allosteric cis-acting DNA element coding strand dominant constitutive mutation coordinate regulation of genes denatured

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1 A B C D E F G H I J K L M N O P Q R S T U V W X Y Z AA BB CC DD EE FF GG HH II JJ KK LL MM NN OO PP QQ RR SS TT UU VV allosteric cis-acting DNA element coding strand codominant constitutive mutation coordinate regulation of genes denatured dominant dominant lethal expressivity F ( F prime ) F+ ( F plus ) first division segregation four-strand double crossover gain-of-function mutation germline mutations hemizygous heterogametic heteromorphic heteroplasmic heterozygous homozygous homogametic independent assortment induced mutations lac operon linkage loss-of-function mutation molecular hybridization penetrance recessive recessive lethal reciprocal translocation Robertsonian translocation second division segregation segregation sex-influenced sex-limited somatic mutations spontaneous mutations template strand trans-acting DNA element triploid trisomic trp operon Turner Syndrome two-strand double crossover X-linked 2

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4 Problem 1 (10 Points) For each yeast strain described below, indicate whether the GAL1 and GAL10 genes would be inducible, uninducible, or constitutive. You must explain your answer to receive full credit. a) A haploid yeast strain with a loss of function mutation in GAL80. (1.5 Points) b) A haploid yeast strain with a loss of function mutation in GAL4. (1.5 Points) c) A haploid yeast strain with a loss of function mutation in GAL3 (1.5 Points) d) A haploid yeast strain with a deletion of all UAS G sequences. (1.5 Points) e) A diploid yeast strain that has a loss of function mutation in GAL4 on one chromosome and a wild type GAL4 on the other. (2 Points) f) A diploid yeast strain that has a deletion of all UAS G sequences on one chromosome and wild type UAS G sequences on the other. (2 Points) 5

5 Problem 2 (10 Points) A population of wild fruit flies has been studied to determine the genetic basis of bristle number. Bristle number ranged from a low of 20 bristles to a high of 120 bristles, with about 1/4000 flies having 20 bristles and 1/4000 having 120 bristles. The population variance for bristle number was 20 bristles 2. and the mean of the population was 70 bristles. You selected flies from this population and allowed them to mate and produce a new generation using the exact same environment. The mean of the selected flies was 100 bristles and the mean of the new generation was 100 bristles. a) How many genes are estimated to contro bristle number in the original population? (2 Points) b) Assuming simple additive genetics, how much does each additive allele contribute to bristle number? (2 Points) c) What is the narrow sense heritability of bristle number in the original population based on the artificial selection experiment? (2 Points) d) What is the broad sense heritability in the original population? (2 Points) e) If you transferred the flies to a new environment and repeated experiments to measure heritability in the next generation, would you expect the broad sense heritability to change? Explain. (2 Points) 6

6 Problem 3. (10 Points) In cross 1, a female fruit fly with knobby knees (kk) was crossed to a male fly with the squinty eyes (se). All the male progeny had knobby knees and normal eyes and all the female progeny were wild type. In cross 2, a female fruit fly with squinty eyes was crossed to a male fruit fly with knobby knees. All the male progeny had squinty eyes and normal knees and all the female progeny were wild type. a) What is the genotype of the F1 females from cross 1? (1 Point) b) What is the genotype of the F1 males from cross 1? (1 Point) c) What is the genotype of the F1 females from cross 2? (1 Point) d) What is the genotype of the F1 males from cross 2? (1 Point) e) Assume that the knobby knee and squinty eye phenotypes assort independently. If F1 from cross 2 were allowed to interbreed to create an F2, what are the expected progeny genotypes and the expected proportions of each genotype? (6 Points) 7

7 Problem 4. (10 Points) A female fruit fly with stiff wings (sw) was crossed to a male fruit fly with bulging bristles (bb). All the F1 were wild type. A test cross of a virgin F1 female was performed and the progeny were analyzed. The results of the analysis are shown below. phenotype number of flies stiff wings and bulging bristles 20 stiff wings and wild type bristles 440 wild type wings and bulging bristles 415 wild type wings and wild type bristles 25 a) What is the genotype of F1 female? (3 Points) b) Are sw and bb genetically linked? If you think they are linked, explain why and estimate the distance between sw and bb in cm. If you don t think they are linked, explain why not. (5 Points) c) If the F1 female s parents had been a stiff winged and bulging bristled mother and a wild type father, would the results of the test cross have differed from that shown in the table above? Explain. (2 Points) 8

8 Problem 5. (10 Points) Use the restriction map of the 10 kb chromosome shown to right to answer the following questions. B refers to BamHI, E refers to EcoRI, and K refers to KpnI. a) Predict the results of restriction digest analysis of the chromosome by drawing the bands in the lanes using the image of a gel to the right. (3 Points) b) Use the image of the blot shown to the right to predict the results of a Southern blot analysis performed using the 3.0 kb KpnI fragment as the probe. (4 Points) c) If you digested the chromosome and puc19 with BamHI and KpnI (both enzymes used together to cut both DNAs), added these DNA fragments to a ligation reaction, and transformed E. coli with a sample of the ligation, how many different recombinant puc19 clones could you isolate? Explain. (3 Points) 9

9 Problem 7. (10 Points) You have isolated four new lac operon mutants. Mutants 1 is inducible for lactose permease synthesis but uninducible for β-galactosidase synthesis. Mutant 2 is uninducible for both β-galactosidase synthesis and lactose permease synthesis. Mutant 3 is uninducible for both β-galactosidase synthesis and lactose permease synthesis. Mutant 4 is constitutive for β-galactosidase and lactose permease synthesis. The four mutants differ from wild type by single nucleotide substitution mutations and all four mutants are different from each other. a) A merodipliod was created by mating a F carrying a wild type lac operon to mutant 1. This merodiploid was like wild type; inducible for synthesis of β-galactosidase and lactose permease. What was the genotype of mutant 1? Briefly explain your answer. (2.5 Points) b) A merodipliod was created by mating a F carrying a wild type lac operon to mutant 2. This merodiploid was like wild type; inducible for synthesis of β-galactosidase and lactose permease. What was the genotype of mutant 1? Briefly explain your answer. (2.5 Points) c) A merodipliod was created by mating a F carrying a wild type lac operon to mutant 3. This merodiploid was like mutant 3; uninducible for synthesis of β-galactosidase and lactose permease. What was the genotype of mutant 3? Briefly explain your answer. (2.5 Points) d) A merodipliod was created by mating a F carrying a wild type lac operon to mutant 4. This merodiploid was the same as mutant 4; constitutive for synthesis of β-galactosidase and lactose permease. What was the genotype of mutant 4? Briefly explain your answer. (2.5 Points) 10

10 Problem 8. A wild type male and wild type female of unknown genotypes are mated and produce all mutant F1 progeny. The mutant progeny are allowed to breed to produce an F2 and all F2 flies are wild type. The F2 are allowed to breed to produce an F3, in which ¼ are mutant and ¾ are wild type. a) Is the mutant phenotype due to a mitochondrial mutation? Explain. (3.3 Points) b) Is the mutant phenotype due to an autosomal recessive mutation? Explain. (3.3 Points) c) Is the mutant phenotype due to a maternal effect mutation? Explain. (3.3 Points) 11

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