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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? ABab\begin{array} { c c } \mathbf { A } & \mathbf { B } \\\hline\\\hline \mathbf { a } & \mathbf { b }\end{array}


A) 0%
B) 15%
C) 30%
D) 35%
E) 70%

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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+180py+r22pyr+19pyr191 Total 412\begin{array} { l l r } p y ^ { + } & r ^ { + } & 180 \\p y ^ { + } & r & 22 \\p y & r ^ { + } & 19 \\p y & r & 191 \\\text { Total } && 412\end{array} 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?


A) 112 with one degree of freedom
B) 265 with three degrees of freedom
C) 367 with four degree of freedom
D) 16.5 with three degrees of freedom
E) 367 with three degrees of freedom

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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 m.u., the distance between Y and Z is 8 m.u. 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 five double-crossover offspring. What is the interference value shown by this cross?


A) 0.008
B) 0) 42
C) 0.12
D) 0.58
E) 0.22

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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 testcross 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 2AB37AB5ab33ab4Ab4Ab35aB6aB36\begin{array} { l l } \text { Female } 1 & \text { Female } 2 \\A B - 37 & A B - 5 \\a b - 33 & a b - 4 \\A b - 4 & A b - 35 \\a B - 6 & a B - 36\end{array} How can you explain the drastic difference between these two crosses?


A) The two genes are assorting independently in female 1 and are linked in female 2.
B) The two genes are linked in female 1 and are assorting independently in female 2.
C) The two alleles are in the coupling configuration in female 1 but in the repulsion configuration in female 2.
D) The two alleles are in the repulsion configuration in female 1 but in the coupling configuration in female 2.
E) The two genes are likely located on a sex chromosome in female 1 and are likely located on an autosome in female 2.

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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?


A) 2.25%
B) 15%
C) 9%
D) 30%
E) 4.5%

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Two linked genes, (A) and (B) , are separated by 18 m.u. 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?


A) 0.168
B) 0.0081
C) 0.032
D) 0.062
E) 0.13

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Consider the following three-factor (trihybrid) testcross: Consider the following three-factor (trihybrid)  testcross:   Calculate the number of individuals of a<sup>+</sup>a bb c<sup>+</sup>c genotype if 1000 progeny result from this testcross. A)  about 102 B)  about 46 C)  about 130 D)  about 65 E)  about 250 Calculate the number of individuals of a+a bb c+c genotype if 1000 progeny result from this testcross.


A) about 102
B) about 46
C) about 130
D) about 65
E) about 250

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In a two-point linkage analysis, genes a and b have been found to be 26 m.u. apart on the same chromosome. A third gene, c, has just been discovered and found to be located between a and b. A three-point linkage analysis with a, b, and c indicates that a and b are actually 33 m.u. apart rather than 26 m.u. Why does the three-point analysis give a different map distance for a and b than does the two-point linkage analysis, and which is more accurate?

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The difference in map distance between g...

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Geneticists often assume that map distances less than 7 to 8 map units (m.u.) are quite accurate. Map distances that exceed this threshold significantly are assumed to be less accurate and the level of accuracy declines as map distances increase. Briefly explain this observation.

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Geneticists use map distances to estimat...

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An individual has the following genotype. Gene loci (A) and (B) are 15 m.u. apart. What are the correct frequencies of some of the gametes that can be made by this individual? An individual has the following genotype. Gene loci (A)  and (B)  are 15 m.u. apart. What are the correct frequencies of some of the gametes that can be made by this individual?   A)  Ab = 7.5%; AB = 42.5% B)  ab = 25%; aB = 50% C)  AB = 7.5%; aB = 42.5% D)  aB = 15%; Ab = 70% E)  aB = 70%; Ab = 15%


A) Ab = 7.5%; AB = 42.5%
B) ab = 25%; aB = 50%
C) AB = 7.5%; aB = 42.5%
D) aB = 15%; Ab = 70%
E) aB = 70%; Ab = 15%

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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 testcrossed 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?


A) 0.09 and one degree of freedom
B) 0.56 and two degrees of freedom
C) 0 and one degree of freedom
D) 9.72 and four degrees of freedom
E) A chi-square test is not the appropriate statistical test to answer this question.

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Compared with a physical map, a genetic map:


A) is more accurate.
B) is less accurate.
C) is equally accurate.
D) measures different things.
E) cannot be made for humans.

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A cell possessing two nuclei derived from different cells through cell fusion is called:


A) a heterokaryon.
B) a haplotype.
C) recombinant.
D) nonrecombinant.
E) None of the answers is correct.

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Use the following to answer questions : 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 makeup: 41with baby blue eyes and pink wings 207with baby blue eyes only 210with pink wings only 42with 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?


A) b+ pw+/b pw × b pw/b pw
B) b+ pw+/b pw × b pw+/b+ pw
C) b+ pw/b pw+ × b pw/b pw
D) b+ pw/b pw+ × b+ pw+/b pw
E) b+ pw+/b pw × b+ pw/b pw+

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


A) One parent is homozygous recessive for the three genes, and the other parent is homozygous dominant.
B) All three genes are located on separate chromosomes, and one parent is homozygous dominant for at least two of these genes.
C) Both parents are homozygous for the three genes.
D) One parent is heterozygous for the three genes, and the other parent is homozygous recessive.
E) 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.

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Interference occurs when:


A) two genes are assorting independently.
B) two genes are far apart on a genetic map.
C) one crossover inhibits another.
D) the number of recombinant progeny classes in the testcross of a heterozygote exceeds the number of parental progeny.
E) a crossover causes the termination of the meiosis event in which the crossover is occurring.

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What is the chi-square test used for, and what does it tell you?

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The chi-square test, also known as the χ...

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Recombination occurs through:


A) 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.

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A study is done on families in Sweden that are segregating a genetic disorder. Lod-score analysis indicates that the gene involved in the disorder shows a strong likelihood of linkage with a particular DNA marker locus. However, a second study done in Italy with other families segregating the same genetic disorder results in lod-score values that strongly indicate the lack of linkage between the gene and the same DNA marker locus. Assuming that both studies were performed appropriately, what is the most likely explanation for the different outcomes?


A) No recombinants were found in the families studied in Sweden.
B) The allele that caused the disorder was in coupling linkage with one of the DNA marker alleles in the Swedish families but was in repulsion linkage in the Italian families.
C) This disorder is caused by mutations in either of two different genes; one of these genes is linked to the DNA marker locus and the other gene is not.
D) In the Italian families, the gene involved with the disorder is near a lethal allele at another locus and most of the parental or nonrecombinant genotypes contain the lethal allele; this reduces the number of nonrecombinants observed.
E) Linkage should have been observed in the Italian families, but there were only two alleles at the DNA marker locus that prevented recombinant offspring from appearing.

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A geneticist finds that a human gene and a particular DNA marker locus are located on chromosome 8 on the basis of somatic-cell hybridization studies. However, when lod-score analysis is done with these two loci using family pedigrees, no evidence for linkage between the two loci can be found. Assuming that both types of studies were done correctly and the results are valid, how would you explain the different outcomes?

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The different outcomes between the somat...

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