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In Drosophila, the genes b, c, and sp are linked and arranged as shown below: b--30 m.u.--c--20 m.u.--sp This region exhibits 90% interference.How many double crossovers would be recovered in a three-point cross involving b, c, and sp out of 1000 progeny?


A) 3
B) 6
C) 54
D) 60
E) 600

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A haploid yeast strain has mating type a and genotype his4 TRP1. A different haploid yeast strain has mating type ? and genotype HIS4 trp1. The two strains mate and produce diploid offspring with genotype his4 TRP1 / HIS4 trp1. The diploid cells undergo meiosis to produce a variety of tetrad types. In the following questions, the genotypes of the spores in a tetrad are shown. For each tetrad, tell what type it is: parental (PD) , non-parental (NPD) , or tetratype (T) . -his4 trp1; his4 trp1; HIS4 TRP1; HIS4 TRP?


A) PD
B) NPD
C) T
D) cannot be determined

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Twin spotting provides evidence of what genetic event?


A) meiotic recombination
B) mitotic recombination
C) linkage
D) mutation
E) biological evolution

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Sectors with a different phenotype in an otherwise uniform yeast colony may be evidence of mitotic recombination.

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Which process(es) can generate recombinant gametes? (Select all that apply. )


A) crossing-over between two linked heterozygous loci
B) independent assortment of two unlinked heterozygous loci
C) segregation of alleles in a homozygote
D) crossing-over between two linked homozygous loci

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If the p value corresponding to a given χ2 value and number of degrees of freedom is lower than 0.05, then the null hypothesis is rejected and it can be said with some confidence that the two genes being evaluated are linked.

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In Neurospora, the white-spore allele (ws) results in white spores rather than the wild-type black spores. Diploid ws / ws+ cells undergo meiosis followed by one round of mitosis to form ordered tetrads. Indicate if the pattern of spores seen in the tetrads are consistent with first-division or second-division segregation. -ws; ws; ws; ws; ws+; ws+; ws+; ws+


A) first-division segregation pattern
B) second-division segregation pattern

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If only 100 progeny had been counted and the same proportions of progeny genotypes observed, how would the p value and the conclusion drawn about linkage change?


A) The p value would increase, and the likelihood of linkage decreases.
B) The p value would decrease, and the likelihood of linkage increases.
C) Neither the p value nor the likelihood of linkage would change.
D) The p value would decrease, and the likelihood of linkage decreases.

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A linkage group includes all of the genes on a chromosome, including genes that are so far apart from each other on a chromosome that they assort independently during meiosis.

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Suppose the map for a particular human chromosome interval is: a——1 m.u.——b--1 m.u.——c--1 m.u.——d——1 m.u.——e——1 m.u.——f In a man heterozygous for all six genes, what fraction of his sperm would be recombinant in the a-f interval?


A) 0%
B) 1%
C) 2.5%
D) 5%
E) cannot be determined

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Females heterozygous for the recessive second chromosome mutations px, sp, and cn are mated to a male homozygous for all three mutations.The offspring are as follows: \begin{array}{|l|r|}\hline p x  s p  c n & 1461 \\\hline p x  s p  c n^{+} & 3497 \\\hline p x  s p{+ c n} & 1\\\hline p x  s p^{+}  c n+ & 11 \\\hline p x^{+}  s p  c n & 9 \\\hline p x^{+}  s p  c n+ & 0 \\\hline p x^{+}  s p+ c n & 3483 \\\hline p x^{+}  s p+ c n+ & 1539 \\\hline & 10,000 \\\hline\end{array} -What is the genotype of the females that gave rise to these progeny?


A) px+ sp cn / px sp+ cn+
B) px+ sp cn+ / px sp+ cn
C) px+ sp+ cn+ / px sp cn
D) px sp cn+ / px+ sp+ cn
E) insufficient data

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Crossing-over takes place in bivalents (tetrads) consisting of ________ chromatids, and one crossover involves ________ chromatids.


A) 2; 2
B) 2; 4
C) 4; 2
D) 4; 4
E) 8; 4

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n Drosophila, the autosomal recessive pr and cn mutations cause brown and bright-red eyes, respectively (wild-type flies have brick-red eyes) .Flies who are homozygous recessive at both pr and cn have orange eyes.A female who has wild-type eyes is crossed to an orange-eyed male.Their progeny have the following distribution of eye colors:  wild-type  8  brown 241 briphit-red 239 orarnge 12500\begin{array} { | l | r | } \hline \text { wild-type } & \text { 8 } \\\hline \text { brown } & 241 \\\hline \text { briphit-red } & 239 \\\hline \text { orarnge } & \underline{12} \\\hline & 500 \\\hline\end{array} -What is the map distance between the pr and cn genes?


A) 20 m.u.
B) 2 m.u.
C) 4 m.u.
D) 46 m.u.
E) 8 m.u.

