Recitation Section 15 Answer Key April 4-5, 2005 Diploid Genetics and Chromosomal Inheritance A. Diploid Genetics For each set of data below, determine the genotype of the parents in cross 1. Where it applies, indicate which phenotypes are dominant and which are recessive. 1. Mice I a) cross 1: red-eyed mouse ____________ X white-eyed mouse ____________ gives F1: all red-eyed cross 2: red-eyed F1 X red-eyed F1 gives F2: 36 red-eyed 13 white-eyed Cross 1: red eyes X white eyes gives all red eyes – a likely model is that eye color is controlled by one gene that has 2 alleles producing two phenotypes, red and white eyes. The red eyed phenotype is likely dominant. Therefore, choose appropriate symbols (one letter per gene, capital letter for allele with dominant phenotype): •R - red allele. Dominant phenotype of red eyes. •r - white allele. Recessive phenotype of white eyes. Then: Cross 1 is: RR X rr ----> all Rr (all red eyes). Cross 2 is: Rr X Rr ----> 3:1 red:white. The predictions agree with the data. b) cross 1: long-eared mouse ____________ X short-eared mouse ____________ gives F1: 12 long-eared 10 short-eared cross 2: long-eared F1 X long-eared F1 gives F2: 36 long-eared 13 short-eared Cross 1: long ears X short ears gives some long and some short ears – a likely model is that ear length is controlled by one gene with 2 alleles resulting in two phenotypes, long ears and short ears. From this data, we cannot determine which allele has the dominant phenotype. Cross 2: long ears X long ears gives some long and some short ears – a likely model is that long ears is the dominant phenotype. Therefore, the appropriate symbols are: L - long ears and l - short ears . Then: Cross 1 is: Ll X llÆ 1:1 long ears: short ears Cross 2 is: Ll X Ll Æ 3:1 long ears: short ears The predictions agree with the data. 2. Flowers cross 1: blue-flowered plant ____________ X white-flowered plant ____________ gives F1: all pale-blue-flowered cross 2: pale-blue F1 X pale-blue F1 gives F2: 27 blue 49 pale-blue 24 white Cross 1: blue flowers X white flowers gives all pale blue flowers – a plausible model is that color is controlled by one gene with 2 alleles, resulting in blue and white flowers, and that neither phenotype is dominant. This is an example of co- dominance. Then, BB- blue, B'B - white, BB'- pale blue phenotypes respectively. Then: Cross 1 is: BB X B'B'Æ all BB', pale blue flowers. Cross 2 is: BB' X BB'Æ 1:2:1 BB, blue flowers: BB', pale blue flowers: B'B', white flowers. The predictions agree with the data.3. Blood Type Blood type is determined by one gene that has three alleles: IA - allele associated with the A phenotype, co-dominant with the B phenotype IB - allele associated with the B phenotype, co-dominant with the A phenotype IO - allele associated with the O phenotype, recessive to both the A and B phenotype a) cross 1: person, type A blood ____________ X person with type B ____________ gives F1: all type AB blood cross 2: type AB F1 X type AB F1 gives F2: 2 type A 4 type AB 1 type B Then: Cross 1 is: IAIA X IBIB Æall type IAIB Cross 2 is: IAIB X IAIBÆ1:2:1 types A:AB:B (note skewed ratio with small number of progeny) b) cross 1: type A blood ____________ X type B ____________ gives F1: 2 type A blood 3 type AB blood 1 type B blood 2 type O blood Then: Cross 1 is: IAIO X IBIO----> 1:1:1:1 types A:B:AB:O (note skewed ratio with small number of progeny) The predictions agree with the data. 4. Mice II cross 1: tail-less mouse ____________ X normal mouse ____________ gives F1: 10 tail-less 9 normal cross 2: tail-less F1 X tail-less F1 gives F2: 10 normal 21 tail-less 9 dead Tail-lessness and lethality are controlled by one gene with two alleles: T allele - dominant tail-less phenotype; recessive lethal phenotype t allele - recessive normal-tail phenotype; dominant viable phenotype therefore: TT - tail-less and dead Tt - tail-less and alive tt - normal-tail and alive Then: Cross 1 is: Tt X ttÆ 1:1 Tt (tail-less) : tt (normal) Cross 2 is: Tt X TtÆ 1:2:1 normal:tail-less:dead The predictions agree with the data. Note that usually in such problems the number of dead animals is not reported, but needs to be inferred from the rest of the data (i.e. “missing” phenotype or numbers of progeny, i.e. ratios that don’t make sense).B. Chromosomes and Recombination 1. What is the physical basis of the genetic inheritance? Genes are fragments of DNA that encode when, where, and what product (protein or functional RNA) is to be made. Genes are assembled together into chromosomes. During cell division chromosomal segregation can be observed with a microscope. Chromosomes are inherited from parents. Thus, sexually reproducing diploid organisms get one copy (allele) of each gene from each parent and pass one allele on to each of their offspring at random. These alleles are passed on the chromosomes inherited from each parent. 2. Why is sexual reproduction a powerful source of variation? Sexual reproduction allows for great diversity and fast change (through bringing together genetic information from two parents). Gamete production in meiosis (2nÆ 4nÆ 2nÆ n) allows for reshuffling of parental genetic information through independent segregation of chromosomes. 3. Why was it evolutionarily advantageous to develop the ability to recombine chromosomes? Recombination—the exchange of parts of chromosomes between homologous pairs of chromosomes—increases the rate of reshuffling of parental genetic information compared with the independent segregation of chromosomes alone. Faster reshuffling of genetic information means more variation in the population. In turn, that leads to better survival for a species because, for any change in the environmental conditions (selective pressure), it increases the chances that an individual will arise that is better suited to the new conditions (selective advantage). This individual and the individual’s offspring will have be more likely to survive and procreate under the new conditions, and, thus, greater proportion of the population will end up with the advantageous allele in the future. Think about the following question for next time (we will talk about it in Section 16): Recombination can occur anywhere along the length of
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