MIDTERM EXAM 1
100 points total (6 questions)
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
husband mailman hemophilia
both
P, R, or ?: _?_ _P_ _P_ _R_ _P_
(a). What is the genotype of the female in generation 2. Show the arrangement of alleles on the X- chromosomes below. hG Hg or Xh,G/XH,G
(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?
g Male = X Y = 50% H,G H,G h,g Female = XgXg = 50% X /Y X /X
What is the probability that this child will have hemophilia?
Male = XhY = 50% 1 Female = XhXh = 0%
2 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 (XO/Xo), and I (I/i) genes? Use a “?” to indicate every allele that you are uncertain about.
Female tabby cat genotype: A? XoXo ii
Male black cat genotype: aa Xo ??
(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: Aa XoXo ii
Male black cat genotype: aa Xo Ii
(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 12 19 49 4.1 Silver tabby 12 6 36 3 Black 24 23 1 0
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 χ2 analysis to evaluate your hypothesis:
χ2 value = _7.1______# of degrees of freedom = _2______P value = __0.025______
(d). What do you conclude about the parentage of the kittens (BE SPECIFIC)? The P value is below 0.05 so the odds of a chance deviation from my hypothesis of this magnitude is extremely remote. Therefore I reject the hypothesis and conclude that some of the kittens were likely the result of matings with the next-door cat.
(e). Why could you NOT have hypothesized that the neighbor’s cat had fathered SOME of the kittens, and test that hypothesis by χ2 analysis?
Because you do not know how many kittens were produced by the next door neighbor’s cat. Without this information you cannot define the expected outcome.
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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.
GM17cM LM35cM
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 0 249 249 Long, gray, half-moon 502 0 502 Long, amber, round 0 253 253 Long, amber, half-moon 498 0 498 Short, gray, round 0 247 247 Short, gray, half-moon 0 0 0 Short, amber, round 0 251 251 Short, amber, half-moon 0 0 0 TOTAL 1000 1000 2000
(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.
Since the body color gene (G) and bristle length gene (L) are both linked to the same molecular marker there are two possible maps that are consistent with the known linkage relationships to the marker:
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GM17cM 35cM L 52cM OR L 18cM GM17cM
35cM
The data above indicate that the G and L genes are segregating independently. Therefore, these genes must be 50cM or more apart. This indicates that the first genetic map is correct.
(b). What does the data in the table above say about the linkage relationship between the genes? This says that the G and L genes are unlinked.
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
5kb G = restriction site g 2kb 3kb
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).
l hG
L Hg
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:
5 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 half- moon 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?
There appear to be six parental types and 4 recombinant types (circled) among the half-moon female progeny. Since the recombinant types are found less often than the parental types these results are consistent with linkage between the molecular marker and the Half-moon trait. Because the RFLP corresponds to the body color trait (G), these results indicate that G and half-moon (H) are linked. The map distance separating the G and H genes is calculated as follows:
4/10(100) = 40cM
(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 this data.
H 40cM GM17cM 35cM L
OR 35cM GM17cM H L
40cM
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
6 Mut 1 - + - - + + + Mut 2 - + + + - + Mut 3 - - + + + Mut 4 - + + + Mut 5 - + + Mut 6 - +
+ 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?
These mutations represent 3 complementation groups
(b). Describe which mutations fall into each complementation group.
One complementation group consists of the mutations 1, 3 and 4 Another group consists of the mutations 2 and 6 The third group consists of mutation 5
(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 7 ------
+ 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?
When you cross Mut 7 to WT you get long-lived mutants. This tells you that Mut 7 is a dominant mutation. Therefore, you would expect to see the long-lived phenotype with whatever you cross this mutant to (i.e., it fails to complement everything it is crossed to). This is why dominant mutations cannot be used in complementation experiments.
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.
7 (1)
DNA damage (2) per kb
(3) Time UV light
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 Experimental Conditions Corresponding Curve (1, 2, or 3)
wild type in blue ( 300-500nm ) light
wild type in dark
uvrA mutant in dark
phr mutant in blue ( 300-500nm ) light
8 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).
Increased methylating activity could decrease the time during which the newly polymerized DNA is hemimethylated. The mismatch repair enzymes would then have less of an opportunity to find a hemimethylated strand of DNA to use as a guide to distinguish between parental and daughter DNA strands. As a result, fewer mismatches would be repaired, and for those that are repaired fewer would be repaired in the correct fashion and the mutation rate would increase.
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