Genetics 371B Practice problems--Autumn 2000 week 3
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Which two questions would you like to see worked out in quiz section? Email Raghu <raghu@u.washington.edu> with your choices by Monday evening.
| 1. | The amount of recombination between the linked genes e and f is found to be 9%. Gene g is linked to e and f. Gene g undergoes recombination with gene e with a frequency of 5%, and with gene f with a frequency of 4%. | |
| (a) |
A cell of a heterozygote (e+ f / e f+) is about to undergo meiosis (the chromosomes all have two sister chromatids each). Pick any one homolog of the chromosomes carrying these two genes, then draw a picture of the homolog and mark the alleles of the two genes on the homolog. Self-test: Click here after you're done with your diagram |
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| (b) | Draw a picture of the relevant homolog pair immediately before and immediately after a single crossover event between the two genes. Be sure to mark the genes on the chromosome. | |
| (c) | Explain in your own words what the first sentence of the question means ("The amount of recombination..."), outlining the cross and the progeny phenotypes/ratios that would lead to the conclusion that e and f are linked. (In this instance, since you aren't told what genes e and f do, you'll just have to call the phenotypes by the name of the allele ("Phenotype e" or "Phenotype f+", etc.) | |
| (d) | What is the order of these genes on the chromosome, and what are the map distances between them? | |
Questions 2-4 are based on the following information:
The progeny phenotypes resulting from five Drosophila crosses are tabulated below. In all cases, the '+' allele is wildtype.
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Cross (i) se+ se h+ h x se se h h se = sepia eyes; h = hairy body |
Cross (ii) se+ se b+ b x se se b b b = black body |
Cross (iii) b+ b rd+ rd x b b rd rd rd = reduced bristle |
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Progeny: 400 se+ h+ 412 se h 4 se+ h 5 se h+ |
Progeny: 210 se+ b+ 222 se b 214 se+ b 208 se b+ |
Progeny: 12 b+ rd+ 15 b rd 421 b+ rd 414 b rd+ |
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Cross (iv) corr+ corr b+ b x corr corr b b corr = corrugated |
Cross (v) corr+ corr rd+ rd x corr corr rd rd |
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Progeny: 380 corr+ b+ 360 corr b 44 corr+ b 51 corr b+ |
Progeny: 70 corr+ rd+ 80 corr rd 419 corr+ rd 440 corr rd+ |
| Q. 2a Hint:
If a gene is located on the X chromosome, how many copies of the gene will the female parent bring to a cross? How about the male parent? |
| 2. | (a) | None of the genes listed in the crosses above is X-linked. How can you deduce that information from the data given above? |
| (b) | Where possible, show the arrangement of alleles in the heterozygous flies that are being crossed (i.e., indicate which alleles go together on which homolog in the heterozygote). Don't worry about the order of the genes on the chromosomes. |
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Q.3 Hints
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| 3. | For this question, the answer is being provided along with the question. Your task is to explain, step by step, how to arrive at the answer using the information provided in the question (including the crosses shown above). Your goal should be to examine the thought process involved in addressing the question, so that you will be comfortable solving similar problems even if the wording or format is changed. |
| The question:
Draw a genetic map for the five genes. Your map should indicate linkage groups, as well as approximate map distances where appropriate. |
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The answer:
se----1.1 cM ---h |---------------| |
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Q.4a Hints
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Q.4b Hints
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| 4. | Again, the answer is being provided along with each question. As before, your task is to explain, step by step, how to arrive at the answer using the information provided. | |||||||||||
| (a) |
The question: Homozygous recessive h h flies were crossed to homozygous recessive rd rd flies (both strains were homozygous normal for all other genes). The resulting F1 flies were crossed to h h rd rd flies, and 1000 F2 progeny were collected. What F2 progeny phenotypes and numbers do you expect in the F2 progeny? |
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The answer:
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| (b) |
The question: The raisin eye (rai+) gene is located 9 map units away from the hairy body gene. The normal (rai+) allele is dominant. Homozygous recessive h h flies were crossed to homozygous recessive rai rai flies (both strains were homozygous normal for all other genes). The resulting F1 flies were crossed to homozygous h h rai rai flies, and 1000 F2 progeny were collected.What F2 progeny phenotypes and numbers do you expect in the F2 progeny? |
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The answer:
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| 5. | The dominant allele O is required for pigment deposition in the iris of the human eye, while its recessive allele o causes ocular albinism. The dominant allele D is required for color perception, while its recessive allele d is associated with color blindness. Both genes are located on the X chromosome. | |
