las progenies son poco numerosas (estadísticamente problemáticas). El mapeo preciso y fino se realiza
Lod = 3 para un par de genes indica que es mil veces más probable que los genes estén ligados respecto a que no lo estén. Programas como el MapMaker tam bién utiliza Lod Scores para comparar diferentes órdenes posibles entre varios genes. Al orden más probable le asigna un 0 y a los restantes le asigna valores negativos. Porque permiten una cobertura del genoma que no permiten los caracteres morfológicos. 13) ¿Qué similitudes y diferencias existen entre los cálculos de distancias usando las fórmulas vistas arri ba y los algoritmos de programas como el Mapmaker? R: Las fórmulas son un distintas porque los algoritmos del Mapmaker están basados en Lod score (cálculo de probabilidades). Estos cálculos probabilísticos están basados, sin embargo, en el mismo concepto y la misma lógica que existe atrás de las funciones de ma peo en pruebas de dos o tres puntos. Sin embargo, son más poderosas, porque al tratarse de una prueba de “muchos” puntos simultáneamente, permite afinar mucho más el cálculo minimizando errores debidos a dobles entrecruzamientos e interferencias.
Problemas:
1) En el gusano de seda Bombix mori, el gen recesivo l produce color amarillo limón en la larva, mientras que el L da color blanco. El dominante B, produce bandas negras transversales, mientras que el recesivo b no las produce. Ambos loci están ligados y su distancia genética es de 30 unidades de recombinación. Se cruzó una hembra homocigota blanca y con bandas con un macho amarillo sin bandas negras. Los machos de la F1 se utilizaron para fecundar hembras amarillas y sin bandas negras. ¿Qué segregación fenotípica y
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13
cruce rosa redonda por púrpura alargada, se obtuvie- P/p A/a R/r ron los resultados indicados: PR =99, =99, Pr=44, Pr=44, pR =44, =44, _l___________l____________l__ pr=3. pr=3. ¿Existe ligamiento entre ambos loci? l__15 um____l___20 um____l 5) En la descendencia de un cruzamiento prueba de un Si el Sr. Spock se casa con una terráquea y no existe triheterocigota se obtuvieron los siguientes resultados: interferencia (génica), ¿Qué proporción de sus hijos Fenotipos Frec. abs. mostrarán: ABD 50 a) Apariencia vulcaneña para los tres caracteres? ABd 32 b) Apariencia terráquea para los tres caracteres? AbD 538 c) Orejas y corazón vulcaneños, pero adrenales terráaBd 600 queas? abD 28 d) Orejas vulcaneñas pero corazón y adrenales terráabd 62 queas? Se desea saber: a) la fase del genotipo parental; 9) Bridges y Morgan realizaron numerosos cruza b) el locus central; c) las distancias genéticas; mientos en Drosophila melanogaster . En cierto exped) el valor de la interferencia. rimento cruzaron machos sin alas ( apterous ( apterous)) con 6) En los ovarios de plantas superiores se forman hembras de cuerpo negro ( black ( black). ). En la F2 obtuvieron núcleos haploides que resultan de una meiosis y suce- 186 individuos salvajes, 85 apterous, 99 black y nin apterous, 99 black sivas mitosis sin citocinesis, formando una célula mul- guno con ambos caracteres. ¿Qué se deduce de los tinucleada llamada gametofito femenino (el núcleo resultados obtenidos?. Razone la respuesta. masculino se fusiona con uno de estos núcleos, que 10) Un genetista quiere mapear los genes A, B, J, D y actúa como óvulo). En algunas coníferas, el gametofi- E mediante dos cruzamientos de tres puntos. El prito femenino puede ser bastante grande; este hecho mer cruzamiento lo realiza con líneas puras para los permite que pueda ser extraído y analizado electro- cinco genes. Después, el investigador cruza los indiforéticamente. Un árbol de pino es heterocigota para viduos de la F1 con individuos homocigotas recesivos, los alelos electroforéticos "rápido" (r ( r) y "lento" (l (l) de y el fenotipo de la descendencia es como sigue: F S las enzimas alcohol dehidrogenasa (Adh /Adh ) y fos- ABJDE 316 ABJdE 31 F S foglucomutasa (Pgm /Pgm ). Se estudió electroforéti- AbJdE 130 AbJDE 17 camente una muestra de 237 gametofitos femeninos y abJdE 314 abJDE 39 F se encontraron los siguientes fenotipos: 23 Adh aBJDE 140 aBJdE 13 F S S F S S F Pgm , 24 Adh Pgm , 93 Adh Pgm y 97 Adh Pgm . Las líneas puras utilizadas en el segundo cruzamiento a) ¿Están estos genes ligados? son las siguientes: AABBJJDDEE x aaBBjjDDee. aaBBjjDDee. b) Calcule la distancia entre los genes Adh y Pgm y en De nuevo, el investigador cruza los individuos de la qué fase están. Otro gen que codifica para la enzima F1 con homocigotas recesivos, obteniendo los si-
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14
observados en los ensayos biológicos con el virus: R (totalmente resistente), P (parcialmente resistente) y S. Fenotipo Padres 1 2 3 4 5 6 7 8 9 Sonda R S P P R R P R R P R PTG9 ----- --- --- --- ---- ---- ---- ----- --- ----- --PCD3 --- --- ----------- --- --- --- --- --- --- --- --PXY ----------- ----- --- ----- --- ----Fenotipo 10 11 11 12 13 14 15 16 17 18 19 20 Sonda R P P R S R S R S P R PTG9 --- --- --- --- --- ----- --- --- --------- --- --- ----- --------- --PCD3 --- --- --- --------- --PXY --- ------- --- --- --- --- ----- ----- ----- ----a) Calcular el porcentaje de cosegregación del gen de resistencia junto a cada gen marcador e indicar cuál de estas 3 sondas (pTG9, pCD3, pXY) utilizaría Ud. como marcador de RFLP para seguir la herencia del gen de resistencia al virus en un proceso de mejoramiento a través de cruzamientos controlados. b) ¿Qué ventajas tendría usar el marcador elegido para testar resistencia al virus, en lugar de hacer el ensayo biológico correspondiente? c) ¿Cuántas generaciones de retrocruza (contra la variedad susceptible) llevaría, en promedio, tener una variedad resistente al virus pero con un 99% del genoma de la variedad agronómicamente interesante? d) ¿Qué individuos F2 elegiría para la primera retro-
su aplicación a este tipo de cuestiones, que tomaron importancia en la restitución de hijos de personas desaparecidas por la última dictadura (1976-1983) a sus legítimas familias. Existen seis niños de los cuales se sabe con seguridad que sólo tres pertenecen a dos familias que no guardan ninguna relación entre sí (son de distintas ciudades). De ambas familias sólo viven los abuelos probables. Utilizando una sonda para un gen con un alto polimorfismo para una enzima de restricción (existen 4 alelos en la población) se revelan los Southern blots de ADN genómico de los seis niños y de los posibles abuelos, que se muestran en el gráfico:
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15
una enzima de restricción E (indicado como +/- en la figura). Mapa de restriccion del gen que hibrida con la sonda:
Ud. aisló DNA de linfocitos de los abuelos y de los miembros vivos de la familia y ha obtenido los correspondientes fragmentos de restricción visualizados en el esquema del Southern. Ud. está ahora en condiciones de hacer un diagnóstico pre-natal a partir de células fetales. ¿Qué patrón de restricción indicaría que el feto heredó la enfermedad (dibújelo)?
15) Se lleva a cabo el análisis de ADN en una familia numerosa en la que algunos miembros están afectados por una enfermedad rara autosómica dominante de manifestación tardía (hacia los 40 años). Se extrae el ADN genómico de todos los miembros de la familia,
16) El mosquito transmisor del dengue ( Aedes aegypti) es una plaga endémica de climas tropicales, pero que desde hace pocos años está extendiendo su distri bución hacia Buenos Aires. Se encontró una variante de mosquito que no transmite el temible virus. El carácter se comportó como dominante mostrando un patrón de herencia mendeliana simple. Con el objeto de introducir este carácter en la población, comenzaron a realizarse cruzamientos. Como el carácter es muy difícil de evaluar porque se requiere (para su análisis) que el mosquito hembra pique huéspedes infectados con el virus, se analizaron en cada generación los patrones de 50 microsatélites con el fin de determinar ligamiento entre alguno de ellos y el carácter de resistencia. Sólo uno de los marcadores (m25) mostró ligamiento a este carácter y a una distancia de 5 cM. Los patrones microsatélites observados cuando se utilizó este marcador en los genotipos homocigotas notransmisor y en el transmisor, respectivamente, fueron como se ve en la figura. Si se cruzan las dos variantes de mosquitos (no-transmiso-res x transmisores: generación F0) se obtienen individuos heterocigotas con fenotipo no transmisor (F1). a) ¿Qué patrones de micro- Tam. Trans NO satélites y fenotipos de mole- misor trans misor transmisión y no- cular transmisión presentarán los descendientes de una F1 258 pb cruzada por la línea parental transmisora? y 236 pb b) ¿en qué proporciones? c) Como se explicó más arriba, para la determinación de los fenotipos sólo se pueden utilizar las hembras.
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16
autosómica dominante letal de manifestación tardía (alrededor de los 40 años). El método aplicado fue el del análisis de segregación de RFLP utilizando la sonda denominada T4. A continuación se muestra un esquema del autorradiograma de los individuos analizados alineado con la genealogía correspondiente. Los individuos afectados se representan con símbolo lleno.
