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Test Bank for Genetics A Conceptual Approach – 5th Edition by Benjamin A. Pierce
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  5. Essay Questions
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Test Bank for

Chapter 7: Linkage, Recombination, and Eukaryotic Gene Mapping

 

Multiple Choice Questions

 

  1. Linked genes always exhibit

 

  1. phenotypes that are similar.
  2. recombination frequencies of less than 50%.
  3. homozygosity when involved in a testcross.
  4. a greater number of recombinant offspring than parental offspring when involved in a testcross.
  5. a lack of recombinant offspring when a heterozygous parent is testcrossed.

 

Answer: b

Section 7.1

Comprehension

 

  1. Linked genes

 

  1. assort randomly.
  2. can’t crossover and recombine.
  3. are allelic.
  4. co-segregate.
  5. will segregate independently.

 

Answer: d

Section 7.1

Comprehension

 

  1. Recombination occurs through

 

  1. crossing over and chromosome interference.
  2. chromosome interference and independent assortment.
  3. somatic-cell hybridization and chromosome interference.
  4. complete linkage and chromosome interference.
  5. crossing over and independent assortment.

 

Answer: e

Section 7.1

Comprehension

 

  1. A genetic map shows which of the following?

 

  1. The distance in numbers of nucleotides between two genes
  2. The number of genes on each of the chromosomes of a species
  3. The linear order of genes on a chromosome
  4. The location of chromosomes in the nucleus when they line up at metaphase during           mitosis
  5. The location of double crossovers that occur between two genes

 

Answer: c

Section 7.2

Comprehension

 

  1. A testcross includes

 

  1. one parent who is homozygous recessive for one gene pair and a second parent who is homozygous recessive for a second gene pair.
  2. one parent who is homozygous dominant for one or more genes and a second parent who is homozygous recessive for these same genes.
  3. two parents who are both heterozygous for two or more genes.
  4. one parent who shows the dominant phenotype for one or more genes and a second parent who is homozygous recessive for these genes.
  5. one parent who shows the recessive phenotype for one or more genes and a second parent who is homozygous dominant for these genes.

 

Answer: d

Section 7.2

Comprehension

 

  1. Recombination frequencies can be calculated by

 

  1. counting the number of recombinant and parental offspring when an individual who is heterozygous for two genes is testcrossed.
  2. 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.
  3. counting the number of offspring who are expressing the dominant phenotype when a heterozygous individual for two genes is testcrossed.
  4. performing a chi-square test of the progeny of a cross between parents who are both heterozygous for the same two genes.
  5. 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 7.2

Comprehension

 

  1. 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?

 

  1. The genes A and B are on the same chromosome and closely linked.
  2. The genes A and B are on the same chromosome and very far apart.
  3. The genes A and B are probably between 10 and 20 map units apart on the same chromosome.
  4. The genes A and B are likely located on different chromosomes.
  5. Either b or d could be correct.

 

Answer: e

Section 7.2

Comprehension

 

  1. Is it possible for two different genes located on the same chromosome to assort independently?

 

  1. No, if two genes are on the same chromosome, they will be linked and the recombination frequency will be less than 50%.
  2. Yes, if the two genes are close enough to each other, there are a limited number of crossover events between them.
  3. No, there will be a very high crossover interference such that the recombination frequency will be reduced significantly.
  4. Yes, if the genes are far enough apart on the same chromosome, a crossover occurs between them in just about every meiotic event.
  5. Yes, but only if the two genes are both homozygous.

 

Answer: d

Section 7.2

Comprehension

 

  1. Genetic distances within a given linkage group

 

  1. cannot exceed 100 cM.
  2. are dependent on crossover frequencies between paired, non-sister chromatids.
  3. are measured in centiMorgans.
  4. cannot be determined.
  5. Both b and c are correct.

 

Answer: e

Section 7.2

Comprehension

 

  1. Crossing over occurs during

 

  1. late anaphase.
  2. prophase.
  3. metaphase.
  4. early anaphase
  5. telophase.

 

Answer: b

Section 7.2

Comprehension

 

  1. What major contribution did Barbara McClintock and Harriet Creighton make to the study of recombination?

 

  1. Genetic recombination of alleles is associated with physical exchange between chromosomes.
  2. Genes were locate on chromosomes and the map distance between them could often be measured by the number of nucleotides in the DNA.
  3. Determining map distances in humans could be done by using pedigrees and calculating lod scores.
  4. Association studies allow genes that have no obvious phenotype to be accurately mapped.
  5. Crossing over does not occur in male Drosophila, so there is no genetic recombination.

