MCB142/IB163 Mendelian and Population Genetics 9/15/02

Transcription

1 MCB142/IB163 Mendelian and Population Genetics 9/15/02 Practice questions lectures 2-4 (Mendel s first law and pedigrees, single gene action and expression, genetic polymorphisms) Question 1 1. having two identical alleles of a given gene a. phenotype 2. the allele expressed in the phenotype of the heterozygote b. alleles 3. alternate forms of a gene c. gametes 4. observable characteristic d. gene 5. reproductive cells containing only one copy of each gene e. segregation 6. the allele that does not contribute to the heterozygote phenotype f. heterozygote 7. the cross of an individual of ambiguous genotype with a g. dominant homozygote recessive individual 8. an individual with two different alleles of a gene h. haplotype 9. the heritable entity that determines a characteristic i. test cross 10. the alleles an individual has j. genotype 11. the separation of the two alleles of a gene into different gametes k. recessive 12. offspring of the P generation l. homozygote 13. the set of polymorphic alleles of closely linked genes m. F1 which are inherited together Question 2 In a species of Mexican mollies, fish with ragged fins and fish with nonragged fins were found. When mollies of a true-breeding (= pure-breeding) ragged strain were crossed with nonragged fish, the F 1 all had ragged fins, and in the F 2 3 ragged appeared to 1 nonragged. Diagram this cross, letting R = ragged and r = nonragged, and give parental phenotypes, genotypes, and germ cells, as well as F 2 genotypes and phenotypes. Question 3 A C/c ch female rabbit is crossed to a c e /c male rabbit. What proportion of the baby rabbits do you expect to be C/c e in genotype? Question 4 A true-breeding strain of blue-eyed bats is developed. In bats, as in humans, the male is the heterogametic (XY) sex. Males of this true breeding blue eyed strain strain are mated to females of a true-breeding brown-eyed strain of bats. The F 1 are all brown-eyed. What would be the phenotypes of the F 1 bats in the reciprocal cross if: (a) The blue-eyed character is an X-linked recessive? (Show your work) (b) The blue-eyed character is an autosomal recessive? (Show your work) Pedigree questions For all the following pedigree questions, unless otherwise stated, assume: (a) The trait is due to the action of a single gene, which has, for the purposes of our study, only two phenotypically distinguishable alleles (one for the trait, one for wild-type if the trait is rare). (b) The trait shows complete penetrance and there are no age of onset effects, i.e., the trait is expressed at birth. (c) The trait shows uniform expressivity. (d) There are no new mutations in any of the individuals in the pedigrees. (e) Mendelian segregation has occurred. (f) If the trait is rare, assume individuals marrying into the pedigree are not carriers, unless there is evidence from the pedigree to the contrary. (g) For maternally (mitochondria) inherited traits, assume there is no heteroplasmy, i.e., individuals are homogeneous for the mitochondrial variation. Page 1

2 Question 5 The following pedigree depicts the inheritance of a rare fully penetrant hereditary disease affecting muscles: For the following possible modes of inheritance, indicate which is (are) compatible with the pedigree, and which are incompatible with the pedigree, and in these cases indicate one individual who is incompatible with that mode of inheritance and why they are incompatible. (a) autosomal dominant (b) autosomal recessive (c) X-linked dominant (d) X-linked recessive (e) Y-linked inheritance (f) Maternal (mitochondrial inheritance) Question 6 A couple without a family history of Tay-Sachs disease (a rare autosomal recessive) have two normal children and an infant affected with Tay-Sachs. The sister of the male parent wants to mate with the brother of the wife. In such a mating, what would be the probability of their first child having Tay-Sachs disease? Question 7 Huntington's disease is an autosomal dominant with a variable age of onset, but most cases occur after the age of 35. A young woman whose great-grandfather had Huntington's disease marries and is considering having a child. Both her father and grandfather were killed in wars while in their twenties. If she consulted you, how would you advise her about the risks to her child? (Make sure you give her probability values for your predictions.) Question 8 A young woman is worried about having a child because her mother's only sister had a son with Duchenne muscular dystrophy (DMD). The young woman has no brothers or sisters. (DMD is a rare X-linked recessive disorder.) (a) Draw the relevant parts of the pedigree of the family described above. (Be sure to include the grandmother, the three women mentioned, and all their mates.) (b) State the most likely genotype of everyone in the pedigree. (c) Calculate the probability that the young woman's first child will be a boy with DMD. Question 9 Drosophila eyes are normally red. Several purple-eyed strains have been isolated as spontaneous variants (mutants) and the purple phenotype has been shown to be inherited as a Mendelian autosomal recessive in each case. Two purple-eyed pure strains were crossed. If the purple mutations in this case are in the same gene (that is they are allelic), the F 1 is expected to have what genotype and phenotype. Page 2

3 Question 10 In the purple penguin, a series of alleles occurs at the p locus on an autosome. All alleles affect the color of feathers: p d = dark-purple, p m = medium-purple, p l = light-purple, and p vl = very pale purple. The order of dominance is p d > p m > p l > p vl. This means that p d is dominant to p m and p l and p vl, and p m is dominant to p l and p vl, and p l is dominant to p vl. If a light-purple female, heterozygous for very pale purple, is crossed to a darkpurple male, heterozygous for medium-purple, the ratio of genotypes and phenotypes expected among the baby penguins would be Question 11 You have gone out into nature and recorded the flower petal color of 250 individual plants of a certain species. You found that 85 had blue petals, 57 had green petals and 108 had yellow petals. From previous experience you know that this is a one locus, two allele system where the heterozygous individuals are green. Calculate the frequencies of the two alleles, denoted B and Y respectively for blue and yellow, and the 95% confidence intervals for these estimates of the population level allele frequencies. Question 12 Four babies were born in a hospital on the same night. ABO and MN blood typing have been done on all the babies and parents. Assign the babies to the correct families. (Each family has only one child.) (In the ABO blood group system, alleles A and B are codominant, and allele i is recessive for the O blood group, while the MN bllod group system is codominant.) Family no. Mother Father Child no. Child 1 O, M O, N 1 O, N 2 AB, MN O, N 2 AB, M 3 A, M B, MN 3 A, MN 4 O, N O, N 4 O, M Page 3