a. Proponga los genotipos más probables para el síndrome y el marcador molecular para cada individuo de la genealogía b.De b. De acuerdo con estos resultados, ¿se podría suponer ligamiento entre el marcador molecular y la enfermedad? Justifique su respuesta. c. Si un investigador demuestra que el “lod score” entre estos dos caracteres (bandas y enfermedad) es mayor que 3, ¿cómo explicaría la ocurrencia de un enfermo que presentara sólo la banda de 5kb? d.Si d. Si otro integrante de la familia de 20 años de edad quisiera saber el riesgo de tener la enfermedad ¿qué estudio le haría y cómo interpretaría los resultados? e. Cuando se analizaron líneas celulares híbridas hom bre-ratón se obtuvieron los siguientes resultados para la sonda T4 y tres productos metabólicos A, B y C. Indique en los casos en que sea posible la ubicación
c) ¿A la zona invertida o al cromosoma completo? d) ¿A las inversiones paracéntricas o a las pericéntricas? 3) Enumere las dos características que considere más importantes en las inversiones paracéntricas y en las translocaciones recíprocas. 4) En un heterocigoto estructural para translocación recíproca, podría originarse algún gameto viable a partir de los productos meióticos originados en una meiosis con orientación adyacente? 5) Los grupos de ligamiento se mantienen constantes en cada especie? 6) El número diploide de un organismo es 2n= 12. ¿Cuántos cromosomas tendrá: a) un monosómico. b) un disómico.c) un tetrasómico d) un doble trisómico. e) un nulisómico.f) un haploide. g) un triploide.h) un autotetraploide. 7) ¿Cuál es la diferencia fundamental entre la aneu ploidía y la euploidía? 8) Indique las diferencias entre auto y alopoliploides. 9) Debido al pequeño tamaño del cromosoma IV de Drosophila melanogaster , las moscas pueden ser monosómicas o trisómicas para este cromosoma y continuar siendo viables. a) Si una mosca disómica para el cromosoma 4 y homocigota para el gen recesivo eyeless de dicho cromosoma se aparea con una mosca monosómica para este cromosoma pero normal, ¿cuál será el aspecto de la F1? b) ¿Cuáles serían las proporciones fenotípicas que se obtendrían al cruzar los distintos fenotipos de la F 1? Problemas 1) En una sección de cromosomas salivales de Dro-
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17
Teniendo en cuenta que todas las deleciones estudia- aspectos, salvo recesiva para los tres caracteres antedas son letales en homocigosis y hemicigosis, y que dichos. Toda la descendencia del cruzamiento fue de las hembras utilizadas en los cruzamientos eran hijas fenotipo dominante. Cuando las hembras de este crude machos de fenotipo normal, deduzca cuál es la po- zamiento se sometieron a un cruzamiento de prueba sición del locus sn en el mapa de cromosomas polité- con machos de la población de laboratorio se obtuvo nicos. la siguiente descendencia: 3) En la descendencia de un cruzamiento entre una Fenotipo Número de individuos hembra normal de Drosophila y un macho, white (w ( w)ABC 4250 Bar (B (B), Bridges observó observó que una hembra no había Abc 490 heredado el carácter Bar del padre, pero era heterociaBC 510 gota para el carácter white. A esta hembra se la retroabc 4480 cruzó con un macho w B, y las proporciones que se a) Explique por qué aparecen sólo cuatro fenotipos. obtuvieron en la descendencia fueron dos hembras a b) De los datos de la tabla se deduce que la fracción un macho. de recombinación entre A,a y B,b es 0,1. ¿Cómo se a) ¿Cuál es la explicación más simple de estas propor- puede explicar esta discrepancia con los datos del ciones anormales? enunciado? b) ¿Cuál es el fenotipo de la descendencia? 7) En hembras de D. melanogaster heterocigotas para Se cruzaron las hembras heterocigotas estructurales de las siguientes 3 mutaciones: Bristle (Bl ( Bl), ), mutación la descendencia con distintos machos que eran hemi- dominante ubicada en el cromosoma 2; Dichaete (D ( D), cigotas para los mutantes rudimentary (r ( r), forked (f ( f ) o mutación dominante ubicada en el cromosoma 3, y fused (fu (fu), ), que están cercanos a Bar ( como lo muestra eyeless (ey (ey), ), mutación recesiva ubicada en el cromoel siguiente mapa): soma 4, fueron cruzados individualmente con machos homocigotas para ey. ey. La mayoría de los cruzamientos
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18
9) Suponga que está estudiando la citogenética e cinco especies íntimamente relaciona as de Dros phila. La figura d e abajo muestra el orde de los gen es (las letras indican genes q e son idénticos en las cinco especies) y los grupos e cromoso as que se e cuentran en cad especie.
1
2
3
4
1+
2
3
4
a) Explique de qué modo, probable ente, estas species evolucionaron una a partir de o ra, esquematizando los cam ios ocurridos en cada p so. (Nota: a egúrese de que comparar el orden de lo genes con uidado). b) Esquem tice una cel la en paquitene y otra anafase
2n=26. Los hí ridos entre estas especies muestran los si uientes arre los cromosómicos dura te la meiosis: H BRIDO Configurac ón meiótica en metafase I G. hirsutum x G. thurberi G. irsutum x G.herbaceum G. hurberi x G. erbaceum
13 II pequeños + 13 I grand s 13 II grandes 13 I pequeñ s 13 I grandes 13 I pequeños
Nota: II bivalent
s, I univale tes a) ¿Cómo pod ía interpret r estos resu tados desde un punto de vista filogenético b) ¿Cómo podría probar ue su interp retación es correcta?. 13) Existen se s especies rincipales del género B as sica al que per tenece la colza (y la canola): B. carinata, B. campest is, B. nigra , B. olerace , B. juncea, B. na us. Las rel ciones entr las especies pueden de du-
cir se a partir d la tabla: a) Deduzca el número de cromosomas de B. cam estri s, B. nigra y B. oleracea. specie o híbr ido F1 Cro osomasBivalentesUnivale tes 34 1 0 B. carinata 38 19 0 B. napus 36 18 0 B. uncea
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19
moleculares, como el mapeo con enzimas de restric- restricción de Southern blot entre distintos individuos, ción (como se vio en la guía 1 –Biología Molecular-) especies, etc. observando diferencias (polimorfismos) usando o no Southern blot, la hibridación in situ, la en sitios de corte. Debido a que se basan en la hibridaalineación de insertos genómicos clonados en BAC y ción molecular lo que permite el reconocimiento entre YAC, caminatas cromosómicas, comparación sinténi- secuencias que no son totalmente homólogas, se los ca de genomas filogenéticamente relacionados, etc. considera marcadores relativamente conservados que Los mapas genéticos se construyen a partir de análisis permiten comparaciones evolutivas entre especies, de ligamiento en poblaciones segregantes posteriores géneros y familias relacionados (dependiendo de la a la recombinación y sumando distancias entre loci secuencia, se puede progresar en la escala taxonómica sucesivos lo más próximos posibles (ver guía 7 de hasta más arriba). Se continúan usando en estudios de mapeo). Ambos mapas no coinciden en cuanto a las sintenia (mapeo comparativo). Como contrapartida, unidades utilizadas (número de nucleótidos o pares de resulta más complicado encontrar variabilidad (poli bases versus unidades de recombinación) ni en las morfismos informativos) intraespecíficos para hacer, distancias relativas (las frecuencias de recombinación por ejemplo, genotipificación (identificación a nivel no son parejas a lo largo de todo el cromosoma), pero de individuo o fingerprinting ). ). sí en el orden de los marcadores. Hoy en día, los RFLPs han sido desplazados en gran 2) Compare el uso de marcadores genéticos molecula- medida por los SNPs (polimorfismo de nucleótido res, bioquímicos y morfológicos. simple –único-, se pronuncia SNiP). Los SNP no tieR: Su uso para mapeo genético es similar, salvo cuan- nen una única metodología como los RFLP sino vado los marcadores morfológicos están determinados rias, aunque todas ellas requieren de PCR en alguno por más de un locus (epístasis) o son de distribución de sus pasos. Cualquier técnica que permite visualizar continua (“cuantitativos”). Pero aún en estos casos un cambio en un nucleótido se clasifica como SNP. existen formas de utilizarlos que se verán en otras gu- Como ejemplo ya se mencionó a los CAPs que impli-
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20
ciador ( primer primer ) en la reacción de PCR. Son polimorfismos para fragmentos de ADN amplificados al azar del genoma, obteniéndose una serie de bandas de distintos tamaños según donde se haya pegado ese primer . Este conjunto de bandas sirve como una huella de ADN que puede emplearse para caracterizar a un individuo. Se dice que es al azar pues se utilizan primers que no se sabe en qué lugar del genoma hibridan. Por eso se dice que son anónimos porque se desconoce la secuencia y (salvo mapeo específico) su localización cromosómica. Como son oligonucleótidos de 10 bases, en general “pegan” en varias regiones del genoma, por lo que al correr el producto de la PCR en geles de agarosa, generalmente se observan varios fragmentos, que miden hasta 2 kpb. Otra desventaja es que son marcadores dominantes (no se puede diferenciar al heterocigota del homocigota) y problemas de reproducibilidad. Su mayores ventajas son los costos y la posibilidad de usarlos con cualquier genoma de cualquier especie aunque se desconozca absolutamente todo de su secuencia genómica. Al igual que los RFLPs, se los puede mejorar secuenciando y convirtiendo en marcadores de PCR específica (en este caso
ficidades de secuencia de 6 pares de bases, como EcoRI? ¿y para una de 4 o de 8 pb? b) ¿Cuántos sitios de restricción y, por lo tanto, bandas de DNA en un gel, se generarían en un genoma humano (10 9 pb), bacteriano (10 6 pb), o mitocondrial humano (10 4 pb) se generarían con EcoRI ? c) ¿Cuántas bandas generarían los RAPD que usualmente se hacen con decámeros y se detectan por PCR? R: a) Un sitio de restricción de 6 bases aparecerá, como promedio, una vez cada 4 6 bases, es decir 4.096 pb. Una de 4 será cada 44 pb = 256 pb, uno de 8 pb será de una cada 65.536 pb. pb. 9 b) 10 dividido por 4x10 3 = 250.000 bandas o sitios para el genoma humano; 250 para una bacteria y entre 2 y 3 para el genoma mitocondrial. c) Por un lado debemos considerar que los sitios com plementarios a los primers tienen que ser “perfectos” (los 10 nucleótidos tienen que ser perfectamente com plementarios, lo que no es necesariamente cierto). Por lo tanto la distancia se podría calcular como para los sitios de restricción 4 10 = 1.048.576. En un genoma humano habría 1000 sitios, en uno bacteriano, uno
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21
ciar “alelos” epigenéticos (como los que están en el genes. Como los genes (a diferencia de las secuencias cromosoma X en mujeres) o diferencias de estados del intergénicas no codificantes) se transcriben, una forma gen entre tejidos expresantes (hipometilados) versus es armar clonotecas de ADNc de distintos tejidos y en silenciados (hipermetilados). distintas situaciones ambientales y fisiológicas y se7) ¿En qué consiste el clonado posicional y el "cami- cuenciarlas en forma grosera y masiva. De esta manenado cromosómico"? ¿Para qué se utiliza? ra se obtiene una colección de secuencias representaClonado posicional: El clonado posicional se utiliza tivas del transcriptoma denominadas EST ( expressed cuando no se dispone de otra información del gen que sequence tags o secuencias expresadas) de las cuales se desea clonar más que su posición cromosómica re- se desconoce mayormente su función. En muchos calativa respecto a un marcador molecular. La idea es sos, esta función puede deducirse en base a la homo poder disponer en el mismo inserto clonado en un logía de las secuencias con las de genes caracterizados vector, tanto el gen deseado como la secuencia del en otras especies mediante búsqueda y comparación marcador molecular o en algún otro vector de secuen- con las bases de datos de los bancos de genes. Estos cia vecina contigua (contig). Este tipo de clonados ESTs pueden usarse para el diseño de micromatrices fueron posibles a partir de la aparición de vectores que (chips de ADN) y para mapeo por RFLP/SNP y servir permiten clonar segmentos de ADN grandes (200.000 como marcas ancladas (STS). bases hasta 1.000.000 de bases) como los BAC y 10) ¿A qué se llama genómica funcional y análisis de YAC. proteoma? En este contexto ¿para qué se usa el etiqueCaminado cromosómico: Los extremos de un frag- tado mediante transposones o “transposon-tagging”? mento clonado se usan como sondas para seleccionar ¿Qué son las micromatrices o microarrays y chips de otros clones de la genoteca. Estas sondas detectan DNA? ¿Para qué se usan? ¿Qué técnicas para la caracclones de regiones del ADN que se superponen con el terización de interacciones proteína-proteína conoce? clon inicial Se aisla un pequeño segmento de ADN de R: La genómica funcional es la rama de la genómica