 

Answer: a

Section 7.2

Comprehension

 

  1. 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 lines yields 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                      82

Green shell, short antenna                  78

Blue shell, short antenna                     37

Green shell, long antenna                    43

Total                                                               240

 

A 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?

 

  1. 27.1 and one degree of freedom
  2. 14.9 and three degrees of freedom
  3. 14.9 and two degrees of freedom
  4. 27.1 and three degrees of freedom
  5. 0.42 and two degrees of freedom

 

Answer: d

Section 7.2

Application

 

  1. 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 lines yields 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                      82

Green shell, short antenna                  78

Blue shell, short antenna                     37

Green shell, long antenna                    43

Total                                                                240

 

Assuming that the genes are linked, what is the map distance between them in cM?

 

  1. 33.3 cM
  2. 25.0 cM
  3. 49.5 cM
  4. 8.0 cM
  5. The genes are actually assorting independently.

 

Answer: a

Section 7.2

Application

 

  1. Dr. Disney has been raising exotic fruit flies for decades. Recently, he discovered a strain of fruit flies that in the recessive condition have baby blue eyes that he designates as bb. He also has another strain of fruit flies that in the recessive condition have pink wings that are designated as pw. He is able to establish flies that are homozygous for both mutant traits. He mates these two strains with each other. Dr. Disney then takes phenotypically wild-type females from this cross and mates them with double recessive males. In the resulting testcross progeny, he observes 500 flies that are of the following make-up:

 

41      with baby blue eyes and pink wings

207    with baby blue eyes only

210    with pink wings only

42            with wild-type phenotype

 

Assuming the wild-type alleles for these two genes are b+ and pw+, what is the correct testcross of the F1 flies?

 

  1. b+ pw+/b pw ´ b pw/b pw
  2. b+ pw+/b pw ´ b pw+/b+ pw
  3. b+ pw/b pw+ ´ b pw/b pw
  4. b+ pw/b pw+ ´ b+ pw+/b pw
  5. b+ pw+/b pw ´ b+ pw/b pw+

 

Answer: c

Section 7.2

Application

 

  1. Dr. Disney has been raising exotic fruit flies for decades. Recently, he discovered a strain of fruit flies that in the recessive condition have baby blue eyes that he designates as bb. He also has another strain of fruit flies that in the recessive condition have pink wings that are designated as pw. He is able to establish flies that are homozygous for both mutant traits. He mates these two strains with each other. Dr. Disney then takes phenotypically wild-type females from this cross and mates them with double recessive males. In the resulting testcross progeny he observes 500 flies that are of the following make-up:

 

41      with baby blue eyes and pink wings

207    with baby blue eyes only

210    with pink wings only

42            with wild-type phenotype

 

What is the relationship with respect to location between the two genes?

 

  1. They are far apart on the same chromosome and assorting independently.
  2. They are linked and the map distance between them is 41.5 cM.
  3. They are on different chromosomes and assorting independently.
  4. They are linked and 16.6 cM apart.
  5. They are linked and 50.0 cM apart.

 

Answer: d

Section 7.2

Application

 

  1. Two linked genes, (A) and (B), are separated by 18 cM.  A man with genotype Aa Bb marries a woman who is aa bb.  The man’s father was AA BB. What is the probability that their first child will be Aa bb?

 

  1. 0.18
  2. 0.41
  3. 0.09
  4. 0.25
  5. 0.50

 

Answer: c

Section 7.2

Application

 

  1. Two linked genes, (A) and (B), are separated by 18 cM.  A man with genotype Aa Bb marries a woman who is aa bb. The man’s father was AA BB. What is the probability that their first two children will both be ab/ab?

 

  1. 0.168
  2. 0.0081
  3. 0.032
  4. 0.062
  5. 0.13

 

Answer: a

Section 7.2

Application

 

  1. You are studying two linked genes in lizards. You have two females and you know that both are the same genotype, heterozygous for both genes (A/a and B/b). You test-cross each female to a male that is fully homozygous recessive for both genes (a/a and b/b) and get the following progeny with the following phenotypes:

 

Female 1                      Female 2

AB – 37                      AB – 5

ab – 33                        ab – 4

Ab – 4                         Ab – 35

aB – 6                          aB – 36

 

How can you explain the drastic difference between these two crosses?

 

  1. The two genes are assorting independently in female 1 and are linked in female 2.
  2. The two genes are linked in female 1 and are assorting independently in female 2.
  3. The two alleles are in the coupling configuration in female 1 but in the repulsion configuration in female 2.
  4. The two alleles are in the repulsion configuration in female 1 but in the coupling configuration in female 2.
  5. The two genes are likely located on a sex chromosome in female 1 and are likely located on an autosome in female 2.