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In Drosophila, the autosomal recessive pr and cn mutations cause brown and bright-red eyes, respectively (wild-type flies have brick-red eyes) .Flies who are homozygous recessive at both pr and cn have orange eyes.A female who has wild-type eyes is crossed to an orange-eyed male.Their progeny have the following distribution of eye colors:  wild-type  8  brown 241 briphit-red 239 orarnge 12500\begin{array} { | l | r | } \hline \text { wild-type } & \text { 8 } \\\hline \text { brown } & 241 \\\hline \text { briphit-red } & 239 \\\hline \text { orarnge } & \underline{12} \\\hline & 500 \\\hline\end{array} -What is the genotype of the wild-type mother of these progeny?


A) pr cn / pr+ cn+
B) pr+ cn / pr+ cn
C) pr+ cn / pr cn+
D) pr cn+ / pr cn+
E) pr cn / pr cn

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In Drosophila, the genes y, f, and v are all X-linked. Females who are homozygous for recessive alleles of all three genes (y f v / y f v) are crossed to wild-type males. The resulting trihybrid F1 females are testcrossed. The F2 are distributed as follows: yfv3210yfv+72yf+v1024yf+v+678y+fv690y+fv+1044y+f+v60y+f+v+322210,000\begin{array} { | l | r | } \hline y f v & 3210 \\\hline y f v ^ { + } & 72 \\\hline y f ^ { + } v & 1024 \\\hline y f ^ { + } v^ + & 678 \\\hline y ^ { + } f v & 690 \\\hline y ^ { + } f v ^+ & 1044 \\\hline y ^ { + } f ^+ v & 60 \\\hline y ^ { + } f ^+ v^ + & 3222 \\\hline & 10,000 \\\hline\end{array} -Which of the following linkage maps correctly shows the order and distance between the y, f, and v genes?


A) f--35 m.u.--y--15 m.u.--v
B) f--22 m.u.--y--15 m.u.--v
C) y--35 m.u.--f--22 m.u.--v
D) y--22 m.u.--v--15 m.u.--f
E) y--15 m.u.--v--22 m.u.--f

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The zipper-like connection between paired homologs in early prophase is known as a


A) spindle fiber.
B) synaptic junction.
C) synaptonemal complex.
D) chiasma.
E) None of the choices is correct.

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Tetrad analysis shows that crossing-over occurs at the four-strand stage (i.e. , after replication) because, when two genes are linked,


A) NPD > T.
B) T > NPD.
C) T > PD.
D) PD > NPD.
E) PD > T.

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A haploid yeast strain has mating type a and genotype his4 TRP1. A different haploid yeast strain has mating type ? and genotype HIS4 trp1. The two strains mate and produce diploid offspring with genotype his4 TRP1 / HIS4 trp1. The diploid cells undergo meiosis to produce a variety of tetrad types. In the following questions, the genotypes of the spores in a tetrad are shown. For each tetrad, tell what type it is: parental (PD) , non-parental (NPD) , or tetratype (T) . -HIS4 trp1; HIS4 TRP1; his4 TRP1; his4 trp1?


A) PD
B) NPD
C) T
D) cannot be determined

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A dihybrid testcross is made to determine if genes C and D are linked.The results are shown in the table.  Parent genotypes: Ccnbsp;Dd×ccddCcnbsp;Dd222Ccnbsp;dd280 Progeny genotypes: ccnbsp;Dd280ccnbsp;dd2181000\begin{array}{|l|c|r|}\hline \text { Parent genotypes: } & C c &nbsp;D d \times c c d d & \\\hline & C c&nbsp; D d & 222 \\\hline & C c &nbsp;d d & 280 \\\hline \text { Progeny genotypes: } & c c &nbsp;D d & 280 \\\hline & c c &nbsp;d d & 218 \\\hline & & 1000 \\\hline\end{array}  A dihybrid testcross is made to determine if genes C and D are linked.The results are shown in the table.   \begin{array}{|l|c|r|} \hline \text { Parent genotypes: } & C c  D d \times c c d d & \\ \hline & C c  D d & 222 \\ \hline & C c  d d & 280 \\ \hline \text { Progeny genotypes: } & c c  D d & 280 \\ \hline & c c  d d & 218 \\ \hline & & 1000 \\ \hline \end{array}    -The chi-square value is the sum for all progeny classes of (observed-expected) <sup>2</sup>/expected.Using the chi-square test for goodness of fit, calculate the chi-square value to test the null hypothesis that genes C and D are unlinked.What is the chi-square value? A) 0 B) 0.0576 C) 10.8 D) 14.4 E) cannot be determined -The chi-square value is the sum for all progeny classes of (observed-expected) 2/expected.Using the chi-square test for goodness of fit, calculate the chi-square value to test the null hypothesis that genes C and D are unlinked.What is the chi-square value?


A) 0
B) 0.0576
C) 10.8
D) 14.4
E) cannot be determined

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Suppose a three-point testcross was conducted involving genes X, Y, and Z.If the most abundant classes of progeny are X Y z and x y Z and the rarest classes are x Y Z and X y z, which gene is in the middle?


A) X
B) Y
C) Z
D) cannot be determined

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