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22
el cual se sometió a corte por una enzima de r estricción apropiada y posterior electrof oresis, cuyo Southern blot s analizó po hibridació con el fag M13 (sonda de Jeffreys, Nature 314 [19 5]: 67-72). En un carril paral lo se corrió , en idénticas condicion s, una muestra de DNA de linfocitos de sospechos Jack Destripado ius. En la fi ura se muestra el result ado de la correspo diente auto radiografía: I.- Si Ud. fuera Watson (no Jame s), cuál de las siguientes afirmaciones creería correcta para ase orar a Sherlock?: a) Jack es presumible ente culpa le porque la sonda hibrida tanto con el D A de la m ncha como con el
se uencias únicas, se sele ci naron 3, l s cuales s e hi ridaron co los YACs obteniéndose los siguiente s re ultados: a) ¿Por qué no se trató d e hacer Southe ns con lo s Y C y realizar un mape o por restricción de esta m nera? ¿Porqué no hibrid r di ectamente l s YAC entr e sí, evitando r currir a una su genoteca e fago lam -
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23
c) Podrá acer una es imación de qué proporción de los genes está implicado, directa o indirect mente con el roceso de e porulación? d) ¿A qué se dedica el resto de los 6000 genes? 4) Pasó sin llamar muc o la atenci n una pelíc la estrenada ha e unos años llamada GATTAC (que no debe su nombre a una ga a muy agresiva si o a una secuencia nucleotídica y la vez
que utilizar pa a esta detección? ¿Podr á detectar todas las mutaciones posibles? Discuta este punto. c) Después de resolver el último problema de la guía de mutaciones ¿le parece ue librar a l humanida de to os los alel s deletéreo es un objetivo alcanz ble p r “mejoram ento”? Para “premiar” la “plani icación genética familiar”, la sociedad de GATTAC se divide n 2 clases b ien diferenciadas. Aquellos que fueron “planificados” ti nen acceso a los mejores empleos, mientras que a uellos que son “hijos el azar” son considerados “i válidos” o “de-gen “de-gen-era -erados” y sólo pueden ser barr enderos o te er profesio es mal pag s. En la pelí cu-
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24
b) ¿Porqué algunas secuencias están en la mayoría de los genes y otros no? ¿Es esta una muestra representa-
pKT25 (anzuelo codifica el dominio T25) y también se construyó una genoteca de M. tuberculosis en el plásmido pUT18C (codifica el dominio T18 en fusión con el gen de la proteína a “pescar”). Para ello, lo primero que se llevó a cabo fue la amplificación del gen P36 por PCR. a) El oligonucleótido 5’ fue diseñado de manera que se respete el marco de lectura del fragmento codificante para T25 presente en el sitio de clonado ¿Por qué? b) Luego de llevada a cabo la ligación y la transformación, los clones positivos se identificaron en la placa por presentar color blanco en presencia de IPTG
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25 Definición detallada de gen (combinación de estructu-
le esquematizar (en el operón lac) un segundo operara y función) dor que es el sitio de unión (reconocimiento) de la Unidad de información (físicamente constituida por proteína CAP, que serviría para que el operón lac un ácido nucleico, DNA o RNA, que puede estar par- constituyera un ejemplo de regulación negativa (reticionado, superpuesto con otros genes o no) codifi- presor) y positiva (activador CAP). No se suele escante para una “unidad” de función (polipéptido o quematizar tampoco el promotor del gen I, ni los terRNA, por ej. ribosomal). Propiedades: tienen capaci- minadores de transcripción correspondientes y otras dad de variar (mutar, recombinar, es decir evolucio- secuencias no codificantes importantes del futuro nar) y de tener una dinámica poblacional (comporta- mensajero (por ej. la secuencia de unión a ribosomas miento frente a la selección natural o artificial) dentro llamada Shine Dalgarno), los codones de iniciación del contexto de especies biológicas. Tienen capacidad (que muchos estudiantes suelen confundir con el nude almacenamiento y reproducción fidedigna de la cleótido +1) y los codones de terminación de lectura. información, así como su decodificación
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26
La degradación mediada por mutaciones terminadoras 5) ¿Cuál es la diferencia entre un operón, que codifica o nonsense mediated decay (NMD), es un proceso de para un mensajero policistrónico, y un gen que codifieliminación de ARNm portadores de una mutación ca para una poliproteína? ¿En qué tipo de organismos terminadora o de una mutación del marco de lectura se presenta cada uno de estos mecanismos? que derivan en la creación de codones de finalización R: Un operón es una unidad transcripcional compues precoz de la síntesis de una proteína. De no eliminarse ta por una zona regulatoria de la transcripción y una estos ARNm se producirían polipéptidos defectuosos serie de genes ubicados hacia 3’ de dicha zona promoo incompletos al ser traducidos. Estos polipétidos tora-reguladora. Este segmento de ADN u operón se pueden retener parte de su función (por ejemplo acti- transcribe como un todo y a partir del mensajero polividad enzimática o capacidad de unirse a otra proteí- cistrónico, se producen varias proteínas, cada una a na) pero no toda (por ejemplo se puede perder la regu- partir de su propio codón de iniciación, como el caso lación de la actividad enzimática o la capacidad de del operón lactosa en bacterias. Un gen que codifica unirse a otra proteína). Por lo tanto si se produjeran para una poliproteína es un gen cuyo producto se cli-
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27
tar una posible actividad endógena. Se utilizan para reemplazar productos génicos difíciles de medir y poder, entonces, determinar la actividad promotora.
se pega al ADN junto con el triptofano. Se trata de un mecanismo de regulación negativa, ya que al igual que el operón lac, el binding de la proteína regulatoria
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28
secuencia). Estas estructuras secundarias se ven afectadas porque la propiedad de interés de estos aptámeros es que tienen una afinidad muy fuerte de unión a
Problemas: 1) La figura muestra la estructura de un plásmido. El gen R del fago lambda codifica un represor que re-
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29
chos investigadores. Con el objeto de encarar, en un futuro, la producción de plantas transgénicas con genes de resistencia a virosis reguladas por luz, se rea-
una de las 2 construcciones se hizo un Northern, todo bajo las mismas condiciones que antes:
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30
Dado que los pacientes X4 y X5 presentaban el mismo patrón de bandas en el Southern, en el northern y en el western, y con el fin de determinar si ambos pa-
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31
bas enzimas son estructural y bioquímicamente idénticas. El gen en cuestión tiene la siguiente estructura: E1
E2
E3
Cuando se expresa tTA se induce la transcripción del gen de interés dado que se une a las zonas regulatorias de pTetOp. El sistema tiene un paso más de regu-
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62
aparece en el huésped como consecuencia de la ingestión de tejido nervioso enfermo. En cambio, en enfermedades hereditarias como la de Kreutzfeld-Jakob ciertas mutaciones en el gen priónico aumentan la chance de aparición espontánea de la conformación infecciosa sin necesidad de ingestión del agente infeccioso. A este último tipo de transmisión se lo considera “herencia infecciosa” (un tipo especial de herencia extranuclear). Hasta hace poco se pensaba que sólo existían priones en mamíferos, pero actualmente se tiene certeza de que también están presentes en levaduras. Uno de estos priones fúngicos (el PSI) tiene una función biológica conocida (es un factor de terminación de la traducción citoplasmático). Esta proteína puede existir en 2 variantes: PSI+ y PSI-. La herencia sigue un patrón de herencia infecciosa como el menmencionado anteriormente. a) Esquematice un cruzamiento entre levaduras Psi+ y psi- y su segregación en F1 y F2. No omita las convenciones correspondientes para la ploidía, tipos de apareamiento y proporciones genotípicas y fenotípicas en cada una de las generaciones. Lea atentamente, para ello, el ciclo de vida de las levaduras en su libro de texto. b) ¿Dónde tendría que estar localizado el gen Psi? ¿en un autosoma, en un cromosoma sexual, en un plásmido o en genoma genoma mitocondrial? ¿por qué? qué? Justifique.
La herencia de esta proteína fue siempre intrigante porque en ninguna de las decenas de mutantes petite se afecta la herencia de la proteína. En cambio el fenotipo priónico PSI + podía ser revertido a PSI - con altísima frecuencia al someter a las levaduras a agentes desnaturalizantes de proteínas como el hidrocloruro de guanidina, no sucediendo así con tratamientos mutagénicos. Los tratamientos mutagénicos sí pueden producir que levaduras PSI- se transformen en PSI +. Las mutaciones responsables son puntuales y el gen determinante ( Psi Psi) pudo ser ubicado en un cromosoma nuclear. Todo esto llevó a cuestionar la herencia infecciosa-materna del gen y llevó a la hipótesis de que el fenotipo PSI+ depende de la conformación de la proteína como en el mal de la vaca loca. c) Teniendo en cuenta lo anterior y disponiendo de: • una sonda del gen Psi (sabiendo que la mutación puntual que condujo a transformar una levadura PSI en PSI+ coincide con el sitio de restricción EcoRI). • anticuerpos monoclonales que reconocen diferencias conformacionales entre PSI + y PSI-. Esquematice un experimento para estudiar la herencia del gen Psi en sus dos variantes alélicas. ¿Cómo serían las proporciones genotípicas y fenotípicas en una progenie resultante del cruzamiento de Psi+ y psi- (recordando que la proteína codificada por Psi+ también tiene propiedades de chaperona “infecciosa”)?
16. GENÉTICA DEL DESARROLLO y EPIGENÉTICA
Guía teórica: 1) ¿Qué similitudes y diferencias tiene el desarrollo embrionario en vertebrados y en moscas? ¿Qué im portancia relativa tienen las gametas masculina y femenina (los óvulos o futuros huevos)? 2) ¿Qué diferencias y similitudes existen entre determinación (o compromiso) y diferenciación? 3) Describa la determinación autónoma versus la determinación en comité o grupal. 4) ¿Qué es una cascada regulatoria y los efectos de posición? ¿Sirven para explicar porqué el producto de un mismo gen tiene efectos opuestos en distintos tejidos? 5) ¿Qué características tienen los genes que condicionan y participan en el desarrollo? ¿Son todos genes para factores de transcripción? 6) ¿Son los genes homeóticos un subgrupo de los mismos? 7) ¿Existen efectos maternos (como los que Ud. vio para el caracol del género Limnaea) para alguno de estos genes? 8) Si la mayoría de estas mutaciones (como la del problema 2) son letales. ¿Cómo hacen para mantenerlas (y visualizar su efecto sobre los embriones)?