 

Answer: c

Section 7.2

Application

 

  1. Assume that A and B are two linked genes on an autosome in Drosophila. A testcross is made where AB/ab flies are crossed to ab/ab flies and the progeny are counted and shown below.   However, it is known that the Aa/bb genotype is lethal before the flies hatch and does not appear among the testcross progeny counted. What is the most precise map distance that can be calculated from these data?

 

Aa Bb = 235

aa bb = 225

aa Bb = 20

  1. 4.2 cM
  2. 4.0 cM
  3. 16.4 cM
  4. 8.0 cM
  5. 50 cM

 

Answer: d

Section 7.2

Application

 

  1. If the recombination frequency between genes (A) and (B) is 5.3%, what is the distance between the genes in map units on the linkage map?

 

  1. 53 cM
  2. 5.3 CM
  3. 0.53 cM
  4. 10.6 cM
  5. 25 cM

 

Answer: b

Section 7.2

Application

 

  1. You are examining the following human pedigree and want to determine if the rare dominant disease allele (D) is linked to a specific DNA sequence location you are using as a molecular marker. Parental and progeny genotypes and phenotypes are indicated. Note that the father is a dihybrid at both loci, but the mother is homozygous at both loci. There is complete penetrance of the trait and a linkage phase of D-R1/d-R2 in the father. Assuming that the marker and the gene are linked, what is the best estimate of the map distance between the two loci?

 

 

D/d, R1/R2                              d/d, R1/R1

 

 

 

 

 

R1 marker detected:                            yes      yes                   yes                  yes                    yes      yes      yes        yes

R2 marker detected:                            no       yes        yes       no         yes        no        no         yes

 

  1. 12 cM
  2. 50 cM
  3. 16 cM
  4. 5 cM
  5. 25 cM

Answer: e

Section 7.2

Application

 

  1. In soreflies (a hypothetical insect), the dominant allele, L, is responsible for resistance to a common insecticide called Loritol. Another dominant allele, M, is responsible for the ability of soreflies to sing like birds. A true-breeding mute, Loritol-resistant sorefly was mated to a true-breeding singing, Loritol-sensitive sorefly, and the singing, Loritol-resistant female progeny were test-crossed with true-breeding wild type (i.e., mute, Loritol-sensitive) males. Of the 400 total progeny produced, 117 were mute and Loritol-resistant, 114 could sing and were Loritol-sensitive, 83 could sing and were Loritol-resistant, and 86 were mute and Loritol-sensitive. A chi-square test is done to determine if there is equal segregation of alleles at the L locus. What will be the chi-square value obtained and how many degrees of freedom would be used to interpret this value?

 

  1. 0.09 and one degree of freedom
  2. 0.56 and two degrees of freedom
  3. 0 and one degree of freedom
  4. 9.72 and four degrees of freedom
  5. A chi-square test is not the appropriate statistical test to answer this question.

 

Answer: c

Section 7.2

Application

 

  1. In soreflies (a hypothetical insect), the dominant allele, L, is responsible for resistance to a common insecticide called Loritol. Another dominant allele, M, is responsible for the ability of soreflies to sing like birds. A true-breeding mute, Loritol-resistant sorefly was mated to a true-breeding singing, Loritol-sensitive sorefly, and the singing, Loritol-resistant female progeny were test-crossed with true-   breeding wild-type (i.e., mute, Loritol-sensitive) males. Of the 400 total progeny produced, 117 were mute and Loritol-resistant, 114 could sing and were Loritol-sensitive, 83 could sing and were Loritol-resistant, and 86 were mute and Loritol-sensitive. What will be the results of a chi-square test for independent assortment?

 

  1. 9.70 with three degrees of freedom
  2. 4.63 with three degrees of freedom
  3. 6.48 with four degrees of freedom
  4. 2.54 with one degree of freedom
  5. Because there are four classes of offspring, the genes must be assorting independently.

 

Answer: a

Section 7.2

Application

 

  1. Why are the progeny of a testcross generally used to map loci?  Why not the F2 progeny from an F1 × F1 cross?

 

  1. Only recombinant offspring would be found in the progeny of an F1 ´ F1 cross.
  2. The progeny of an F1 ´ F1 cross would be found in a 9:3:3:1 ratio when two genes are involved, whereas the progeny of a testcross would result in a 1:1:1:1 ratio.
  3. It is easier to classify recombinant and parental offspring of a testcross than with the progeny of an F1 ´ F1 cross.
  4. In a testcross more of the progeny would be expected to display the dominant phenotype than in the progeny of an F1 ´ F1 cross.
  5. A testcross is more useful for mapping genes that are located near each other but when genes are quite far apart on the same chromosome, an F1 ´ F1 cross actually is more useful.

