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1 MIDTERM EXAM points total (6 questions) 8 pages PLEASE PUT YOUR NAME ON ALL PAGES, SINCE THEY WILL BE SEPARATED DURING GRADING. PLEASE NOTE: YOU MUST ANSWER QUESTIONS 1-4 AND EITHER QUESTION 5 OR 6. RAISE YOUR HAND IF A QUESTION IS UNCLEAR χ 2 table P df χ2 =Σ (O-E) 2 E Interference = 1-COC I Yellow/orange Precursor (tyrosine) C gene (tyrosinase) Y Intermediate B T (Tabby gene) A (agouti) 1 Black O Orange
2 Problem 1. (20 points) In this pedigree, colorblindness is represented by horizontal hatching, and is determined by an X-linked recessive gene (g); the dominant allele for color-vision is (G). Hemophilia is represented by vertical hatching and is determined by the X-linked recessive gene (h); the dominant allele for normal blood clotting is (H). colorblind hemophilia husband mailman both P, R, or?: (a). What is the genotype of the female in generation 2. Show the arrangement of alleles on the X- chromosomes below. (b). Indicate whether each child in the third generation received a parental combination of alleles (with a P) or a recombinant combination of alleles (with an R) from the mother. If it cannot be determined, indicate with a (?) (c). When the husband walks in on his wife and the mailman, he becomes so infuriated that he immediately files for divorce and takes his three kids to live in another state with him; he forbids his three children to see or even talk with their cheating mother. The mother, finding out during the divorce trial that she is pregnant with the mailman s child, rushes to marry him. Together, the new couple has a total of two children (as shown in the above pedigree). Yet, the mother and mailman s always keep the secret of the mother s previous family from their children. However, when the female child (offspring of the mother and the mailman/2nd husband) meets the colorblind male (offspring of the mother and her first husband) at college, they fall in love, elope, and are currently expecting their first child. What is the probability that this child will be colorblind? What is the probability that this child will have hemophilia? 2
3 Problem 2 (20 points) You acquire a female tabby cat (not orange or silver; see the projection in class) and a solid black male cat. (a). What can you say about the genotypes of the two cats with respect to the Agouti (A/a), orange (X O /X o ), and I (I/i) genes? Use a? to indicate every allele that you are uncertain about. Female tabby cat genotype: Male black cat genotype: (b). Your two cats mate with each other and the first litter of kittens produces 2 tabby kittens that look just like the mother, 2 silver tabby kittens and 1 black kitten. Now what can you say about the genotypes of the parent cats? Female tabby cat genotype: Male black cat genotype: (c). Your two cats produce a total of 48 more kittens over the years. Shortly after the first litter of kittens, however, you got a new neighbor who owns a male black cat that you find out is an aa ii homozygote. As your tabby cat produces litters year after year, you begin to wonder if she has been having an affair with the next-door cat. Tabulating all the kittens your tabby cat has produced gives data shown in the table below: Kitten phenotype Expected (E) Observed (O) (O-E) 2 (O-E) 2 /E Standard tabby 19 Silver tabby 6 Black 23 Fill in the Expected column in the table above based on the hypothesis that your tabby has been mating exclusively with YOUR black male cat. Be sure to show how you arrived at these values. Use χ 2 analysis to evaluate your hypothesis: χ 2 value = # of degrees of freedom = P value = (d). What do you conclude about the parentage of the kittens (BE SPECIFIC)? 3
4 (e). Why could you NOT have hypothesized that the neighbor s cat had fathered SOME of the kittens, and test that hypothesis by χ 2 analysis? Problem 3. (25 points) From experiments discussed in quiz section 5 you learned the following about the inheritance pattern of bristle length, eye shape, and body color in Drosophila: Bristle length: The long bristle phenotype (L) is dominant to short bristles (l). Eye shape: The half-moon eye phenotype (H) is dominant to round eyes (h). Body color: The gray body color phenotype (G) is dominant to amber body color (g). Additionally, you learned that the genes responsible for all three of these traits reside on the X- chromosome and that the body color gene (G) and bristle length gene (L) are linked to a polymorphic DNA marker (here designated M). The linkage relationships are shown below. G M 17cM L M 35cM Both the body color gene (G) and the bristle length gene (L) are linked to the marker, but it is not known if the two genes are linked to each other, and their order relative to the marker (M). To further evaluate their linkage relationship, you examine the progeny of the same female/male pair that was examined in lab. (Recall that the female was a long, round, gray fly and the male was a long, half-moon, amber fly.) Suppose that when these flies were mated, the progeny had the phenotypes for bristle length, body color, and eye shape as listed below. Cross: long, round, gray female x long, half-moon, amber male Phenotype # Females # Males Total Long, gray, round Long, gray, half-moon Long, amber, round Long, amber, half-moon Short, gray, round Short, gray, half-moon Short, amber, round Short, amber, half-moon