del gen myoDI (gen que codifica para un factor de transcripción de activa expresión en mioblastos: células musculares) y un promotor constitutivo del virus SV40 (virus animal cuyo promotor funciona en un amplio rango de células de mamíferos). Como resultado, se observó la brusca formación de un mioblasto incluyendo la formación de un sincisio a partir de células uninucledas. Gen quimérico
Gen virus SV40 prom Sec. cápside cápside viral viral promotor
Gen myoDI promSec. prom Sec. estructural
Sec. estructural
a) Explique cómo tan abrupta "transdetermina-ción” pudo ser causada por un solo gen. b) ¿Por qué no se utilizó el gen myoDl con su propio promotor para realizar el experimento y se lo cambió por uno constitutivo? c) ¿Podría ocurrir lo mismo con cualquier célula de cualquier otro tejido, como ser una célula neuronal? ¿Por qué? 2) Para estudiar la regulación del gen homeótico "hairy" de Drosophila, se clonó la región promotora del
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63
coloreados de azul. A continuación se hicieron deleciones en la región 5' de la región promotora para analizar la contribución de cada parte de esta región promotora. Como resultado, se identificaron secuencias * Región promotora gen hairy Región 5'
CAT y TATA boxes
regulatorias específicas que resultaron ser responsa bles, individualmente, de la formación de cada uno de los anillos. Embrión transfectado
Sec. estruct. lac Z
transcripción
1234567 1 2 6
Para estudiar la acción de estas secuencias, se tomó la construcción responsable de la formación del anillo 6, y con ella se transformaron embriones de moscas mutantes homeóticas deficientes para el gen "knirps" y para el gen "Krüppel", y otra que expresa el gen "Krüppel" en forma constitutiva, con los siguientes resultados: Salvaje - knirps - Krüppel expresión constitutiva Krüppel Con anticuerpos específicos e hibridacción " in situ" del mRNA, se ensayó la concentración relativa de los productos de los genes knirps y Krüppel en el embrión de las moscas salvajes, observándose la siguiente distribución: proteína Krüppel proteína knirps
prot. knirps + prot. Krüppel
Anillo 6
a) ¿Por qué se utilizó una secuencia estructural de ß-galactosidasa en lugar de la secuencia estructural del mismo gen, pudiendo disponer de anticuerpos para detectar la expresión de la proteína "hairy"? b) ¿Por qué no se eliminó la región promotora conteniendo las CAT y TATA boxes en los experimentos de deleción? c) ¿Cómo explica la disminución en número, de 7 a 1, de los segmentos "teñidos" con X-gal en las deleciones? d) En base a los experimentos con mutantes y de detección de productos de los genes knirps y Krüppel ¿qué función cumplen estos productos? En su contestación considere la posibilidad de que sean receptores de mem brana, hormonas, factores de transcripción, reguladores positivos, negativos, segundos mensajeros y/o proteínas específicas de tejidos diferenciados como podría ser la hemoglobina de los glóbulos rojos de vertebrados. e) Proponga un modelo de regulación (un esquema al estilo del operón lactosa) para explicar la función de la secuencia regulatoria estudiada en la expresión del gen “hairy”
INTERFERENCIA/SILENCIAMIENTO INTERFERENCIA/SILENCIAMIENTO E IMPRONTA 3) La impronta genómica ("genomic imprinting") es una rara propiedad que presentan algunos genes de transcribir y expresar sólo el alelo derivado del progenitor de un determinado sexo, es decir o del padre exclusivamente, o de la madre exclusivamente y no del otro; mientras que el alelo proveniente del progenitor
impronta o marca a nivel molecular para distinguir el alelo paterno del materno (para una revisión del tema: Cell 77, 473-476 [1994]). El gen del factor de crecimiento "insuline-like growth factor" tipo 2 (IGF2) humano, humano, que se expresa en las etapas embrionaria y fetal, está sujeto a "imprinting".
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64
de borrarse en la línea germinal de la descendencia del minutos y medir los niveles de mensajero de GFP se sexo opuesto. observaron los siguientes resultados: Para estudiar este fenómeno en el gen del IGF2, se 2 aprovechó uno de sus polimorfismos -que cae en una Sin flag22 región del exón 9 que corresponde a la región 3' no Con flag22 codificante-, que da dos alelos posibles: uno que presenta el sitio para la enzima Apa I , y otro que carece de él (Nature Genetics 4, 98-101 [1993]). [1993]). Usando dos 1 oligonucleótidos específicos del exón 9, los autores del trabajo amplificaron una región de 236 pb que Cantidad relativa abarca a dicho sitio. La enzima Apa I corta al frag- de mRNA mento amplificado a partir del primer alelo menciona- GFP do, en dos fragmentos: uno de 173 pb y otro de 63 pb, como muestra la figura: Ud. es un biólogo muy riguroso y crítico (eso está miR39 TIR1 ABF ABF bien) y duda de mucho de lo que le dicen sus docentes (eso no está tan bien) con respecto a la existencia del b) ¿Cómo interpreta los resultados del experimento 2 "imprinting", queriendo revisar Ud. mismo lo hecho teniendo en cuenta lo observado en el experimento 1? por los autores del trabajo. Diseñe un protocolo expe- ¿Qué tipo de regulación estaría siendo ejercida en carimental para determinar si el gen IGF2 está sujeto a da caso? "imprinting" y si éste es de tipo materno o paterno, El análisis de secuencia de la región codificante de TIR1, ABF2 y ABF3 reveló que los tres genes poseen indicando: una región conservada de 25 pb que puede ser reconoa) ¿Qué individuos analizaría? b) ¿Qué genotipo (homocigota o heterocigota con res- cida por microRNAs. pecto al sitio Apa I) eligiría en los individuos a anali- c) Proponga un mecanismo para explicar los cambios observados en los niveles de transcriptos de los cuatro zar? ¿Cómo determinaría esos genotipos? c) ¿Qué técnicas usaría? ¿Qué tipo de muestra usaría genes en respuesta a flag22. 5) La ingeniería genética de plantas para conferir rede cada individuo (¿DNA o RNA?). 4) Las plantas poseen diversas estrategias de defensa sistencia a enfermedades virales tiene larga data. Las para contrarrestar el ataque de patógenos. En Arabi- primeras transgénicas resistentes a virosis, lo fueron dopsis, por ejemplo, la presencia de un péptido co- por expresión de funciones (proteínas) virales (como rrespondiente a los últimos 22 residuos de la flagelina la cápside) las cuales interfirieren la estrategia de mulde bacterias ( flag22) sirve de señal para desencadenar tiplicación de los virus. También se intentó bloquear grandes cambios en los niveles de transcriptos de cier- la traducibilidad de los mensajeros virales expresando tos genes. Navarro et al. (2006) hallaron que la res- secuencias de ARN complementario (llamadas de puesta a flag22 involucraba principalmente cuatros orientación antisentido). En los últimos tiempos se genes, cuyos patrones de de abundancia de transcriptos enfocó en el aprovechamiento del mecanismo de defensa llamado silenciamiento post-transcripcional mese muestran a continuación: diante la expresión de secuencias nucleotídicas viraSin flag22 les. En un trabajo reciente, Vazquez Rovere y col. Con 1.25 (Arch. Virology 146:1337, 2001), obtuvieron papas Cantidad 1 transgénicas que expresaban las siguientes 3 construcRelativa 0.7 ciones genéticas conteniendo el gen de la replicasa del de mRNA 0.50 virus PLRV (ORF2b): Aclaraciones: 0.25 ATG-rep: Construcción que contiene el gen de la re plicasa con el codón de iniciación (ATG AUG); TIR1 ABF ABF miR39 No codific-rep: codific-rep: Idem sin ATG; Anti-rep: construcción antisentido (la secuencia se TIR1, ABF2, ABF3: receptores de auxinas encuentra invertida); miR393: microRNA 393 RB y LB: LB: bordes derecho e izquierdo del segmento T *Todas las diferencias son significativas de Agrobacterium tumefaciens; a) ¿Cómo se modificaron los niveles de mRNA en
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65
a) ¿Para qué se usó la resistencia a kanamicina? Las plantas transgénicas fueron desafiadas con el virus y se encontraron plantas con resistencia (y también con susceptibilidad) con cualquiera de las 3 construcciones. Para caracterizar el mecanismo que media esta resistencia, se bombardearon hojas de las papas transgénicas utilizando un cañón génico y una construcción que codifica para una proteína de fusión GUS::replicasa (Fig. 2, la GUS –ß-glucouronidasa– es detectable en forma similar a la ß-lac). La fusión, sin embargo, afecta parcialmente la actividad enzimática. Es decir, la enzima funciona un poco peor, pero fun-
ciona. De esta forma se consigue la expresión (transitoria) de las construcciones genéticas bombardeadas en las células impactadas y su fácil visualización en forma de puntos azules (fig.2). Nota: No Nota: No se grafican las transgénicas con la construcción ATG-rep porque ATG-rep porque dan casi igual que las no codific-rep. fic-rep. b) ¿Para qué se hicieron los experimentos de la Fig. 2? Interprete los resultados: i) ¿Cómo me doy cuenta de cuánto afecta a la actividad GUS la fusión con la replicasa? ¿Qué resultados comparo para evaluarlo? ii) ¿Para qué sirven, además, los experimentos realizados con la construcción 2 (GUS sólo)? iii) ¿A qué atribuye las diferencias entre A1 y A2 y entre S1 y S2? ¿Qué relación tiene con la resistencia? iv) ¿Cuál mecanismo (mediado por proteína, mediado por bloqueo de la traducción –antisentido-, mediado por silenciamiento posttranscripcional) media la resistencia? Aclaraciones: GUS-rep: GUS-rep: construcción genética capaz de expresar una proteína (y su mRNA) de fusión GUS+replicasa, GUS: GUS: Idem sólo GUS. uid-A: uid-A: gen codificante para GUS; A1 y A2 son 2 plantas transgénicas representativas para la construcción antisentido (Anti-rep); S1 y S2 plantas S2 plantas con la construcción “sentido” (no codific-rep). (S) y (R) representan plantas susceptibles y resistentes. Ken (NT) es un control de la misma variedad de papa que el resto (Kennebec) pero no transformada.
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66
16. EVOLUCIÓN, BIODIVERSIDAD Y MEJORAMIENTO
1) La epidemia global del SIDA es provocada por el HIV, un retrovirus que infecta linfocitos. Se conocen al menos 2 variantes llamadas HIV-1 y HIV-2, siendo el HIV-1 el causante principal de la epidemia) Se es peculó mucho sobre su origen y ha habido muchas versiones como que el virus fue diseñado en un laboratorio de ingeniería genética hasta otras en las que médicos “a la antigua” se inyectaron suero de monos para probar protección (vacunas) contra determinadas enfermedades y permitieron así el ingreso del HIV al huésped humano ayudando a cruzar barreras interes pecíficas.
Mono verde africano Mono sootey mangadey Mono sykes HIV-2 Mandril
chimpancé ZR59
HIV-1 Independientemente de esta última hipótesis, siempre se especuló acerca de la relación evolutiva con la familia de SIVs (virus de inmunodeficiencia de simios) que presentan chimpancés, mandriles y otros simios, los cuales no son patogénicos en sus huéspedes pero
tocondriales de sus hospedantes. De esta manera se puede tener una relativa precisión para evoluciones que, se supone, tienen menos de 100 años de antigüedad. A partir de un estudio utilizando RT-PCR de secuencias nucleotídicas virales de distintos aislamientos de HIV y sus variantes en humanos, así como los virus del tipo SIV en monos, se identificó una región que muestra un gran polimorfismo, la cual se quiere usar para hacer un análisis filogenético y un estudio poblacional de cuantificación de la variabilidad genética. a) ¿Por qué se utilizó RT-PCR y no clonado molecular de cDNA (y posterior secuenciación de los clones)? ¿Cómo se relaciona esto con el concepto de las cuasiespecies en virología? b) Se muestran las secuencias de cuatro aislamientos:
1. 2. 3. 4.