 

Answer: c

Section 7.2

Application

 

  1. In corn, small pollen (sp) is recessive to normal pollen (sp+) and banded necrotic tissue, called zebra necrotic (zn), is recessive to normal tissue (zn+).  The genes that produce these phenotypes are closely linked on chromosome 10.  If no crossing over occurs between these two loci, give the types of progeny expected from the following cross:

 

×

 

  1. sp+ zn+/sp  zn; sp zn/sp zn
  2. sp+ zn/sp zn; sp zn+/sp zn
  3. sp+ zn+/sp+ zn+; sp+ zn+/sp  zn; sp zn/sp zn
  4. sp+ zn/sp+ zn;sp+ zn/sp zn+;sp zn+/sp+ zn; sp zn+/sp zn+
  5. sp+ zn+/sp zn; sp+ zn/sp zn

 

Answer: c

Section 7.2

Application

 

  1. In flower beetles, pygmy (py) is recessive to normal size (py+), and red color (r) is recessive to brown (r+). A beetle heterozygous for these characteristics was test crossed to a beetle homozygous for pygmy and red.  The following are progeny phenotypes from this testcross.

 

py+     r+        180

py+     r           22

py                   r+        19

py                   r           191

Total               412

 

 

Carry out a series of chi-square tests to determine if there is equal segregation of alleles at the py locus.What is the correct chi-square value and how many degree(s) of freedom should be used in its interpretation?

 

  1. 0.16 with one degree of freedom
  2. 0.16 with three degrees of freedom
  3. 0.48 with one degree of freedom
  4. 0.48 with two degrees of freedom
  5. 4.56 with one degree of freedom

 

Answer: a

Section 7.2

Application

 

  1. In flower beetles, pygmy (py) is recessive to normal size (py+), and red color (r) is recessive to brown (r+). A beetle heterozygous for these characteristics was test crossed to a beetle homozygous for pygmy and red.  The following are progeny phenotypes from this testcross.

 

py+     r+        180

py+     r           22

py                   r+        19

py                   r           191

Total               412

 

Carry out a series of chi-square tests to determine if the two loci are assorting independently. What is the correct chi-square value and how many degree(s) of freedom should be used in its interpretation?

 

  1. 112 with one degree of freedom
  2. 265 with three degrees of freedom
  3. 367 with four degree of freedom
  4. 16.5 with three degrees of freedom
  5. 367 with three degrees of freedom

 

Answer: b

Section 7.2

Application

 

  1. A series of two-point crosses among fruit flies is carried out between genes for brown eyes (bw), arc wings (a), vestigial wings (vg), ebony body color (e), and curved wings (cv). The following number of nonrecombinant and recombinant progeny were obtained from each cross:

 

Genes in crossProgeny (NR)Progeny (R)

a, bw2224117

a, cv2609823

a, e32003200

a, vg51722379

bw, cv46141706

bw, e41504150

bw, vg27961434

cv, e31163116

cv, vg2102305

vg, e45594559

 

Using these data from two-point crosses, what it the best genetic map (in cM) that can be developed?

 

  1. cv 5  bw 13  a 34 vg with e assorting independently
  2. bw 5 cv 24 vg 32 a with e assorting independently
  3. a 5 bw 13 vg 24 e with vg assorting independently
  4. cv 13  bw 5 a 27 vg with e assorting independently
  5. bw 5 a 24 cv 13 vg with e assorting independently

 

Answer: e

Section 7.2

Application

 

  1. An individual has the following genotype.  Gene loci (A) and (B) are 15 cM apart. What are the correct frequencies of some of the gametes that can be made by this individual?

 

A                   b

 

a                    B

 

  1. Ab = 7.5%; AB = 42.5%
  2. ab = 25%; aB = 50%
  3. AB = 7.5%; aB = 42.5%
  4. aB = 15%; Ab = 70%
  5. aB = 70%; Ab = 15%

 

Answer: c

Section 7.2

Application

 

  1. Two genes, A and B, are located 30 map units apart. The dihybrid shown below is mated to a tester aa bb. What proportion of the offspring is expected to be dominant for both traits?

