Test Bank forChapter 7: Linkage, Recombination, and Eukaryotic Gene MappingTest Bank for Chapter 7: Linkage, Recombination, and Eukaryotic Gene MappingMultiple Choice Questions1. Linked genes always exhibita. phenotypes that are similar.b. recombination frequencies of less than 50%.c. homozygosity when involved in a testcross.d. a greater number of recombinant offspring than parental offspring when involved in a testcross.e. a lack of recombinant offspring when a heterozygous parent is testcrossed.Answer: bSection 7.1Comprehension 2. Linked genesa. assort randomly.b. can’t crossover and recombine.c. are allelic.d. co-segregate.e. will segregate independently.Answer: dSection 7.1Comprehension 3. Recombination occurs througha. crossing over and chromosome interference.b. chromosome interference and independent assortment.c. somatic-cell hybridization and chromosome interference.d. complete linkage and chromosome interference.e. crossing over and independent assortment.Answer: eSection 7.1Comprehension4. A genetic map shows which of the following?a. The distance in numbers of nucleotides between two genesb. The number of genes on each of the chromosomes of a speciesc. The linear order of genes on a chromosomed. The location of chromosomes in the nucleus when they line up at metaphase duringmitosise. The location of double crossovers that occur between two genesAnswer: cSection 7.2Comprehension5. A testcross includesa. one parent who is homozygous recessive for one gene pair and a second parent who is homozygous recessive for a second gene pair.b. one parent who is homozygous dominant for one or more genes and a second parent who is homozygous recessive for these same genes.c. two parents who are both heterozygous for two or more genes.d. one parent who shows the dominant phenotype for one or more genes and a second parent who is homozygous recessive for these genes.e. one parent who shows the recessive phenotype for one or more genes and a second parentwho is homozygous dominant for these genes.Answer: dSection 7.2Comprehension6. Recombination frequencies can be calculated bya. counting the number of recombinant and parental offspring when an individual who is heterozygous for two genes is testcrossed.b. performing a chi-square test on the F2 progeny when an individual who is homozygous for two genes is initially crossed to an individual who is homozygous recessive for these two genes.c. counting the number of offspring who are expressing the dominant phenotype when a heterozygous individual for two genes is testcrossed.d. performing a chi-square test of the progeny of a cross between parents who are both heterozygous for the same two genes.e. counting the number of offspring that are found in the cross of an individual who is heterozygous for two genes with another parent who is homozygous dominant for one of these genes and homozygous recessive for the other gene.Answer: a Section 7.2Comprehension7. Assume that an individual of AB/ab genotype is testcrossed and four classes of testcross progeny are found in equal frequencies. Which of the following statements is TRUE?a. The genes A and B are on the same chromosome and closely linked.b. The genes A and B are on the same chromosome and very far apart.c. The genes A and B are probably between 10 and 20 map units apart on the same chromosome.d. The genes A and B are likely located on different chromosomes. e. Either b or d could be correct.Answer: eSection 7.2Comprehension8. Is it possible for two different genes located on the same chromosome to assort independently? a. No, if two genes are on the same chromosome, they will be linked and the recombination frequency will be less than 50%.b. Yes, if the two genes are close enough to each other, there are a limited number of crossover events between them. c. No, there will be a very high crossover interference such that the recombination frequency will be reduced significantly. d. Yes, if the genes are far enough apart on the same chromosome, a crossover occurs between them in just about every meiotic event. e. Yes, but only if the two genes are both homozygous.Answer: dSection 7.2Comprehension9. Genetic distances within a given linkage group a. cannot exceed 100 cM.b. are dependent on crossover frequencies between paired, non-sister chromatids.c. are measured in centiMorgans.d. cannot be determined.e. Both b and c are correct.Answer: eSection 7.2Comprehension10. Crossing over occurs duringa. late anaphase.b. prophase.c. metaphase.d. early anaphasee. telophase.Answer: bSection 7.2Comprehension11. What major contribution did Barbara McClintock and Harriet Creighton make to the study ofrecombination?a. Genetic recombination of alleles is associated with physical exchange between chromosomes.b. Genes were locate on chromosomes and the map distance between them could often be measured by the number of nucleotides in the DNA.c. Determining map distances in humans could be done by using pedigrees and calculating lod scores.d. Association studies allow genes that have no obvious phenotype to be accurately mapped.e. Crossing over does not occur in male Drosophila, so there is no genetic recombination.Answer: aSection 7.2Comprehension12. You are doing lab work with a new species of beetle. You have isolated lines that breed true for either blue shells and long antenna, or green shells and short antenna. Crossing these linesyields F1 progeny with blue shells and long antenna. Crossing F1 progeny with beetles that have green shells and short antenna yield the following progeny:Blue shell, long antenna 82Green shell, short antenna 78Blue shell, short antenna 37Green shell, long antenna 43Total 240A chi-square test is done to test for independent assortment. What is the resulting chi-square value and how many degree(s) of freedom should be used in its interpretation?a. 27.1 and one degree of freedomb. 14.9 and three degrees of freedomc. 14.9 and two degrees of freedomd. 27.1 and three degrees of freedome. 0.42 and two degrees of freedomAnswer: dSection 7.2Application 13. You are doing lab work with a new species of beetle. You have isolated lines that breed true for either blue shells and long antenna, or green shells and short antenna. Crossing these linesyields F1 progeny with blue shells and long antenna. Crossing F1 progeny with beetles that have green shells and short antenna yield the following progeny:Blue shell, long antenna 82Green shell, short antenna 78Blue shell, short antenna