AGGCC AAGCC ATAGC TGTCC AGGCA AGGCA AAGA AAGAC ATAC ATACC TGA TGACC AGGCC AAGA AAGAC ATAGC TGTCC AGGCA AGGCA AAGA AAGAC ATAC ATACC TGTCC
Construya una matriz de distancia genética simple expresada como el número de diferencias nucleotídicas por sitio polimórfico para cada par de secuencias comparadas (por ejemplo, las secuencias 1 y 2 tienen 20 nucleótidos y tienen 4 diferencias 4/20 = 0,2). c) ¿Qué secuencias se encuentran más relacionadas entre sí? d) Construya un fenograma con estos datos e) ¿Podría proponer una posible secuencia evolutiva de cambios? ¿Qué tiene que ver esto con el principio de parsimonia? f) ¿Podría asignar a una de las secuencias el papel de “primitiva” y otras de evolucionadas? ¿Qué necesitaría para poder hacerlo? En un trabajo publicado por Wain-Hobson (Nature 391:531, 1998) se propone un árbol filogenético como el que se muestra a continuación. g) ¿Opina que la 1 2 3 4 clasificación en HIV-1 y 1 0,2 HIV-2 refleja 2 correctamente la taxonomía de este virus? 3 ¿El origen del HIV, es 4 monofilético o polifilético? Discuta. h) En 1998 causó mucho asombro el análisis de una muestra de sangre conservada desde 1959 perteneciente a un individuo africano de sexo masculino (ZR59). Si bien no se pudo secuenciar completamente, fue suficiente para ubicar su relación evolutiva en el centro de HIV-1. Como Ud. sabe, el SIDA se hizo conocido (por afectar
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67
Paciente C Paciente A Paciente G Paciente B Paciente E Dentista CL 1 CL 2 Paciente F CL 3 CL 4 Paciente D
b) Sobre la base de los datos, los autores concluyen que el chimpancé es el primate más cercano al humano. ¿Está Ud. de acuerdo? c) Los autores no incluyeron "gaps" (huecos que aparecen en alguna de las secuencias al alinear todas) en el análisis porque dicen que uno no está seguro de que los "gaps" fueron consecuencia de eventos simples únicos de mutación? ¿Ud. esta de acuerdo con esa decisión de los autores? d) ¿Conoce algún otro método molecular que pudiera aportar más datos para confirmar la teoría concluida en b? 3) Otro caso evolutivo de interés epidemiológico es el de la BSE (encefalopatía espongiforme bovina o “mal de la vaca loca” causada por priones) y que ya se comentó en un problema de la guía de herencia no mendeliana. Se sabe que la proteína normal debe estar involucrada en la señalización nerviosa dado que se encontraron alteraciones en los niveles de GABA en ratones “knock-out” para el gen correspondiente. Una de las particularidades epidemiológicas de esta enfermedad es que la misma es transmisible desde ovejas enfermas de “scrappie” a bovinos que se alimentan de tejidos ovinos infectados. Los humanos desarrollan la enfermedad de Creutzfeld-Jacobs (o kuru) si consumen tejidos bovinos infectados, pero nunca pudo demostrarse la infección por consumo de carne ovina. Los genes para priones de 33 especies distintas han sido secuenciados y permitieron construir un fenograma como el de la figura.
i) ¿Cuán altas le parece que pueden ser las probabilidades de contagiarse de SIDA de un chimpancé infectado con SIV? ¿Porqué? La rapidísima velocidad de evolución se puso de manifiesto en un sonado estudio a nivel de epidemiología molecular con pacientes de un dentista en Florida, Estados Estados Unidos, en la década del ‘90 (ver figura, CL son controles locales de pacientes de las inmediaciones, con SIDA, pero que no tuvieron ningún contacto o relación con el dentista y sus pacientes). a) La topología del árbol es bastante inusual en j) ¿Podría Ud. confirmar con estos datos que las algunos aspectos de relaciones entre primates prácticas dentales hechas sin las debidas precauciones deben ser consideradas como un factor de riesgo? Orangután ¿Considera que todos los pacientes se enfermaron por Humano esta causa? ¿Para qué se utilizaron controles? ¿Porqué fueron locales y no externos? ¿Le permite esto inferir Gorila que todos los casos analizados provienen de un origen común (monofilético)? Chimpancé 2) En un trabajo de un grupo de investigación intereGibón sado en filogenia de hominoides, publicado en Mol. Biol. Evol. 7, 203 (1990), se muestran las secuencias Cabra alineadas de una región transcripta (3500 pb) del gen
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68
caso de este gen, vacas y humanos tienen un origen filogenético común? ¿Surge de estos resultados alguna nueva interpretación epidemiológica? d) Opine si el resultado experimental observado: que ratones y ovejas pueden ser infectados con BSE (aunque carecen de las 2 modificaciones mencionadas) contradice las conclusiones de c). 4) Se postula que los cromosomas sexuales de los humanos evolucionaron a partir de autosomas. Se cree que los actuales (y escasos) pares de genes que aún recombinan entre su posición en el cromosoma X e Y son fósiles moleculares de una homología mucho más extensiva del pasado evolutivo. En un trabajo de Lahn y Page (aparecido en Science 286:964, 1999) se estudió la evolución comparando secuencias genes autosómicos ancestrales (llamados ancestrales porque su homología con genes autosómicos actuales es muy baja) que persisten en los cromosomas X e Y. Se tomaron en cuenta 2 variables: 1) la posición de los genes en el mapa y 2) la divergencia de la secuencia nucleotídica. a) El mapeo se realizó mediante la utilización de híbridos somáticos irradiados. El mapa se ilustra en la figura de la izquierda. En el cromosoma X estos genes se concentran en la región distal del brazo corto. En cambio, aparecen mucho más dispersos en el cromosoma Y. b) La comparación de secuencias se hizo analizando las mutaciones silenciosas (también llamadas sustituciones sinonímicas) que no cambian la secuencia de aminoácidos de la proteína. A partir de este análisis se hizo una cuantificación mediante el parámetro Ks (promedio de sustituciones sinonímicas por sitio, estadístico que sirve para estimar el tiempo evolutivo transcurrido desde que las secuencias que se comparan empezaron a divergir por ausencia de recombinación homóloga; en este caso, entre cromosomas X e Y). i) ¿Por qué se considera que las sustituciones sinonímicas o silenciosas son prácticamente neutras respecto a la selección? ii) ¿Porqué se eligieron estas mutaciones en particular (y no otras) para este cálculo de "edad" de genes?
combinación homóloga, porque los alelos de los a genes homólogos dejan b de aparearse en la meio- 4 sis. c iii) ¿le ayudan estos co- 3 nocimientos a entender las diferencias de “edades” de los genes contenidos estos 4 segmentos 2 cromosómicos? Por lo tanto, se postula a que estos resultados reflec jan 4 eventos evolutivos puntuados de inversión cromosómica (y eventual b pérdida de DNA de los segmentos) del cromosoma Y, que llevaron a la 1 pérdida de ordenamiento por grupo de edades de estos genes en el cromosoma Y (ver segunda figura). Esto sugiere que la inversión más reciente (grupo 4) ocurrió hace 30-50 millones de años (en la línea de los primates), la del grupo 3 hace 80-130 Ma, etc. Desde el punto de vista de la evolución de los grandes grupos, esto significa que el sistema de determinación del sexo basado en cromosomas X e Y apareció muy tempranamente en la línea evolutiva de los mamíferos. iv) Indique si estos datos corroboran (o no) que el sistema de determinación Z-W de las aves es independiente (o no) del de los mamíferos. ¿Qué podría decir del sistema de determinación del sexo de los reptiles de los que se se originaron mamíferos y aves? Otro resultado de interés es que los genes SOX 3 y SRY (genes que primariamente determinan el sexo) tienen valores de Ks que los ubican entre los genes más antiguos (es decir, genes originalmente homólogos y que
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69 s - 0 e o 4 l a r c 2 u n 0 x ó 2 e 3 s i s r ( s e a v m n Y o i a s o m o t o m n e s o o r v c e m r s e o r o l m c i e r l e d p d l a i e a c c n d s i e é g m r u ó e p s ) s m e o a E d m M
1
Autosomas Autosomas reptiles
) a M 0 7 1 0 3 1 ( Y l e d n ó i s r e v n i a d n u g e S
a c i m ó s o t u a o d u e s p a n o z a l e ) d a n M ó 0 i s 3 1 n a p 0 x 2 E 1 (
2
1
X Y XY monotremas
X
) a M 0 3 1 0 8 ( Y l e d n ó i s r e v n i a r e c r e T
2
1
Y XY marsupiales
X
Y
3 2
1
4 3
) a M 0 5 0 3 ( Y l e d n ó i s r e v n i a t r a u C
2
1
X Y X Y XY en mamíferos placen- XY humanos tarios no simios
MEJORAMIENTO 5) La κ (kappa) -caseína es una proteína presente en la leche bovina y juega un papel importante en el proceso de elaboración de quesos. Un alelo del gen de la κcaseína, llamado “A”, produce un tipo de caseína que da un queso de calidad superior a la que da otro alelo, gtgctggag(t/c)aggtatcctag
llamado “a”. La única diferencia diferencia entre ellos radica en una base que abarca el sitio de restricción para la endonucleasa Hind III , lo cual hace que el sitio esté presente en “A” y ausente en “a”:
sitio Hind III +/
región codificante codificante DNA amplificado (874 pb)
Al hacer PCR con iniciadores (" primers") específicos (indicados en la fig.) sobre muestras de sangre de cinco toros y posterior tratamiento con Hind III , se obtuvieron los siguientes resultados (Animal Genetics 22, 11 [1991]):
1
2
región 3’ no codificante
3
4
5
pb 874 521
ggtcacaacaa cgtg agatg
d) ¿Qué ventajas tiene este método sobre el clásico de RFLP? e) ¿De qué otros tejidos se podría extraer DNA de los toros? f) ¿Se podría determinar el genotipo de vacas, haciendo PAGE no desnaturalizante de proteínas en muestras de leche? 6) Varios trigos argentinos portan la translocación 1BL/1RS trigo-centeno (es decir son “transgénicos” con muchos genes de centeno). El segmento de centeno translocado lleva genes de resistencia a roya de la hoja, del tallo y a oidio. Además, algunos autores en-
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70
i) descripción de cómo sería el modus operandi de la A la papa común ( Solanum tuberosum spp tuberosum) técnica involucrada se le pueden aplicar muchas técnicas biotecnológicas ii) complejidad del equipamiento y del personal invo- como son: lucrado (es decir, ¿qué instrumentos y nivel de capaci- i) cultivo de anteras (es decir, polen) para reducción tación se requiere?) del número cromosómico iii) posibilidad iii) posibilidad de masificar el análisis (relación tiem- ii) fusión de células somáticas (de protoplastos) po empleado por muestra) iii) cruzamientos amplios (cruzamientos interespecífi• Citológicas: recuento cromosómico, análisis ca- cos con otras especies de número cromosómico com patible) riotípico, bandeo C • Bioquímicas: electroforesis de proteínas de reser- Además, tiene especies silvestres relacionadas con va (secalinas vs. gluteninas-gliadinas), isoenzimas, genes interesantes para el mejoramiento pero que, en general, son diploides y autógamas (por lo tanto con ELISA de secalinas. un alto grado de homocigosis), pero que también ad• Moleculares: Microsatélites, hibridación in situ miten la realización de cruzamientos interespecíficos • Inoculación con roya mediante castración e inoculación de polen. c.- ¿Cuál recomendaría? ¿Por qué? Nota: el satélite del brazo corto 1RS del cromosoma Ud. dispone genotipos caracterizados de 4 especies de centeno no se expresa en el fondo genético del tri- (Solanum tuberosum tuberosum -4n-, Solanum tube-2n-, Solanum chacoense -2n- y Sogo. El cultivar de trigo que lleve centeno tendrá una rosum andigena -2n-, todas cultivadas en distintos paíconstitución 42-2 (cromosomas-satélites) en lugar de lanum phureja ses) con varios genes agronómicos de interés y quiere 42-4 de un cultivar sin genoma de centeno. combinarlos de alguna manera. 7) El vigor híbrido o heterosis es una interacción que a) Diseñe un método para la obtención de tetrahíbrise produce en genotipos heterocigotas muy buscada dos de estas 4 especies. en mejoramiento animal y vegetal por su asociación con caracteres productivos como el rendimiento. b) Sabiendo que existe un mapa genético de RFLPs Además, en muchos casos se producen relaciones muy bien caracterizado (con sondas que detectan bandas en todos los cromosomas) en Solanum tuberosum epistáticas positivas importantes cuya base bioquímica y que hay mucho polimorfismo para este o molecular está lejos de ser entendida. Las especies tuberosum tipo de marcadores, indique una forma sencilla de autotetraploides como la papa ofrecen la posibilidad evaluar el grado de éxito que obtuvo con la estrategia de obtener tetrahíbridos (heterocigosis con hasta 4 diseñada en a). alelos diferentes).