 

  

 

 

 

 

 

 

 

  1. a.0%
  2. b.15%
  3. c.30%
  4. d.35%
  5. e.70%

 

Answer: d

Section 7.2

Application

 

  1. You just bought two black guinea pigs, one male and one female, of the same genotype from the pet store that are known to be heterozygous (Bb). You also know that black fur (BB) is dominant over white fur (bb), and that a lethal recessive allele is located only one cM away from the recessive b allele and your animals are both heterozygous for this gene also. You decide to start raising your own guinea pigs, but after mating these animals several times, you discover they produce only black offspring among the first 12 progeny. How would you best explain this result?

 

  1. The B locus is on the X chromosome, so it can never produce a white phenotype.
  2. The B allele is actually codominant with the b allele, so a white phenotype cannot be produced.
  3. The recessive l allele is in tight coupling linkage with the b allele, so almost all of the bb offspring will also be ll and thus die before they can be observed.
  4. The dominant L allele is in tight repulsion linkage with the B allele, so it will be impossible to produce the Bb genotype that would express the white phenotype.
  5. Normally, it would be expected that 25% of the offspring would be white, but in this case, random deviations from the expected resulted in no white offspring.

 

Answer: c

Section 7.2

Application

 

  1. You just bought two black guinea pigs, one male and one female, of the same genotype from the pet store that are known to be heterozygous (Bb). You also know that black fur (BB) is dominant over white fur (bb) and that a lethal recessive allele is located only one cM away from the recessive b allele, and your animals are both heterozygous for this gene also. What is the probability of finding a white individual among the progeny if you cross these two animals?

 

  1. 0.25
  2. 0.002475
  3. 0.000025
  4. 0.004975
  5. 0.495

 

Answer: d

Section 7.2

Challenge

 

  1. What does lod stand for?

 

  1. Linkage over DNA
  2. Linkage of dihybrids
  3. Long overall distances (with respect to map distances)
  4. Linker of DNA
  5. Logarithm of odds

 

Answer: e

Section 7.3

Comprehension

 

  1. Lod scores measure

 

  1. the relatedness of two individuals.
  2. the number of crossover events that occur along an entire chromosome.
  3. how often double crossovers occur.
  4. the length of a linkage group.
  5. the likelihood of linkage between genes.

 

Answer: e

Section 7.3

Comprehension

 

  1. Three-factor testcrosses are only informative in gene mapping when which of the        following occurs?

 

  1. One parent is homozygous recessive for the three genes, and the other parent is homozygous dominant.
  2. All three genes are located on separate chromosomes, and one parent is homozygous dominant for at least two of these genes.
  3. Both parents are homozygous for the three genes.
  4. One parent is heterozygous for the three genes, and the other parent is homozygous recessive.
  5. One of the genes must be located on a sex chromosome and be heterozygous, and the other two genes must be located on an autosome and be homozygous.

 

Answer: d 

Section 7.3

Comprehension

 

  1. A low coefficient of coincidence indicates that

 

  1. far fewer double-crossover recombinant progeny were recovered from a testcross than would be expected from the map distances of the genes involved.
  2. crossing over has been enhanced for genes that are located near the centromere of chromosomes because there is less interference of one crossover on the occurrence of a second crossover event.
  3. single-crossover recombinant classes in the progeny have been increased because the genes involved produce lethal phenotypes when in parental gene combinations.
  4. there is a large map distance between one of the outside genes in the heterozygous parent and the middle gene, while there is a short map distance between the middle gene and the other outside gene.
  5. the physical distance between two genes is very short compared with the genetic map distance between these two genes.

 

Answer: a 

Section 7.3

Comprehension

 

  1. The map distances for genes that are close to each other are more accurate than map distances for genes that are quite far apart because

 

  1. with genes that are far apart, double crossovers and other multiple-crossover events often lead to lethal recombinants that reduce the number of recombinant progeny.
  2. with genes that are far apart, double crossovers  and other multiple-crossover events often lead to nonrecombinant or parental offspring and thus reduce the true map distance.
  3. crossover interference will cause more double crossovers and other multiple crossover events to occur than would be expected and thus result in a higher number of recombinant progeny than expected to occur with genes that are far apart.
  4. double crossovers and other multiple-crossover events occur more often when genes are close to each other and can be readily detected, so these map distances are more accurate than those for genes that are far apart.
  5. when genes are far apart, single-crossover recombinant classes are more difficult to detect than when genes are close together.

 

Answer: b

Section 7.3

Comprehension

 

  1. Interference occurs when

 

  1. two genes are assorting independently.
  2. two genes are far apart on a genetic map.
  3. one crossover inhibits another.
  4. the number recombinant progeny classes in the testcross of a heterozygote exceeds the number of parental progeny.
  5. a crossover causes the termination of the meiosis event in which the crossover is occurring.