5 TOTAL (a). Using the data in the table above, draw out the genetic order of the body color gene (G), bristle length gene (L), and the marker (M), and indicate the distances between each gene and the marker, and the distance between the two genes. (b). What does the data in the table above say about the linkage relationship between the genes? (Problem 3 continues on the following page) 5
6 Information about the location of the half-moon gene (H) on the X-chromosome is still not known. Suppose you want to see if it is linked to the body color gene (G). You identify the mutation causing the recessive short bristle trait, and observe that it creates a restriction enzyme site (see below). You create a probe that is complementary to this gene that can be used in Southern blot analysis. probe 2kb 5kb 3kb G g = restriction site You take one of your long, gray, half-moon female progeny (from the previous cross) that is heterozygous for all genes and mate it to a test cross male. (c). What is the genotype of the FEMALE used in this testcross? Indicate dominant and recessive genes on each X chromosome (you should be able to deduce this genotype from the cross that produced this female). You extract DNA from the female and testcross male used in this cross, as well as all female offspring from this cross that have half-moon eyes. You proceed to perform the restriction digest and probe for the body color gene. Representative results are shown below: parents half-moon female progeny derived from the cross 5kb 2kb (d). Assuming the results shown above are representative of results obtained with the remaining halfmoon female offspring, what does this information suggest about the linkage relationship between the G and H genes? If they are linked, what map distance separates the genes? (e). Draw the genetic map that is consistent with this data and that from the first part of this problem. Be sure to show the L, G, and H genes and M marker on the map and the linkage distances separating these genes/markers. Indicate the two possible locations for the half-moon gene (H) that are consistent with 6
7 this data. 7
8 Problem 4. (15 total points) You are studying aging in fruit flies and have generated six different homozygous long-lived fly mutants (you may assume that each of these mutant strains bears a mutation affecting only ONE gene). You now wish to determine how many genes these six mutants represent and proceed to set up pairwise crosses with all of the homozygous mutants. Results of this analysis are shown in the table below (where the intersection represents the phenotype of the offspring resulting from a particular cross): Mut 1 Mut 2 Mut 3 Mut 4 Mut 5 Mut 6 WT Mut Mut Mut Mut Mut Mut indicates all offspring have normal lifespan. - indicates all offspring are long-lived. WT = a wild type strain of flies. (a). How many complementation groups do these mutations represent? (b). Describe which mutations fall into each complementation group. (c). In more recent experiments you isolate another long-lived fly mutant (Mut 7) and proceed to cross this mutation to your previously characterized long-lived mutants with the following results: Mut 1 Mut 2 Mut 3 Mut 4 Mut 5 Mut 6 WT Mut indicates all offspring have normal lifespan. - indicates all offspring are long-lived. WT = a wild type strain of flies. What are these results telling you? 8
9 ANSWER EITHER ONE OF THE FOLLOWING TWO QUESTIONS (if you answer both, you will receive the statistical average of the two scores): Problem 5. (20 total points) The following graph illustrates the time course of DNA repair in E. coli following UV-light induced DNA damage. Each curve on the graph represents the result of a particular experiment carried out using either wild type or UV-sensitive E. coli mutants. (1) DNA damage per kb (2) (3) UV light Time Given your knowledge of UV-induced DNA damage repair in E. coli, choose the curve (1, 2, or 3) from the above graph that most closely resembles the outcome you would expect from the E. coli cell types and experimental conditions described in the table below (you can use the same curve multiple times): E. coli Cell Type wild type wild type uvra mutant phr mutant Experimental Conditions in blue ( nm) light in dark in dark in blue ( nm) light Corresponding Curve (1, 2, or 3) 9
10 Problem 6. (20 points) E. coli strains bearing a mutation of the deoxyadenosine methylase (dam - ) show elevated mutation rates relative to wild type cells, owing to a defect in mismatch repair (for reasons discussed in class). However, the production of very high levels of the wild type enzyme in E. coli cells also results in increased mutation rates relative to wild type E. coli strains. From your knowledge of the normal role this enzyme plays in mismatch repair processes, propose an explanation for this finding (a diagram may help immensely). DO NOT WRITE BELOW THIS LINE problem 1 problem 2 problem 3 problem 4 problem 5/6 /20 /20 /25 /15 /20 total /100 10
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