BIODIVER SIDAD
8) Ud. es un biólogo muy preocupado por la conservación de la biodiversidad y le han encargado tratar de armar una colección de germoplasma de una especie vegetal o seleccionar un grupo de animales en peligro de extinción para su conservación en un área protegida. Como está interesado en mantener la mayor cantidad de diversidad genética pero el número de individuos que puede seleccionar es pequeño, está tratando de establecer parámetros objetivos para su selección. Decidió probar con los marcadores RAPD y debe eva-
sencia o ausencia de banda. ¡Felicitaciones! Acaba de construir una MBD (matriz básica de datos). b) Calcule la frecuencia de presencia de cada banda (alelo). c) Calcule el contenido de información polimórfica (PIC) de cada una mediante la fórmula de Anderson (1993):
PIC i = 1 − ∑ j =1 pij2 n
donde p es la frecuencia del alelo j para el marcador i.
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71
Anexo: Compendio de preguntas de parciales anteriores (cursos 2000-2006 de verano):
Nota: Nota: la inclusión de las siguientes preguntas y, particularmente, las respuestas esperadas que sirvieron de guía para la corrección de los parciales por parte de los docentes, no tienen otro propósito didáctico que el e l de orientar a los estudiantes acerca de cómo son los parciales de la materia y de cómo se corrigen. Aquellas preguntas de parciales que los docentes consideramos que aportaban a la mejor comprensión de algún capítulo de la materia, fueron incluidas en la guía de problemas y no se repiten aquí. Algunas de las preguntas de parciales integran más de un capítulo de las clases de problemas. Las preguntas de los parciales de la presente cursada tendrán el mismo espíritu pero serán diferentes en sus enunciados. El orden de los temas no es el mismo que en la guía. Regulación e Ingeniería Genética 1) Los operones mce de Mycobacteβ-gal rium tuberculosis (bacteria que produce la tuberculosis) codifica para 500 pP3Rv 200 100 proteínas potencialmente involucra Rv das en la virulencia de este patógePromotor mce no. En el genoma de M. tuberculosis hay 4 operones mce similares en pP3 β-gal secuencia y organización, que codifican para proteínas secretadas o de membrana y una proteína de tipo β-gal pJEM15 invasina. Estos genes son exclusivos de las micobacterias y no se encuentran en otras bacterias mejor caracA terizadas como Escherichia coli. Santángelo y col (del INTA CasteB lar) publicaron en Microbiology (2002, 148:2997–3006) un trabajo C en el que estudian la regulación de este operón en base a la información genómica dispo- a) ¿Qué tipo de efecto tiene la presencia del gen Rv nible. Por un lado conocían las secuencias estructura- sobre la regulación del gen quimérico? les de los genes mce pero no tenían bien caracterizado b) Si la construcción que interesa analizar es pP3Rv su promotor. Por otro lado, les llamó la atención un ¿para qué se hicieron los experimentos con pP3 y con gen (llamado Rv) ubicado hacia 5’ de uno de los ope- pJEM15? rones dado que codificaba para una proteína cuya es- c) ¿Por qué se reemplazó la secuencia codificante de tructura predicha por los modelos bioinformáticos era mce por la de la β-gal? similar a la de algunos factores de transcripción. Para Para investigar la localización del sitio de unión de la Rv hicieron gel-shifting en los estudiar la función de este gen realizaron las siguien- proteína codificada por Rv que se diferencian electroforéticamente los DNAs tes construcciones quiméricas: pP3Rv contiene el gen Rv completo (incluyendo el “sueltos” de los que tienen “pegados” una proteína porque ésta les re promotor de Rv), la región intergénica completa hasta A B C el codón de iniciación del primer gen mce (es decir, 1 2 3 1 2 3 1 2 3 - trasa su migración en el gel. Los donde debería ubicarse el promotor de mce). La seDNAs se detectan cuencia codificante para mce fue reemplazada por el
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72
e) ¿Por q é se utili aron extra tos de E. coli (transformadas con la s construcc ones) y n de micobacterias?
7 kb
E1 pb
pro otor 21 de ctina de omate
12 kb
3 kb
E2 370 b
8
3 pb
E 130 pb
vector
en un Southe n de DNA genómicos digeridos con HindIII prove ientes de: a) tomate normal, b) loe ve a, c) tomat transgénic ; hibridand con una so da de cDNA de longitud com leta. 2) ¿Por qué h abrá elegido tomate y no tabaco, por ej mplo? Respuest s esperad s a) Activa or 3) Analizando la expresió de los mR As de aloe ve b) La co paración ntre la actividad de p3 y rina en distint s tejidos de l tomate tra sgénico por la p3Rv es l que perm te determi ar que Rv es un té nica de Nor thern, usan o como sonda al cDNA de elemento ctivador (o represor si lo hubiese sido). al everina de longitud c mpleta, se observó lo si pJEM15 e un control negativo. RNA de RNA e tomate trans génico c) Porque es fácil de detectar enzi áticamente y los Aloe era: hoja r íz hoja aíz fruto flor resultados o son afectados por la expresión d e los _ genes mce endógenos ( cromosómi os) de las icomRN I bacterias tr nformadas. d) 100-200 mRN II e) Porque necesito extr er de una c lula que ex rese la proteína Rv, porque si no, nunca podría est diar su efecto. 2) El Dr. C reta, en su terna búsqueda de la pó cima guiente: de la juventud eterna, e tudió la estr uctura y ex resión a) ¿Cuáles son los tamaños del mRNA I y mRNA I I? del gen de la aloeverina que produ ce una proteína de b) En el primer esquema bique (con flechas) la romaravillosas propiedad s curativas y rejuvenecedoras ba ble ubicación del codó de iniciación y la del de en Aloe ve a (planta n comestible). Este gen a ori- ter minación.
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73
tejido de interés para alimentación. d) Aparentemente, Con técnicas más modernas se aislaría el ADN de el sistema que permite reconocer al exón 3 como tal, Blade y se lo clonaría en BACs (se puede aceptar que es específico de hoja y de especie y, tal vez, de gen sea en otros vectores como cósmidos o en fagos (pero eso no lo podemos determinar con estos datos). lambda). Con mezclas (“pooles”) de clones recombie) Usar una construcción SIN intrones para evitar el nantes se transfectan vampiroblastos para identificar procesamiento incorrecto. células luminoresistentes (por electroporación o usan3) Para el operón lactosa de Escherichia coli, indique do liposomas). A las mezclas que producen clones el genotipo de al menos uno de los diploides parciales positivos se las separa en clones individuales hasta posibles capaces de producir beta-galactosidasa (codi- identificar el clon conteniendo el gen de interés. Una ficada por el gen Z) constitutivamente y permeasa vez identificado el clon se subclona y secuencia el (codificada por el gen Y) en forma normal (es decir inserto para ver si hay ORFs candidatos, los cuales inducible con análogos de la lactosa) usando la no- son testados funcionalmente de la misma manera. menclatura practicada en las clases de problemas. 4) La Dra. Romántica está intentando encontrar genes Respuesta: Respuesta: lacI+ lacOc lacZ+ lacY- / lacI+ lacO+ que aceleren la floración. Con este objetivo en mente puede procede a transformar plantas de Arabidopsis thaliana lacZ- lacY+ (en alguno de los 2 genotipos lacI puede ser lacI-) mediante la utilización de Agrobacterium, con la si3) En la película Blade II, el héroe (Blade) es un “se- guiente construcción: mi”vampiro que lucha a favor de la humanidad con los vampiros “malos” que quieren dominar el mundo. En una escena trascendental de la película, Blade es capturado para poder extraer su sangre (y su ADN) porque los vampiros estaban diseñando nuevas varian- LB y RB: bordes izquierdo y derecho del T-DNA, tes transgénicas clonadas con mejor adaptación. Si 35S: enhancer fuerte y constitutivo, nptII : gen de re bien ya habían logrado obtener vampiros resistentes a sistencia a kanamicina Pnos: promotor constitutivo, tla plata modificando el gen de la metalotionina, quer- nos: terminador ían extraer de Blade un gen de resistencia a la luz so- Entre las miles de plantas generadas encuentra una lar. Ud. ha sido contratado por los vampiros para mutante hacer este trabajo (le han prometido vida eterna como (M) que florece en la mitad de tiempo que las plantas forma de pago) y dispone de un laboratorio de inge- salvajes (wild type = wt). A partir de una hoja de esta niería genética excelentemente equipado que le permi- planta extrae DNA, lo digiere con BamHI, lo corre en te, entre otras cosas, cultivar in vitro vampiroblastos un gel de agarosa, transfiere a una membrana y lo re(células equivalentes a los fibroblastos humanos). Los vela con una sonda radiactiva correspondiente al Tvampiroblastos extraídos de vampiros comunes se DNA completo. Cuando cruzó esta planta con una
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74
te PCR, secuenciación y finalmente comparación con ver un retraso o ausencia de floración. Si repito el la base de datos del genoma de Arabidopsis, identifi- Northern vería menos mensajero que en el normal. car el sitio exacto de inserción del T-DNA. Este sitio g) Otra respuesta alternativa que pueden pensar los de inserción se encuentra 1 kpb río arriba (5’) de un alumnos sería realizar un knock out de ese ORF. En marco de lectura abierto aún no estudiado (ORF4329). realidad teóricamente no sería correcto pues en plane) ¿Qué efecto cree que tiene la inserción del T-DNA tas no suelen realizarse knock outs, pero ellos probasobre el ORF4329? ¿Qué experimento haría para blemente no lo sepan (conceptualmente no importa, comprobarlo? Muestre mediante un diagrama los re- técnicamente, lo knock outs en la actualidad, salvo en sultados esperados. bacterias, se hacen por interferencia transformando f) ¿Cree que el ORF4329 es un gen inductor o repre- transitoriamente con RNAi o establemente con conssor de la floración? Diseñe un experimento comple- trucciones que expresen iRNA). Además está la aclamentario al realizado por la Dra. Romántica que, me- ración de que la integración del T-DNA no es por rediante ingeniería genética, le permita comprobar su combinación homóloga y sí saben que para hacer un hipótesis, indicando los resultados esperados. Recuer- knock out dirigido se requiere recombinación homóde que en la transformación mediante Agrobacterium loga. el T-DNA no se integra mediante recombinación h) Haría un Southern usando como sonda al homóloga (en biobalística tampoco). ORF4329 y condiciones de baja rigurosidad en la g) Como Arabidopsis es un yuyo y sus flores son bas- hibridación. Para clonarlo haría una genoteca y localitante feas, un estudiante de doctorado de la Dra. zaría el gen homólogo mediante hibridación con la Romántica está interesado en ver si los rosales tam- sonda de Arabidopsis. bién tienen un gen homólogo al ORF4329. ¿Cómo lo 5) Los ritmos circadianos que corresponden a ciclos haría experimentalmente sabiendo que el genoma del de aproximadamente 24 horas permiten a los orgarosal no está secuenciado? En caso de que haya un nismos regular su actividad con la alternancia díagen homologo al ORF4329 presente, ¿cómo clonaría noche. Aunque estos ritmos son autónomos, algunos el gen homólogo completo? estímulos externos, tales como la luz, pueden “poner Respuestas esperadas en hora” este reloj interno. En mamíferos, existe un a) Como gen marcador selectivo que permite selec- marcapasos central, ubicado en el núcleo supracionar las plantas transgénicas de las no transgénicas quiasmático (NSQ) del hipotálamo anterior. Varios durante la regeneración de plantas en un medio conte- genes están involucrados en el mecanismo molecular niendo el antibiótico kanamicina del reloj interno, existiendo entre ellos lazos de retroa b) Porque el objeto del experimento es que este en- limentación positiva y negativa a nivel de la expresión hancer estimule constitutiva y fuertemente los promo- y de la actividad de las proteínas que codifican. Se tores cercanos al sitio de inserción del segmento T con sabe que la fase de este marcapasos central se puede