 

Answer: c

Section 7.3

Comprehension

 

  1. A situation where the coefficient of coincidence greater than 1.0 would indicate that

 

  1. the interference is high and one crossover suppresses the occurrence of a second one.
  2. no double crossovers were found in the progeny of a testcross, even though some were expected based on probability.
  3. double crossovers were found in the progeny of a testcross, but there were fewer of them than would be expected based uon probability.
  4. there were more double crossovers in the progeny than would be expected based on probability.
  5. the genes involved were actually assorting independently.

 

Answer: d

Section 7.3

Comprehension

 

  1. In addition to determining genotypes, two- and three-factor testcrosses can be used to

 

  1. map gene loci.
  2. screen recessive mutants.
  3. measure heritability.
  4. determine parental origin.
  5. determine the physical location of genes.

 

Answer: a

Section 7.3

Comprehension

 

  1. A physical map often measures _____, whereas a genetic map measures ____.

 

  1. distances between chromosomes; distances between genes map units between genes;
  2. physical distances along the chromosome centiMorgans; base pairs distances in base
  3. pairs along the chromosome; centiMorgans
  4. between genes map units between genes; centiMorgans

 

Answer: d

Section 7.3

Comprehension

 

  1. What is a major difference in using lod-score analysis compared to using association studies in determining gene locations in humans?

 

  1. Lod-score analysis relies on family or pedigree data, while association studies use population data.
  2. Lod-score analysis requires that the loci being mapped must be on different chromosome arms, while association studies can map genes on different chromosomes.
  3. Association studies compare genotypes between parents and their children, while lod-score analysis compares genotypes between siblings of the same family.
  4. Lod-score analysis requires isolated human populations, while association studies require very large family pedigrees.
  5. Lod-score analysis requires a large number of genes with multiple alleles, while association studies can use genes that have only two alleles.

 

Answer: a

Section 7.3

Comprehension

 

  1. The results of linkage analysis for DNA marker A and the p53 gene are shown below. What is the best estimate for the approximate genetic distance between marker A and the p53 gene in humans?

 

Recombination values (cM)       1              5           10           20        30        40

lod score                          2.13         2.54       3.14       4.10    4.96     3.22

 

  1. 1 cM
  2. 5 cM
  3. 10 cM
  4. 20 cM
  5. 30 cM

 

Answer: e

Section 7.3

Application

 

  1. In maize (corn), assume that the genes A and B are linked and 30 map units apart. If a plant of Ab/aB is selfed, what proportion of the progeny would be expected to be of ab/ab genotype?

 

  1. 2.25%
  2. 15%
  3. 9%
  4. 30%
  5. 4.5%

 

Answer: a

Section 7.3

Application

 

  1. A testcross is performed on an individual to examine three linked genes.  The most frequent phenotypes of the progeny were Abc and aBC, and the least frequent phenotypes were abc and ABC. What was the genotype of the heterozygous individual that is testcrossed with the correct order of the three genes?

 

  1. Abc aBC
  2. BAC/bac
  3. bcA/BCa
  4. aBc/AbC
  5. bAc/BaC

 

Answer: e

Section 7.3

Application

 

  1. In Drosophila melanogaster, cut wings (ct) is recessive to normal wings (ct+), sable body (s) is recessive to gray body (s+), and vermilion eyes (v) is recessive to red eyes (v+). All three recessive mutations are X-linked. A female fly with cut wings, sable body, and vermilion eyes is crossed to a male with normal wings, gray body, and red eyes. The F1 females produced by this cross were mated with cut, sable, vermilion males in a testcross. The following are the progeny resulting from the testcross.

 

v          ct         s          510

v+        ct         s          1

v+        ct+       s          14

v+        ct+       s+        500

v+        ct         s+        73

v          ct         s+        20

v          ct+       s          81

v          ct+       s+        1

Total   à                    1200

 

What is the correct genetic map with respect to gene order and distances (in cM) for these three genes?

 

  1. s   13              ct      3        v
  2. s   3       v      13                ct
  3. v   13              ct      3        s
  4. s   26               v      3        ct
  5. ct   13              s      3        v

 

Answer: b

Section 7.3

Application

 

  1. In Drosophila melanogaster, cut wings (ct) is recessive to normal wings (ct+), sable body (s) is recessive to gray body (s+), and vermilion eyes (v) is recessive to red eyes (v+). All three recessive mutations are X-linked. A female fly with cut wings, sable body, and vermilion eyes is crossed to a male with normal wings, gray body, and red eyes. The F1 females produced by this cross were mated with cut, sable, vermilion males in a testcross. The following are the progeny resulting from the testcross.

