Name KEY total=107 pts 1. Genes G and H are on one chromosome; gene F is on another chromosome. Assume the organism is diploid and that there is no crossing over in this species. You are examining the cells of a triple-heterozygote with genotype G h / g H F/f The circles below represent nuclei of two dividing cells (one in mitosis, one in meiosis). Diagram those two chromosomes (showing centromeres, each chromatid, and labeling each marker) in: (8 pts) metaphase of mitosis metaphase 1 of meiosis four chromosomes ` four chromosomes NOT paired homologous pairs Two chromatids each two chromatids each correct markers (sister chromatids identical!) correct markers eg. F F and f f and G h G h and g H g H (F F paired with f f Gh Gh paired with gh gh,) b. List all possible gametes that can be produced by this individual. (4 pts) Gh; F only 4 gamete types because the problem assumes no crossing over gh; F Gh; f gh; f 2. This pedigree shows a disease D, with two alleles D and d. Filled symbols are affected individuals. a) Which modes of inheritance are possible (circle Y/N)? b) For each possible mode of inheritance, give the genotypes of the 3 labeled individuals. (12 pts) 1 2 3 Mode Possible? Genotype #1 Genotype #2 Genotype. #3 Autosomal recessive Autosomal dominant Yes / No dd Dd Dd Yes / No Dd dd dd X-linked recessive Yes / No dd DY Dd X-linked dominant Yes / No Dd dy dd 1
3. Three linked genes control flower color, petal length and leaf shape in this imaginary species. Long-narrow petals is dominant to short-fat petals, jagged leaves is dominant to smooth leaves, and dark flowers is dominant to white flowers. A plant that was heterozygous for all three genes was testcrossed and 1000 progeny plants were scored for the three traits. The results are shown below. a. Define symbols for those genes. Long: L/l Jagged: J/j Dark: D/d b. What genotype would you mate the heterozygote with in order to do the testcross? Tester Genotype: ljd/ljd (2 pts) c. Fill in the genotypes of the offspring in the table. (below) (4 pts) d. Label the phenotypes as parental, single, or double recombinants (below) (4 pts) Offspring phenotype Number (total=1000) Genotype 300 350 65 70 100 105 6 4 LJd/ljd ljd/ljd Ljd/ljd ljd/ljd LJD/ljd Ljd/ljd ljd/ljd LjD/ljd Parental, single, or double recombinant P P Single J-L Single J-L Single L-D Single L-D double double e. Draw a map of the chromosome, showing the correct gene order and the distance between the genes. (6 pts) 14.5 cm 21.5 cm J-------------------------------L----------------------------D J-L distance = 65+70+6+4 / 1000= 0.145 L-D distance = 100+105+6+4/1000 = 0.215 2
4. A geneticist crossed two pure breeding strains of mice, one with black fur (BB) and one with brown (bb) to make an F1. The F1 (all black) are crossed among themselves to make the F2 generation, which showed a mixture of black and brown mice. a) What are the expected proportions of black and brown mice in the F2? (4 pts) BBxbb Bb BbxBb BB, Bb, bb, bb 3 Black : 1 brown b) Several Black F2 mice were chosen at random and crossed with some of the brown F2 mice. What proportions do you expect in the new (F3) offspring? (8 pts) 2/3 black, 1/3 brown In the F2, there are two kinds of black F2 mice Black mice are BB with prob=1/3 and Bb with prob=2/3 (The brown mice are all bb) Black homozygotes will produce all black progeny Black heterozygotes will produce 50% black and 50% brown progeny So, the F3 will be Black= 1/3*1.0 + 2/3*1/2 = 2/3 Brown = 2/3*1/2 = 1/3 5. In the nematode C. elegans, the recessive mutations dpy-21 (dumpy) and unc-34 (uncoordinated) identify linked genes that affect body type and movement pattern. I ll just call them d and u. The recombination distance between them is 24%. A double heterozygote d u / + + undergoes self-fertilization (the normal mode of reproduction in this species). a) List all types of gametes that will be produced by that heterozygote and give the expected proportion of each. (6 pts) du 0.38 d+ 0.12 +u 0.12 ++ 0.38 b) What proportion of the offspring will be both dumpy and uncoordinated? (6 pts) Both eggs and sperm will be produced with the same gamete types as above. The only way to produce double recessives phenotypes (dumpy and uncoordinated) is to combine two double recessive gametes: du and du. The combined probability will be 0.38*0.38 = 0.1444 3