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75
Esta relación se determinó realizando Western Blot presentan a los cobayos negros y los blancos a los cocon anticuerpos anti-CREB y anti-CREB-P. bayos de fenotipo blanco. Se dispone del siguiente Para poder confirmar el rol de CREB se construyó el árbol genealógico, en el cual los individuos II1 y II4 siguiente sistema indicador (reportero) y se transfectó pertenecen a líneas altamente endocriadas (asumir que pertenecen a líneas puras). en células del NSQ en cultivo. a) Represente I 2 1 Prom: Prom: promotor del gen Per 1. 1. el genotipo probable de cada indiviProm Sec estr.gen luc cre II 1 2 3 4 duo (en los casos de que 1 2 La luciferasa (codificada por la secuencia estructural III ? pueda ser más luc) es una enzima que reacciona con el agregado del luc) de uno, reprede luciferina produciendo luminiscencia. Se ensaya la actividad de estas células en las siguientes condicio- sente con un guión bajo, i.e. B_) b) Calcule la probabilidad de que un descendiente del nes: cruzamiento entre III1 y III2 sea blanco. Para contes+ inhibidor de AC - (baja expresión) tar es suficiente que escriba los genotipos al lado de los símbolos en el esquema y las probabilidades (sólo - glutamato - sin inhibidor - (baja expr.) de las que se usarán para hacer los cálculos) sobre el + inhibidor de AC + (expr. media) esquema (puede dibujarlo o escribir sobre este enunciado). De lo contrario, explique su razonamiento + glutamato - sin inhibidor +++ (alta expr.) Respuestas esperadas: AC (adenilato ciclasa): enzima que sintetiza cAMP a I Bb 1 2 Bb partir de ATP, responsable del aumento de cAMP durante la transducción de señales de diversos estímulos. 2/3 BB BB Interpretar estos resultados. ---Bb Se sospecha que el glutamato ejerce su acción vía II 1 2 bb 3 4 cAMP pero que esta podría no ser la única vía en que lo hace. ¿Qué opina acerca del rol de CREB en los dos III Bb 1 ½ b ? ½ b 2 ½ BB ½ Bb tipos de estímulo? Justificar. Respuestas a) Esta respuesta implica un mínimo de En caso de respuesta con explicación: explicación : detalle de descripción de para qué se hicieron los 3 I1 y I2 tienen que ser heterocigotas (Bb) porque si no, experimentos y qué datos aporta cada uno. Conclusio-
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76
b) el marido está afectado por la enfermedad pero la mujer no tiene antecedentes familiares de ACF? I) Para cada caso construya las genealogías posibles para esta familia indicando los genotipos y fenotipos de todos los individuos involucrados. II) ¿Sobre qué fundamento(s) incluido(s) en la/s ley(es) de Mendel) se apoyan sus predicciones en los casos anteriores? Nomenclatura: S alelo determinante de hemoglobina falsiforme, N de Hb normal, _ puede ser cualquiera de los 2 alelos (N o S) Respuesta: Respuesta: a) ¼ b) 0 I) a) abuelos SS (enfermo) x NN (sano) padres SN (sanos en ambos casos) hijos posibles NN, SN o SS b) abuelos en un caso: S_ x S_ (ambos sanos o enfermos) padre SS (enfermo), abuelos en otro caso NN x NN padre NN (sanos todos) II) En el concepto de segregación de los alelos al formar las gametas que darán origen a la progenie explicitada en la primera ley de Mendel. 3) La corea de Huntington es una enfermedad rara, mortal, que aparece normalmente a mediana edad. Se debe a un alelo dominante. Un hombre fenotípicamente normal, de poco más de 20 años, advierte que su padre ha desarrollado la corea de Huntington. a) ¿Cuál es la probabilidad de que más tarde él mismo desarrolle la enfermedad? b) ¿Cuál es la probabilidad de que la desarrolle su hijo al cabo del tiempo? (Nota: Para sus cálculos asuma que la frecuencia del alelo que produce la enfermedad en la población humana humana es insignificante). Respuesta Suponiendo que el padre que desarrolla la enfermedad fuese heterozigota para el alelo mutante
b) ¿Qué proporciones genotípicas y fenotípicas esperaría si se realizase el cruzamiento prueba de la F1 del cruzamiento E1 x E2 con el progenitor E1? Respuesta a) Dos genes afectan al tamaño A (normal)> a (enanismo 1) B (normal)> b (enanismo 2) [Es un caso de epístasis complementaria] P E1=AAbb x E2=aaBB F1 AaBb (altas) F2 9 A_B_ (altas) vs 7 aa__ o __bb __bb (enanas) b) AaBb x AAbb AABb ¼, AaBb ¼, AAbb ¼, Aabb ¼. Proporciones fenotípicas: A- B- 1/2 altas; A- bb 1/2 enanas 4) Un joven de 35 años se acerca a un consultorio médico con el fin de investigar cuáles son las chances de desarrollar la enfermedad que aque- Pedigrí de la familia. El individuo 4 ja generación es el que se acercó al consultorio. tras generación a su familia. Su abuelo materno y su madre murieron de una rara enfermedad neurodegenerativa que los sor prendió a la edad de 45-50 años, y que ahora también aqueja a su hermana mayor, pero no así a sus otros dos hermanos, ambos ya en sus 50’s. 50’s. Lo primero que hace el médico es analizar el árbol genealógico de la familia: a) ¿Qué tipo de herencia tiene esta enfermedad?
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77
d) ¿Qué sonda se utilizó para revelar el Southern? Es- Respuesta: Respuesta: Cc = Frecuencia de dobles observadas quematizar la región cromosómica correspondiente y (FO) / Frecuencia de dobles esperados (FE) = 0.6 marcar dónde híbrida la sonda. ⇒ FO= 0.6 x FE= 0.6 x 0.2 x 0.1=0.012 e) A la luz de estos resultados, ¿qué le diría al mucha- Nº de dobles observados = 0.012 x 100 100 = 1,2% indicho? Justifique su respuesta analizando los resultados viduos dobles recombinantes: AbC y aBc del Southern. distancia. A-B = Nº de Individuos recombinantes enRespuesta: a) Debida a un alelo dominante tre estos marcadores + Nº de dobles recombinates x b) Aa x aa 100 / nº total de individuos = 20 ⇒ nº ind. rec. entre ↓ estos marcadores + 1,2 = 18,8 Aa x aa Es decir, 18,8 % son recom. Abc y aBC por lo tanto aproximadamente tendremos la mitad mitad de individuos ↓ (1) Aa , (2) aa, (3) aa, (4) ? (por lo que se ve más aba- de cada tipo: 9,4 % jo es Aa) 2) Se está estudiando una nueva especie de plantas c) Estar localizado (mapear) lo más cerca posible (es- descubierta recientemente. Se sabe que el alelo Q cotrechamente ligado) al locus a diagnosticar. Me tiene difica para hojas verdes, y el q para hojas moteadas (Q que permitir distinguir no solamente al enfermo (o que domina sobre q); por su parte el alelo R codifica para potencialmente se va a enfermar) del no enfermo (fe- hojas redondas, y el r para puntiagudas (R domina notipo) sino también homocigotas de hetrocigotas sobre r). Todavía no se sabe la función del gen E/e. A (sobre todo en enfermedades recesivas, que no es este estos investigadores les resultó raro que, a pesar de caso). haber muestreado una gran región en la que este tipo d) El marcador PV123 tiene 2 alelos: uno que se vi- de planta habita, no se encontró ninguna planta con sualiza como una banda de 5 kb (alelo “sano”) y otro hojas verdes y puntiagudas simultáNro como 2 bandas de 3 y 2 kb (alelo “enfermo”), respec- neamente. Para dilucidar esta cues- clases
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78
Siendo 54,25 el nro de dobles rec esperados (0,369 * cas células se convertirían en “Hfr”). Sin embargo, 0,1327 * 1108) tendrá el mismo problema que la cepa b). Siendo 1108 el total de individuos 2) Se realizaron experimentos de conjugación inc) Una posibilidad es que, como sugería el enunciado, terrumpida entre cepas Hfr AmpS leu+mal +trp+ y la combinación de alelos Q/q r/r sea letal, siempre y F - Amp R leu-mal -trp-. Los resultados se exponen cuando a su vez el genotipo sea E/e. En el campo en la siguiente figura: (mal: capacidad de metaboabierto podría presentarse el mismo fenotipo genera- lizar la maltosa) do por el genotipo anterior por ejemplo por la combinación de QQ Ee rr, Qq EE rr ò Qq eE rr (pues se pueden dar todas las combinaciones posibles, al no tratarse de cruzamientos prueba). Dado que en el enunciado se menciona que a campo abierto no se observa ese fenotipo (hojas verdes y puntiagudas) se puede pensar que tanto Q/Q r/r como Q/q r/r son com binaciones letales en presencia de por lo menos un alelo E. (y no en presencia de ee, ya que se observan parentales del cruzamiento prueba que son Qq ee rr) d) Hay que recordar que para calcular distancias en base a frecuencias fenotípicas, se supone que todas las gametas son igualmente viables. En este caso vemos 1) ¿Por qué los recombinantes para leu y que no es así. mal aparecen con mayor número de reRespuestas esperadas a) Las gametas de los caballos combinantes? tendrán 32 cromosomas (la mitad de 64) y las de los 2) Explique en qué medios se hicieron crecer las
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B
3) 10’, 5’ S19 4) una prueba de recombinación equivalente a una prueba de 3 puntos con los 3 marcadores de este experimento. 5) Di bujito 3) Un tema de interés para la medicina humana y veterinaria es el desarrollo de vacunas más confiables. En el caso de la brucelosis (que es una zoonosis, es decir, enferma a animales y a humanos) se desarrollaron vacunas para el ganado vacuno que consisten en bacterias vivas atenuadas por diferentes procedimientos de cultivo. En este contexto, en la Argentina se utiliza la cepa S19 que tiene algunos problemas porque no está suficientemente atenuada. Con el advenimiento de la ingeniería genética se pudieron caracterizar genes como el bp26 y el bmp18 que, al ser mutados, prácticamente eliminan la virulencia, tal como lo publican Campos y col (Veterinary Microbiology, 2002). En un trabajo pre-
B
B1
/
bp26 696 bp
p26f
B
luc
79
N
1
bmp18 990 bp
p26r
p18f
p18r
B
pKS26L
Primer evento de entrecruza-
KnR
sac
Integración del plásmido suicida R
Kn
luc a
/
bp26 bp26
sacB
b
Segundo evento de entrecruza-
Selección con sacarosa
N
B
/
luc
1837 bp p26f
3
bmp18 p26r
990 bp p18f bmp18 pSD18
Primer evento de entrecruza-
2
bmp18
p18r ApR
sacB
Integración del plásmido suicida luc
/
bmp18
sacB
ApR
4 bmp18
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80
problema. Al utilizarse el sistema de contraselección basado en SacB podría no estar presente, sin afectar con su ausencia el resultado. En este caso, su única función es ser un marcador selectivo en E. coli (otorga al plásmido una ventaja selectiva que ayuda a mantenerlo, evitando su pérdida, durante su multiplicación en E. coli). En el caso del segundo plásmido conteniendo ApR , esta aproximación es mucho más difícil (aunque no imposible) porque no tengo un gen indicador y sólo puedo distinguir dobles recombinantes (construcción 5) de salvajes (construcción 3) mediante PCR de las colonias, por ej. ii) Este gen sirve para distinguir las mutantes “knock out” para el gen bp2 que quiero seleccionar (construcción 3) de las revertantes salvajes (construcción 1), dado que ambas sobreviven en el medio conteniendo sacarosa. iii) Como dice el enunciado, este gen sirve para contraseleccionar (seleccionar en forma negativa) las construcciones 2 y 4 en presencia de sacarosa. c) Básicamente, el que el plásmido sea suicida me permite seleccionar más eficientemente al recombinante. Suponiendo que la selección de las construc-
PCR. En la realidad, esta última variante no es practicable, porque la cepa vacunal sobrevive muy poco a las defensas del animal y resulta muy lento, peligroso y poco práctico el aislamiento de bacterias patogénicas. 4) Bacterias de una cepa de E. coli Hfr rec+ bio+ his+ Str S son mezcladas junto a bacterias de otra cepa de E. coli F rec+ bio- his- Str R. A diferentes tiempos, se inter-rumpieron las conjugaciones, se plaqueó en los siguientes medios selectivos (todos conteniendo Str) y al día siguiente se contó el N° de colonias. (MM = medio mínimo) Tiemp (min) 5 10 15 20 25 30
Nº de colonias en M MM+b MM+h M io is 0 0 0 0 0 34 0 0 44 0 0 50 0 0 59 20 30 80
MM+bio+ his 503 487 499 503 498 501
a) El experimento anterior se aprovechó para mapear los genes involucrados en la síntesis de biotina.