| (a) | Assuming no crossing over, what would you predict should be the results of a cross between a woman with ocular albinism who is homozygous normal for color vision, and a man normal pigmentation of the iris but who is colorblind because of the recessive d allele? | |
| (b) | Assuming no crossing over (and assuming no aberrant events), list all possible results of a cross between a woman who is heterozygous for both traits and a man who is normal with respect to both traits. | |
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Q.6 Hints
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| 6. | In Great Northwestern mice, black nose (B) is dominant over pink nose (b), while rounded ears (R) are dominant over pointy ears (r). The two loci are linked at map distance of 20 cM. True-breeding mice that have pink noses and rounded ears are crossed to true-breeders that have black noses and pointy ears. The resulting F1 progeny are crossed to each other. Using Punnett squares, predict the proportions of progeny phenotypes resulting from this F1 x F1 cross. | |
| 7. | In maize, A, B, and D are three linked genes, with a, b, and d being recessive alleles of these three genes, respectively. A
homozygous normal strain was crossed to a triple homozygous recessive
strain. The resulting F1 progeny were crossed to the triple recessive
again; the resulting progeny phenotypes were:
Show the correct order of the genes A, B, and D. Construct a map showing the distances between the genes, and calculate the coefficient of coincidence. |
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| 8. | SP1 is a membrane protein found in all human cells. To identify the chromosome that has the gene for SP1, you establish a set of mouse-human hybrid cell lines (A-E). Then, to identify the human chromosomes present in each cell line, you do hybridizations using chromosome-specific probes--probe 1 recognizes a portion of chromosome 1, probe 2 recognizes a portion of chromosome 2, etc. Your results are as follows:
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| On which human chromosome do you think the SP1 gene is located, and why? |
| 9. | A true-breeding hot pepper plant of genotype GGDD that produces yellow, round peppers is crossed to a true-breeding hot pepper plant of genotype ggdd that produces green, wrinkled peppers. The F1 progeny are of genotype GgDd and all bear yellow, round peppers. F1 plants were then test crossed to ggdd plants and the following progeny plants were produced:
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| (a) | How many progeny are expected for each phenotype class, assuming independent assortment? | |||||||||||||
| (b) | Use the chi-squared test to decide if there is evidence of independent assortment versus linkage between the two genes. Give the Chi square value, the approximate P value (just give the range of P values from the chi-squared table, as we did in class), and the appropriate degrees of freedom (df). State whether your findings are consistent or inconsistent with the hypothesis of independent assortment. | |||||||||||||
| (c) | If you wanted to be more stringent** about whether to reject or accept your hypothesis of independent assortment based on the P value obtained, would you raise or lower the cut-off P value?
**i.e., you are less willing to be tolerant of deviation from the expected values |
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| (d) | When we are assessing the data for independent assortment vs linkage, why do we assume independent assortment and then test whether the data fit independent assortment, as done above? Why do we not assume linkage and then test whether the data fit the linkage model? | |||||||||||||
| 10. | The fabled Great Northwestern Guava plants produce fruit that have thick or thin skins, tart or sweet fruit, and large or small seeds. Thick skins, tart flavor and large seeds are dominant. You have obtained, by some miraculous means, two of these mythical plants. One produces thick-skinned, tart fruit with large seeds, and the other produces thick-skinned, tart fruit with small seeds. On performing a cross between these two plants, you obtain the following progeny:
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| (a) | What are the most likely genotypes of the two plants you started out with? Explain your reasoning. [Assign your own symbols for the alleles.] | |||||||||
| (b) | Assuming independent assorment of the genes concerned, do the observed numbers of progeny deviate at all from the ratios expected for the genotypes you derived in (a)? If so, what is the probability that the deviation is just due to chance? (No need to interpolate, just give the range of probabilities from the Chi-square table.) | |||||||||
| (c) | Challenge question
Suppose you have a bin of tart fruit. You are told that it either contains only true-breeding tart fruit, or it contains a 50:50 mix of true-breeding and heterozygous tart fruit -- you don't know which. You test them by growing up a couple of plants from the bin and selfing them. Suppose the first plant that you self produces only tart fruit. That result could either mean that there were no heterozygotes in the bin, or that there were heterozygotes mixed with homozygotes, and you just happened to pick a homozygote to self. How many fruit would you have to self in order to be at least 95% certain that if there were heterozygotes in the bin, you would have found at least one? |
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