 

v          ct         s          510

v+        ct         s          1

v+        ct+       s          14

v+        ct+       s+        500

v+        ct         s+        73

v          ct         s+        20

v          ct+       s          81

v          ct+       s+        1

Total   à                    1200

 

What the interference value shown by this cross?

 

  1. 0.42
  2. 0.25
  3. 0.58
  4. −0.42
  5. 0.13

 

Answer: c

Section 7.3

Application

 

  1. You are studying three genes X, Y, and Z that are linked (in that order) in the Imperial Scorpion Pandinus imperator. The distance between X and Y is 10 cM, the distance between Y and Z is 8 cM. You conduct a testcross by crossing a heterozygous female        with a homozygous recessive male and obtain 1500 testcross progeny. When the progeny are analyzed, you find 5 double-crossover offspring. What is the interference value shown by this cross?

 

  1. 0.008
  2. 0. 42
  3. 0.12
  4. 0.58
  5. 0.22

 

Answer: d

Section 7.3

Application

 

  1. Consider the following three-point (trihybrid) testcross:

 

+                      +                               c

(x)  abc/abc

a                       b                               +

 

10.6 cM   

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Test Bank for Genetics A Conceptual Approach – 5th Edition by Benjamin A. Pierce
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INSTANT DOWNLOAD What student Can Expect From A Test Bank? A test bank will include the following questions: True/False Multiple Choice Questions Matching Questions Fill In The Blanks Essay Questions Short Questions Description Test Bank for Chapter 7: Linkage, Recombination, and Eukaryotic Gene Mapping Multiple Choice Questions Linked genes always exhibit phenotypes that are similar. recombination frequencies of less than 50%. homozygosity when involved in a testcross. a greater number of recombinant offspring than parental offspring when involved in a testcross. a lack of recombinant offspring when a heterozygous parent is testcrossed. Answer: b Section 7.1 Comprehension Linked genes assort randomly. can’t crossover and recombine. are allelic. co-segregate. will segregate independently. Answer: d Section 7.1 Comprehension Recombination occurs through crossing over and chromosome interference. chromosome interference and independent assortment. somatic-cell hybridization and chromosome interference. complete linkage and chromosome interference. crossing over and independent assortment. Answer: e Section 7.1 Comprehension A genetic map shows which of the following? The distance in numbers of nucleotides between two genes The number of genes on each of the chromosomes of a species The linear order of genes on a chromosome The location of chromosomes in the nucleus when they line up at metaphase during mitosis The location of double crossovers that occur between two genes Answer: c Section 7.2 Comprehension A testcross includes one parent who is homozygous recessive for one gene pair and a second parent who is homozygous recessive for a second gene pair. one parent who is homozygous dominant for one or more genes and a second parent who is homozygous recessive for these same genes. two parents who are both heterozygous for two or more genes. one parent who shows the dominant phenotype for one or more genes and a second parent who is homozygous recessive for these genes. one parent who shows the recessive phenotype for one or more genes and a second parent who is homozygous dominant for these genes. Answer: d Section 7.2 Comprehension Recombination frequencies can be calculated by counting the number of recombinant and parental offspring when an individual who is heterozygous for two genes is testcrossed. 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. counting the number of offspring who are expressing the dominant phenotype when a heterozygous individual for two genes is testcrossed. performing a chi-square test of the progeny of a cross between parents who are both heterozygous for the same two genes. 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 7.2 Comprehension 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? The genes A and B are on the same chromosome and closely linked. The genes A and B are on the same chromosome and very far apart. The genes A and B are probably between 10 and 20 map units apart on the same chromosome. The genes A and B are likely located on different chromosomes. Either b or d could be correct. Answer: e Section 7.2 Comprehension Is it possible for two different genes located on the same chromosome to assort independently? No, if two genes are on the same chromosome, they will be linked and the recombination frequency will be less than 50%. Yes, if the two genes are close enough to each other, there are a limited number of crossover events between them. No, there will be a very high crossover interference such that the recombination frequency will be reduced significantly. Yes, if the genes are far enough apart on the same chromosome, a crossover occurs between them in just about every meiotic event. Yes, but only if the two genes are both homozygous. Answer: d Section 7.2 Comprehension Genetic distances within a given linkage group cannot exceed 100 cM. are dependent on crossover