BIL 250 Study Guide - Quiz Guide: Wild Type, Dna Replication, Meiosis

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15 Feb 2020

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Workshop #1 questions with answer key in bold: a mutation occurs in a germ cell of a pure-breeding, wild-type male mouse prior to dna replication. The mutation is not corrected, and the cell undergoes dna replication and a normal meiosis producing four gametes. How many of these gametes will carry the mutation: 1, 2, 3, 4, it is impossible to predict, dumpy is a commonly used mutant phenotype in the nematode worm c. elegans. Dumpy individuals are crossed to each other, and this cross produces 210 dumpy and 68 wild-type individuals. The f1 is comprised of five rabbits with a c phenotype, two with cch phenotype, and three with c phenotype. Of the phenotypically c rabbits, two are females and are backcrossed to their father. This cross produces only rabbits with cch phenotype. Both plants are true breeding in flower size and color.

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During the course of the infamous Drosophila experiment, two students noticed a female fly that behaved in a peculiar manner. She crawled very slowly and did not fly. Realizing that this fly, collected as a virgin female, was abnormal, they did not use her in their experiment, but decided to investigate the inheritance of this tired trait. They crossed the tired female with a wild-type male and were surprised to find that all the F1 flies, male and female, exhibited the tired trait. They allowed these flies to mate and observed that all the F2 flies were tired. They generated an F3 and F4 generation. All flies were tired. Satisfied that they had a true-breeding stock, they performed reciprocal crosses with wild type. Wild-type males by tired females yielded all tired flies in the F1 and succeeding generations. However, when tired males were crossed to wild-type females, the F1 progeny were all wild type. Succeeding generations were wild type.

A. What type of inheritance is exhibited by tired? Explain. Suggest a likely location in the cell for this mutation. Explain.

During their studies, an apparently wild-type fly, a male, appeared in one of their tired stocks. Since the fly carried several other markers, these two students concluded--after much argument--that this fly was not a contaminant. They crossed this revertant male with a tired female from the pure-breeding stock. The progeny were phenotypically 50% tired and 50% wild type. Males and females were represented equally between both phenotypes.

B. Was the mutation that produced the wild-type phenotype in the unusual male dominant or recessive? Explain. What is the most likely origin of this mutation? Explain.

greyox830

A three point cross was performed using fruit flies. The three loci are physically linked on the same autosome. The d allele, in the homozygous state, results in the recessive âdouble dorsalâ phenotype. Similarly, the r allele results in the recessive âround earsâ phenotype and the e allele in the recessive âendless eyeâ phenotype. In contrast, the D, R, and E alleles encode for the dominant wild type phenotypes at each of these loci.

True breeding (homozygous) flies were crossed to produce F1 offspring. Due to poor book-keeping, the genotypes of the parent flies were not recorded â but one parent fly had one mutant phenotype, while the other parent had the other two mutant phenotypes.

Phenotype Number of F2 Flies

Ears Dorsal Eyes

1) Round ears Double Dorsal Endless Eyes 230

2) Wild-type Wild-type Wild-type 260

3) Round ears Wild-type Wild-type 4,390

4) Wild-type Double Dorsal Endless Eyes 4,100

5) Round ears Double Dorsal Wild-type 15

6) Wild-type Wild-type Endless Eyes 15

7) Round ears Wild-type Endless Eyes 510

8) Wild-type Double Dorsal Wild-type 480

10,000 total

a) What is the phenotype of the F1 flies?

b) What is the genotype of the F1 flies? Your answer should indicate the phase of linkage.

c) What are the genotypes and phenotypes of the original parent flies (previously not recorded)? Your answer should indicate the phase of linkage.

d) Calculate the map distance in cM between each of the three loci and determine the gene order. Show all work for full credit. (6 points)

You are studying mutations in a bacterial gene that codes for an enzyme whose amino acid sequence is known. In the wild type protein, proline is the 5th amino acid from the amino terminal end. In one of your mutants with nonfunctional enzyme, you find a serine at position number 5. You subject this mutant to further mutagenesis and recover 3 different strains. Strain A has a proline at position 5 and acts just like the wild-type strain. Strain B has tryptophan at position number 5 and also acts like wild type. Strain C has no detectable enzyme function at any temperature, and you can't recover any protein that resembles the enzyme. You mutagenize strain C and recover a strain ( C-1) that has enzyme function. The second mutation in C-1 that is responsible for the recovery of enzyme function does not map at the enzyme locus. a) What is the nucleotide sequence in both strands of the wild-type gene at this location? b) Why does strain B have a wild-type phenotype? Why does the original mutant with serine at position 5 lack function? c) What is the nature of the mutation in strain C?

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BIL 250 Study Guide - Quiz Guide: Wild Type, Dna Replication, Meiosis

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