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In Drosophila, singed bristles (sn) and carnation eyes (car) are both caused by recessive X-linked alleles.The wild-type alleles (sn+ and car+) are responsible for straight bristles and red eyes, respectively.A sn car female is mated to a sn+ car+ male and the F1 progeny are interbred.The F2 are distributed as follows: sn car55sn car+45sn+car45sn+car+55200\begin{array} { | l | r | } \hline s n~ c a r & 55 \\\hline s n~ c a r^+ & 45\\\hline s n ^{+ c a r} & 45 \\\hline s n ^ { + c a r +} & \underline{55} \\\hline & 200 \\\hline\end{array}  In Drosophila, singed bristles (sn) and carnation eyes (car) are both caused by recessive X-linked alleles.The wild-type alleles (sn<sup>+</sup> and car<sup>+</sup>) are responsible for straight bristles and red eyes, respectively.A sn car female is mated to a sn<sup>+</sup> car<sup>+</sup> male and the F<sub>1</sub> progeny are interbred.The F<sub>2</sub> are distributed as follows:   \begin{array} { | l | r | }  \hline s n~ c a r & 55 \\ \hline s n~ c a r^+ & 45\\ \hline s n ^{+ c a r} & 45 \\ \hline s n ^ { + c a r +}  & \underline{55} \\ \hline & 200 \\ \hline \end{array}    -What is the p value from this test? (Pick the most accurate choice. )  A) p > 0.5 B) 0.1 < p < 0.5 C) p < 0.1 D) p < 0.05 E) p < 0.01 -What is the p value from this test? (Pick the most accurate choice. )


A) p > 0.5
B) 0.1 < p < 0.5
C) p < 0.1
D) p < 0.05
E) p < 0.01

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The R and S genes are linked and 10 map units apart.In the cross R s / r S × r s / r s what percentage of the progeny will be R s / r s?


A) 5%
B) 10%
C) 25%
D) 40%
E) 45%

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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} -What is the coefficient of coincidence in this region?


A) 0
B) 0.2
C) 0.4
D) 0.6
E) 0.8

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In peas, tall (T) is dominant to short (t) , red flowers (R) is dominant to white flowers (r) , and wide leaves (W) is dominant to narrow leaves (w) .A tall plant that has red flowers and wide leaves is crossed to a short plant that has white flowers and narrow leaves.The resulting progeny are shown in the table.  tall, red, wide 381 tall, white, wide 122 short, red, wide 118 short, white, wide 3791000\begin{array}{|l|r|}\hline \text { tall, red, wide } & 381 \\\hline \text { tall, white, wide } & 122 \\\hline \text { short, red, wide } & 118 \\\hline \text { short, white, wide } & 379 \\\hline & 1000 \\\hline\end{array} -What is the genotype of the tall plant that has red flowers and wide leaves?


A) T R W / t r w
B) T R W / t r W?
C) T R W / T R W?
D) T R W / T r w?
E) T r W / t R W?

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What does the data analysis allow you to conclude about linkage between sn and car? (Select all that apply. )


A) There is a high probability that the deviations from the expected number of F2 in each genotype class are due to chance.
B) The data do not allow rejection of the null hypothesis.
C) The p value is high meaning that the data is significant.
D) There is good evidence that cn and car are linked.

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The diploid garden pea plant has 14 chromosomes.The haploid fungus Neurospora crassa has 7 chromosomes.Neither organism has separate male and female individuals.Therefore, the number of linkage groups in these two organisms is


A) the garden pea has 14 linkage groups, and Neurospora has 7.
B) the garden pea has 7 linkage groups, and Neurospora has 7.
C) the garden pea has 8 linkage groups, and Neurospora has 8.
D) the garden pea has 15 linkage groups, and Neurospora has 8.

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A female mouse from a true-breeding wild-type strain was crossed to a male mouse with apricot eyes (ap) and grey body (gy) .The F1 mice were wild-type for both traits.When the F1 were interbred, the F2 were distributed as follows: Females  all wild type 200\begin{array} { | l | l | } \hline \text { all wild type } & 200 \\\hline\end{array} Males  wild type 91 apricot 11 grey 9 apricot, grey 89\begin{array} { | l | r | } \hline \text { wild type } & 91 \\\hline \text { apricot } & 11 \\\hline \text { grey } & 9 \\\hline \text { apricot, grey } & 89 \\\hline\end{array} Which of the following statements is correct?


A) ap and gy are unlinked
B) ap and gy are linked on an autosome and 10 map units apart
C) ap and gy are linked on an autosome and 20 map units apart
D) ap and gy are X-linked and 10 map units apart
E) ap and gy are X-linked and 20 map units apart

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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}    -Given this data, use Table 5.2 to find the most accurate range within which the p value falls. A) 0.001 < p < 0.01 B) 0.01 < p < 0.05 C) 0.05 < p < 0.10 D) 0.10 < p < 0.50 -Given this data, use Table 5.2 to find the most accurate range within which the p value falls.


A) 0.001 < p < 0.01
B) 0.01 < p < 0.05
C) 0.05 < p < 0.10
D) 0.10 < p < 0.50

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The hypothesis that predicts no linkage between genes is known as the null hypothesis.

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Individuals heterozygous for the RB+ and RB alleles can develop tumors as a result of (Select all that apply. )


A) a mitotic crossover that leads to homozygosity for RB+ in some cells and RB in other cells.
B) a somatic mutation in the RB+ allele that leads to homozygosity for RB.
C) a somatic mutation in the RB allele that leads to homozygosity for RB+.
D) the fact that RB is dominant to RB+.

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The R and S genes are linked and 10 map units apart.In the cross R s / r S × r s / r s what fraction of the progeny will be R S / r s?


A) 5%
B) 10%
C) 25%
D) 40%
E) 45%

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The cross L p q / l P Q × l p q / l p q is carried out.If the L gene is in the middle, between genes P and Q, what would be the genotypes of the double crossover gametes in this cross?


A) L P Q and l p q
B) L p Q and l P q
C) l p Q and L P q
D) L p q and l P Q
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 &nbsp;s p &nbsp;c n & 1461 \\\hline p x &nbsp;s p&nbsp; c n^{+} & 3497 \\\hline p x&nbsp; s p{+&nbsp;c n} & 1\\\hline p x&nbsp; s p^{+}&nbsp; c n+ & 11 \\\hline p x^{+} &nbsp;s p &nbsp;c n & 9 \\\hline p x^{+} &nbsp;s p &nbsp;c n+ & 0 \\\hline p x^{+}&nbsp; s p+&nbsp;c n & 3483 \\\hline p x^{+}&nbsp; s p+&nbsp;c n+ & 1539 \\\hline & 10,000 \\\hline\end{array} -Which of the three genes is in the middle?


A) px
B) sp
C) cn
D) insufficient data

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Which type of tetrad contains two recombinant and two parental spores?


A) PD
B) NPD
C) T
D) ordered tetrads
E) None of these types contain two recombinant and two parental spores.

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The map of a chromosome interval is: A--10 m.u.--B--40 m.u.--C From the cross A b c / a B C × a b c / a b c, how many double crossovers would be expected out of 1000 progeny?


A) 5
B) 10
C) 20
D) 40
E) 80

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In tetrad analysis, which result would indicate that two genes are linked?


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

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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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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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