6. A mother (call her Jane) had a baby son who she claimed was fathered by a famous actor (call him John). To settle the dispute, they all agreed to have their DNA tested. Here are the results for four genes, each on a separate chromosome: Jane a/a B/b C/c D/d John A/a B/B C/C D/d Husband A/A B/b C/C d/d Baby a/a B/b C/C d/d a) What is the probability that Jane and John would have a child with the observed genotype of the baby? (4 pts) Prob= 1/64 Do each locus separately Prob(baby is aa) = ¼ Prob(baby is Bb) = ½ Prob(baby is CC) = ½ Prob(baby id dd) = ¼ Product across loci = 1/64 b) What is the probability that Jane s husband was the father of the baby? (4 pts) Prob = 0 Do this the same as before AA x Aa Immediately you see that the Prob(baby is aa) = 0 therefore, her husband cannot be the baby s father. 7. In humans, red/green colorblindness is caused by a gene on the X chromosome. Blue/green colorblindness is caused by a different gene on an autosome (chromosome 7). Both forms of colorblindness are rare so you can assume people are not carriers unless you are told otherwise. John has red/green colorblindness, even though both of his parents had normal vision. His wife Mary has blue/green colorblindness, even though both of her parents had normal vision. They are considering having a child. What is the probability that their first child will be colorblind (of either kind)? (10 pts) Zero. The fact that all parents were normal tells you that the traits are recessive. Then consider each type separately. For red/green, John must be ry and Mary is assumed to not be a carrier. That means that all children will inherit one R from Mary and will not be colorblind. For blue/green, Mary must be bb and John is assumed to not be a carrier. Therefore all children must inherit a B from John and will not be colorblind. 4
8. In Neurospora, mutant allele d produces dark sprores and D makes light spores. The mutant allele m produces miniature spores while M produces normal spores. The two genes are on the same chromosome, as shown here: ---------o-------------- D-------------M--------- (-o- signifies the centromere) You have a strain is a double heterozygote: DM/dm. In one of the meiotic cells there has been a crossover between genes D and M. Draw or diagram the arrangement of gamete genotypes in the resulting ascus. (10 pts) There is no crossover between D and the centromere, so it will have a first division segregation pattern. There IS a crossover between M and the centromere so it will have a second division segregation pattern. They gametes should be arranged linearly in an ascus with 8 spores Several patterns are possible; you can draw whichever one you want. One possibility is: dm dm dm dm DM DM Dm Dm 9. A beginning geneticist is testing growth and recombination in of E. coli, so she grows several strains on agar that has all essential nutrients except methionine and leucine. For the first thee samples the strains were streaked directly on media lacking met and leu. For the last threee samples, the two strains were incubated together for several hours and then plated on media lacking met and leu. Based on what you have learned in this class, you quickly notice that some of plates will produce colonies and some will not. On which plates will she see colonies of bacteria? Which colonies are the result of genetic exchange between strains? (10 pts) Strain or mixture Outcome (colonies or no colonies) Are the colonies evidence of recombination? Hfr leu + met + str R Y No F - leu + met - str R N -- F + leu - met - str S N -- Y Yes Y No (both F+); colonies Hfr leu + met - str R and F - leu - met + str S F + leu - met - str R and F + leu + met + str R are just strain 2 Hfr leu + met + str R and F - leu - met - str R Y No- most colonies will be the Hfr parent strain, though there may be some rare recombinants 5
10. Your book discusses several topics that we did not cover in detail during lecture. Choose ONE of the following questions from the book: (5 pts) a. Some varieties of the plant Mirabilis have variegated (green and white patterned) leaves because some chloroplasts produce defective chlorophyll. A) What is the mode of inheritance for that chloroplast phenotype? B) Sometimes a branch is variegated and sometimes a branch can be either all green or all white. If a flower on an all white branch is pollinated using pollen from another flower on the same plant, but from an all green branch, what kind of leaves will the offspring have? Chloroplasts are maternally inherited through the cytoplasm of the ovule. Because the seeds are produced on a white branch, from white ovules, All offspring will have white leaves. See p 56-7 and Fig 2.31 -orb. By what mechanism do chromosomes separate during mitosis or meiosis (include the word kinetochore in your answer)? See p 92 and figs 3-26 and 3-27 Briefly, spindle fibers grow out from the two poles (centrioles) and attach to the chromosomes at the kinetochore (a special protein complex that binds to centromeric DNA sequences). The kinetochore depolymerizes the tubulin of the spindle fiber, and as it gets shorter the chromosome follows along toward one of the poles. (saying it gets pulled is ok, even though it is technically not correct). -orc. Diagram double crossovers between genes A and B involving two chromatids, three chromatids, and all four chromatids, For each type of double crossover, what proportion of the resulting gametes will be recombinant and what proportion will be parental? They only need to show one of the possible crossovers of each type. See Fig 4.18 2 chromatid double crossover: 0% recombinants, 100% parental 3 chromatid double crossover: 50% recombinants 4 chromatid double crossover: 100% recombinants 6