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81
multánea de al menos 2 bases distintas y cuya reversión no necesariamente requieren del mismo mecanismo de mutación Respuesta alternativa que, que, en caso de aparecer, debe ser considerada correcta: Sí, si utilizo MM+ bio (o MM+ arg) 4) Un investigador está interesado en mapear y estudiar 3 genes bacterianos a, b y c involucrados en la biosíntesis de aminoácidos. Se sabe que los productos de los genes a, b y c catalizan las siguientes reacciones: C Leu
Val
donde X es un compuesto que puede ser sintetizado por ambas cepas de bacteria creciendo en un medio mínimo (MM). Para realizar el estudio dispone de 2 cepas de bacterias: Hfr a + b+c+ Str S y otra F- a- b- c- Str R con las cuales realizó un experimento de conjugación interrumpida obteniendo los siguientes resultados: s e t n
se
muestran.
Alu I P
AmpR
AmpR
P = promotor del operón Vector 1
X
que
Alu I
Ile
A B
vectores
lac
Vector 2
B) Luego se transformaron bacterias a - b- c+ AmpS con uno de los vectores y se plaquearon las bacterias transformantes en ágar + MM + ampicilina + un aminoácido. Cada una de las colonias que crecieron en estas placas fueron luego cultivadas en ágar + MM + ampicilina obteniéndose los siguientes resultados: Vector 1: Todas las bacterias que crecieron en MM+Amp+Leu crecieron también en MM+Amp. -De las bacterias que crecieron en MM+Amp+Val el 60% crecieron en MM+Amp.
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82
salen primero debe ser bastante mayor que la de los de mutantes auxótrofas incapaces de utilizar galactosa genes que salen al final. como nutriente. El mapeo de las mutaciones respond) MM+Amp+Leu = a- b+ AmpR y a+ b+ AmpR. Es sables de esta característica permitió clasificarlas en 2 decir, b+ AmpR grupos de ligamiento distintos (mendelianamente inMM+Amp+Val = a+ b- AmpR y a+ b+ AmpR. Es dependientes). decir, a+ AmpR a) ¿Significa este resultado que los genes involucrados MM+Amp = a+ b+ AmpR se encuentran en 2 cromosomas distintos? ¿Por qué? La lactosa no importa, sólo se usa para inducir la ex- b) ¿Cuál será el fenotipo resultante del diploide resul presión mediada por el promotor lac. tado del cruzamiento entre mutantes (haploides) pere) De las bacterias transformadas con el vector 1 todas tenecientes a distintos grupos de ligamiento? ¿Por las bacterias b+ también eran a+, pero sólo el 60% de qué? las a+ eran b+. c) ¿Cómo serán las proporciones fenotípicas de las En cuanto a las bacterias transformadas con el vector levaduras (haploides) derivadas de las ascosporas una
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83
un parental grande, la progenie era toda petite o toda grande. ¿Cómo se explican estos resultados? Solución: el EMS afecta, fundamentalmente a genes nucleares (que pueden afectar a las funciones mitocondriales para resultar en un fenotipo petite -no realizan fosforilación oxidativa- porque éstas importan ciertas proteínas codificadas en el núcleo) y el EtBr tiene mayor especificidad por los genes mitocondriales que se heredan uniparentalmente . 3) En un estudio sobre mutaciones inestables de maíz, un investigador propuso los siguientes genotipos para dos plantas distintas de esta especie: I) C/cd ; Ac+/Ac-
LTR ⇑ Prom Promot otor or
Sec abierta lectura ⇑ Intr Intrón ón
LTR
Mediante sondas específicas para el intrón y para la secuencia abierta de lectura de Ty, es posible seguir, mediante Southern blots, la transposición de Ty en el genoma de levaduras. Cuando las levaduras se cultivaron usando glucosa (es decir en ausencia de galactosa) no se evidenciaron cambios importantes (es decir no hubo -o hubo muy poca- aparición de nuevas copias de Ty). En presencia de galactosa (inductor), en cam-
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84
decir, se detecta una banda con la sonda para P y no se detecta nada con la sonda para el intrón. f) ¿Cómo explica Ud. este resultado? Respuestas: a) Hubo transcripción inducible por galactosa del ORF, splicing con pérdida de intrón, lo cual permitió la retrotransposición retrotransposición (RNAcDNAintegración cromosómica) del Ty mediada por la RT endógena o codificada por el mismo gen. b) Para poder convertir el sistema de expresión de la RT en inducible y poder atribuir diferencias de com portamiento de la transposición a la transcripción inducible del gen.
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85
i) ¿Qué le indican esto resultados en cuanto a la determinació bioquímica del color de las lores?
ii) ¿Cuál fue el objetivo de los experi entos mostrados en a y b? iii ¿Para qué se realizó la ibridación on F3’H? iv) ¿Qué explicación molecular podría dar a los c m bi s en el perfi l de los tran scriptos? A continuación, a partir de ADN g nómico de las pl ntas ampli ican fragm ntos conte iendo el RF F3’5’H (ver p rte c de la figura anterio ): v) ¿Cuál fue el objetivo de este experi ento? vi) ¿Qué evidencian estos esultados? Los fragmentos de 4 kb 3,3 kb fueron clonado s y se uenciados. El fragmento e 4 kb de l a línea 13-9 presentab en
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86
mecanismo/s involucrado/s eran desconocido/s hasta tanto no se sintetizan el pigmento que produce la coloel momento. ración azul (delfinidina). i) Estos resultados indican que hay diferencias en la vii) Los resultados claramente muestran que GsTRIM1 acumulación de las distintas antocianinas en ambas y LjTRIM se agrupan juntos y apartados del resto de plantas (azules y rosadas), en particular, las plantas los TRIMs reportados previamente, indicando que rosadas carecen de delfinidinas. Pero la acumulación estos dos transposones son evolutivamente diferentes de cianidinas es similar para ambas. de los otros elementos TRIM de plantas. ii) Como los autores saben que la F3’5’H es la enzima vii) Al inicio del problema, en el enunciado, se indica clave en la síntesis de delfinidina y viendo los resulta- que también disponen de G. triflora cv. Macyri. Por lo dos anteriores, analizan si hay diferencias en cuanto a tanto, el experimento sería hacer Southern blot de ésta la expresión del gen para esta enzima entre los dos conjuntamente con las 3 plantas analizadas de G. scatipos de plantas y analizan si esto podría explicar los bra e hibridar con dTgs1 y GsTRIM1 (Southerns sefenotipos observados. parados, obviamente) y observar la aparición de mu-
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87
cromosomas (disómicas, ver columna 1) y a su patrón de bandas (columnas 2 a 5): Descendencia de los 5 cruCromosoma en
sionar gametas desbalanceadas. Luego repitió los cruzamientos arvejas de flores blancas por arvejas de flores rojas. ¿Qué proporciones de plantas con flores
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88
d) En este trabajo solo se examinaron mutaciones de tipo estructural y no aquellas de tipo puntual? ¿Se le ocurre algún modelo experimental (no humano) para
ter producción en dicha población a lo largo del proceso selectivo serán a) creciente, b) decreciente, c) primero creciente y luego decreciente d) constante.
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89
argentino) aunque de productividad de lana muy inferior. Indeciso, compra esas llamas y vuelve al establecimiento a realizar nuevos cruzamientos entre ellas y
dor con mi característica de interés para así usarla en selección asistida por marcadores. Respuesta alternativa: llamas transgénicas con el gen de la oveja.
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90
riancia genética) que puede segregar y de la influencia Xs - Xo ambiental (no heredable) o variancia ambiental. b) a la varianza ambiental e interacción genotipo b) La selección diferencial fue de 8 – 5 = 3 mm y la ambiente (carácter cuantitativo, con alta compo-
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91
los mismos contribuyen a la construcción de posibles Solución: Datos: 1024 de los 1028 insectos machos modelos teóricos que pueden ayudar a manejar el de- colectados fueron SS (porque mediante el cruzamiento sarrollo de estas resistencias y, por lo tanto, el impacto pruba con las hembrasdoble recesivas rr, el 100% de
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