4 Answers Question 1 1 l, 2 g, 3 b, 4 a, 5 c, 6 k, 7 i, 8 f, 9 d, 10 j, 11 e, 12 m, 13 h Question 2 R dominant ragged, r recessive nonragged, P generation: RR (ragged) x rr (nonragged), gametes all R and all r respectively from the 2 parents, F1 generation, all Rr (ragged), gametes ½ R and ½ r from both parents, F2 generation, ¾ R- (ragged) and ¼ rr (nonragged), which breaks down to ¼ RR, ½ Rr, ¼ rr. Question 3 The four equally likely genotypes in the offspring are C/c e, C/c, c ch / c e, and c ch /c, so answer is ¼. Question 4 (a) X-linked recessive trait: We are given that pure breeding blue eyed males mated to pure breeding brown eyed females results in F1 individuals all with brown eyes, males and females. If the trait is X-linked recessive, then blue is recessive and denoting the allele by b, this cross was female (X+/X+) (brown eyes) x male Xb/Y (blue eyes) with F1 females (X+/Xb) (brown eyes) and F1 males (X+/Y) (brown eyes). In the reverse cross the female is pure breeding blue eyed (Xb/Xb), and the male is pure breeding brown eyed (X+/Y), so that the F1 females will all be (Xb/X+) (brown eyes) and the males will all be (Xb/Y) (blue eyes). (b) Autosomal recessive trait: The original cross would be female (++) (brown eyes) x male (bb) (blue eyes) with all F1 individuals +/b (brown eyes). In the reverse cross the female is bb (blue eyes) and the male is ++ (brown eyes) and all the F1 individuals will be as in the first cross, i.e., all +/b (brown eyes). Question 5 (a) autosomal dominant, incompatible, either I-1 or I-2 would have to be affected since they have two affected children, and II-2 would have to be affected since she has two affected children. (b) autosomal recessive, compatible, although II-1 marrying in would have to be a heterozygous carrier. The fact that all affected individuals are male is slightly more in favor of an X-linked recessive, but autosomal recessive cannot be excluded. (c) X-linked dominant, incompatible, I-1 would have to be affected since she has two affected sons, and all the daughters of II-6, i.e., III-4-7 would have to be affected. (d) X-linked recessive, compatible. (e) Y-linked inheritance, incompatible, I-2 would have to be affected, also II-3 and III-8-9, and III-2-3 would not be affected. (f) Maternal (mitochondrial inheritance), incompatible, individual I-1 would have to be affected to have two affected sons, and also all of her children (II-2-6) would then have to be affected, etc. Question 6 The couple are each heterozygous so in both sets of their phenotypically normal parents (the grandpararents of the affected child) there must have been one heterozygote, i.e., they were T/t x T/T (we assume since the disease is rare that only 1 was a heterozygote carrier) and the brother and sister each have a 1/2 chance of being T/t. So, in the proposed mating, there is a ½ x ½ chance of them both being heterozygous carriers, in which case ¼ on average of their children would be affected, so overall probability = (1/2)(1/2)(1/4) = 1/16. Question 7 Apparently the father and grandfather did not show symptoms of Huntington's disease, but since it is a lateonset disease, they probably would not have expressed it in their early twenties. There is a 1/2 chance that the grandfather inherited the allele, a 1/2 chance that the father received it from him, a further 1/2 chance that the woman received it, and then a further 1/2 chance that her future child would receive it. The overall chance that the woman's child will develop the disease is 1/2 1/2 1/2 1/2 = 1/16 Page 4

5 Question 8 (c) The grandmother must have been D/d. There is a 1/2 chance that the mother of the young woman is D/d and, if so, a further 1/2 chance that the woman herself is D/d. If she is, 1/2 of her sons will have DMD. Since the probability of a son is also 1/2, the overall probability is 1/2 1/2 1/2 1/2 = 1/16. Question 9 Denoting the two alleles as p1 and p2 in the same gene, the parental cross is p1/p1 x p2/p2 (purple x purple) so the F1 are all p1/p2 and will show the recessive phenotype of all purple. Question 10 The cross is p l /p vl x p d /p m, so that the 4 equally likely genotypes and their phenotypes in parentheses are p l /p d (dark purple), p l /p m (medium purple), p vl /p d (dark purple), and p vl /p m (medium purple), so that the phenotypic ratio is 1: 1 dark to medium purple. Question 11 The allele frequency of B is (2 x )/500 = 0.454, and of Y is ( x 108)/500 = 0.546, check f(b) + f(y) = 1. The variance for both alleles is p(1-p)/(2n) = (0.454)(0.546)/500 = , sd = 0.022, and 2 x sd = 0.044, so 95% confidence interval for B allele frequency is (0.410, 0.498) and for the Y allele is (0.502, 0.590). Question 12 Family no. Child no This is the only way all 4 families and children can be correctly assigned. Page 5