frequencies between paired, non-sister chromatids. are measured in centiMorgans. cannot be determined. Both b and c are correct. Answer: e Section 7.2 Comprehension Crossing over occurs during late anaphase. prophase. metaphase. early anaphase telophase. Answer: b Section 7.2 Comprehension What major contribution did Barbara McClintock and Harriet Creighton make to the study of recombination? Genetic recombination of alleles is associated with physical exchange between chromosomes. Genes were locate on chromosomes and the map distance between them could often be measured by the number of nucleotides in the DNA. Determining map distances in humans could be done by using pedigrees and calculating lod scores. Association studies allow genes that have no obvious phenotype to be accurately mapped. Crossing over does not occur in male Drosophila, so there is no genetic recombination. Answer: a Section 7.2 Comprehension 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 lines yields 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 82 Green shell, short antenna 78 Blue shell, short antenna 37 Green shell, long antenna 43 Total 240 A 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? 27.1 and one degree of freedom 14.9 and three degrees of freedom 14.9 and two degrees of freedom 27.1 and three degrees of freedom 0.42 and two degrees of freedom Answer: d Section 7.2 Application 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 lines yields 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 82 Green shell, short antenna 78 Blue shell, short antenna 37 Green shell, long antenna 43 Total 240 Assuming that the genes are linked, what is the map distance between them in cM? 33.3 cM 25.0 cM 49.5 cM 8.0 cM The genes are actually assorting independently. Answer: a Section 7.2 Application Dr. Disney has been raising exotic fruit flies for decades. Recently, he discovered a strain of fruit flies that in the recessive condition have baby blue eyes that he designates as bb. He also has another strain of fruit flies that in the recessive condition have pink wings that are designated as pw. He is able to establish flies that are homozygous for both mutant traits. He mates these two strains with each other. Dr. Disney then takes phenotypically wild-type females from this cross and mates them with double recessive males. In the resulting testcross progeny, he observes 500 flies that are of the following make-up: 41 with baby blue eyes and pink wings 207 with baby blue eyes only 210 with pink wings only 42 with wild-type phenotype Assuming the wild-type alleles for these two genes are b+ and pw+, what is the correct testcross of the F1 flies? b+ pw+/b pw ´ b pw/b pw b+ pw+/b pw ´ b pw+/b+ pw b+ pw/b pw+ ´ b pw/b pw b+ pw/b pw+ ´ b+ pw+/b pw b+ pw+/b pw ´ b+ pw/b pw+ Answer: c Section 7.2 Application Dr. Disney has been raising exotic fruit flies for decades. Recently, he discovered a strain of fruit flies that in the recessive condition have baby blue eyes that he designates as bb. He also has another strain of fruit flies that in the recessive condition have pink wings that are designated as pw. He is able to establish flies that are homozygous for both mutant traits. He mates these two strains with each other. Dr. Disney then takes phenotypically wild-type females from this cross and mates them with double recessive males. In the resulting testcross progeny he observes 500 flies that are of the following make-up: 41 with baby blue eyes and pink wings 207 with baby blue eyes only 210 with pink wings only 42 with wild-type phenotype What is the relationship with respect to location between the two genes? They are far apart on the same chromosome and assorting independently. They are linked and the map distance between them is 41.5 cM. They are on different chromosomes and assorting independently. They are linked and 16.6 cM apart. They are linked and 50.0 cM apart. Answer: d Section 7.2 Application Two linked genes, (A) and (B), are separated by 18 cM. A man with genotype Aa Bb marries a woman who is aa bb. The man’s father was AA BB. What is the probability that their first child will be Aa bb? 0.18 0.41 0.09 0.25 0.50 Answer: c Section 7.2 Application Two linked genes, (A) and (B), are separated by 18 cM. A man with genotype Aa Bb marries a woman who is aa bb. The man’s father was AA BB. What is the probability that their first two children will both be ab/ab? 0.168 0.0081 0.032 0.062 0.13 Answer: a Section 7.2 Application You are studying two linked genes in lizards. You have two females and you know that both are the same genotype, heterozygous for both genes (A/a and B/b). You test-cross each female to a male that is fully homozygous recessive for both genes (a/a and b/b) and get the following progeny with the following phenotypes: Female 1 Female 2 AB – 37 AB – 5 ab – 33 ab – 4 Ab – 4 Ab – 35 aB – 6 aB – 36 How can you explain the drastic difference between these two crosses? The two genes are assorting independently in female 1 and are linked in female 2. The two genes are linked in female 1 and are assorting independently in female 2. The two alleles are in the coupling configuration in female 1 but in the repulsion configuration in female 2. The two alleles are in the repulsion ...
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