Deck 5: Linkage, Recombination, and the Mapping of Genes on Chromosomes

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Question
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}  <strong>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}     -Using the chi-square test for goodness of fit, how many degrees of freedom are in this data set?</strong> A)1 B)2 C)3 D)4 <div style=padding-top: 35px>

-Using the chi-square test for goodness of fit, how many degrees of freedom are in this data set?

A)1
B)2
C)3
D)4
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Question
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?
Question
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}  <strong>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?</strong> A)0 B)0.0576 C)10.8 D)14.4 E)cannot be determined <div style=padding-top: 35px>

-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
Question
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
Question
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}  <strong>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}     -What is a reasonable conclusion based on the chi-square analysis?</strong> A)There is a high probability that the deviation from the expected results is due chance. B)One can say with a high degree of confidence that genes C and D are linked. C)The analysis supports the null hypothesis. D)Genes C and D are most likely unlinked. <div style=padding-top: 35px>

-What is a reasonable conclusion based on the chi-square analysis?

A)There is a high probability that the deviation from the expected results is due chance.
B)One can say with a high degree of confidence that genes C and D are linked.
C)The analysis supports the null hypothesis.
D)Genes C and D are most likely unlinked.
Question
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}

-If two or more of the genes are linked, what map distance separates them?

A)4 m.u.
B)12 m.u.
C)24 m.u.
D)50 m.u.
E)None of the genes are linked to each other.
Question
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}  <strong>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.</strong> A)0.001 < p < 0.01 B)0.01 < p < 0.05 C)0.05 < p < 0.10 D)0.10 < p < 0.50 <div style=padding-top: 35px>

-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
Question
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
Question
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}

-What is the coefficient of coincidence in this region?

A)0
B)0.16
C)0.33
D)0.5
E)0.66
Question
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}

-This cross is not useful to determine if one of the genes is linked to the others.Which gene?

A)gene T
B)gene R
C)gene W
D)This cross shows that all three genes are linked.
Question
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.
Question
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}

-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
Question
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 following linkage maps correctly shows the order and distance between the px, sp, and cn genes?

A)sp--0.21 m.u.--px--30.01 m.u.--cn
B)sp--30.01 m.u.--px--0.21 m.u.--cn
C)sp--0.2 m.u.--px--30 m.u.--cn
D)px--0.2 m.u.--sp--30.2 m.u.--cn
E)px--30.2 m.u.--sp--0.2 m.u.--cn
Question
Suppose the L and M genes are on the same chromosome but separated by 100 map units.What fraction of the progeny from the cross L M / l m × l m / l m would be L m / l m?

A)10%
B)25%
C)50%
D)75%
E)100%
Question
The pairwise map distances for four linked genes are as follows: A-B = 22 m.u. , B-C = 7 m.u. , C-D = 9 m.u. , B-D = 2 m.u. , A-D = 20 m.u. , A-C = 29 m.u.What is the order of these four genes?

A)ABCD
B)ADBC
C)ABDC
D)BADC
E)CADB
Question
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
Question
In Drosophila, singed bristles (sn)and cut wings (ct)are both caused by recessive, X-linked alleles.The wild-type alleles (sn+ and ct+)are responsible for straight bristles and intact wings, respectively.A female homozygous for sn and ct+ is crossed to a sn+ct male.The F1 flies are interbred.The F2 males are distributed as follows: snct13snct+36sn+ct39sn+ct+12\begin{array} { | l | l | } \hline \operatorname { sn } c t & 13 \\\hline \operatorname { sn } c t ^ { + } & 36 \\\hline s n ^ { + } c t & 39 \\\hline s n ^ { + } c t ^ { + } & 12 \\\hline\end{array} What is the map distance between sn and ct?

A)12 m.u.
B)13 m.u.
C)25 m.u.
D)50 m.u.
E)75 m.u.
Question
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.
Question
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%
Question
If the map distance between genes A and B is 20 map units and the map distance between genes B and C is 35 map units, what is the map distance between genes A and C?

A)15 map units
B)55 map units
C)More information is needed to distinguish between 15 and 55 map units.
D)Gene C must be located on a different nonhomologous chromosome.
Question
Suppose an individual is heterozygous for alternate alleles of gene A (Aa).Under what conditions would a crossover in a somatic cell of this individual lead to a clone of cells that are homozygous for a? (Pick the most precise answer. )

A)The crossover would have to occur between the A locus and the centromere and involve two homologous (nonsister)chromatids.
B)The crossover would have to occur between the A locus and the end of the chromosome and involve two homologous (nonsister)chromatids.
C)The crossover would have to occur between the A locus and the end of the chromosome and involve two nonhomologous chromosomes.
D)The crossover would have to occur between the A locus and the centromere and involve two sister chromatids.
Question
The measured distance between genes D and E in a two-point testcross is 50 map units.Where are genes D and E in relation to each other? (Select all that apply. )

A)D and E are on different homologous chromosomes.
B)D and E are on the same chromosome, at least 50 map units apart.
C)D and E are on the same chromosome, exactly 50 map units apart.
D)D and E are on the same chromosome, less than 50 map units apart.
Question
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
Question
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.
Question
If the recombination frequency between two genes is close to 50%, what could be true about the location of the two genes? (Select all that apply. )

A)They are on nonhomologous chromosomes.
B)They are far apart on the same chromosome.
C)They are very close together on the same chromosome.
D)None of the choices could be true.
Question
When analyzing octads (ordered tetrads), second-division (MII)segregations result from

A)single crossovers between linked genes.
B)double crossovers between linked genes.
C)single crossovers between a gene and a centromere.
D)independent assortment of unlinked genes.
E)nondisjunction of homologs.
Question
In Drosophila, the genes y (yellow body)and car (carnation eyes)are located at opposite ends of the X chromosome.In doubly heterozygous females (y+ car+ / y car), a single chiasma is observed somewhere along the X chromosome in 90% of the examined oocytes.No X chromosomes with multiple chiasmata are observed.What percentage of the male progeny from such a female would be recombinant for y and car?

A)5%
B)10%
C)45%
D)55%
E)90%
Question
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
Question
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
Question
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%
Question
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
Question
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+.
Question
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.
Question
A dihybrid testcross is made between genes H and I.Four categories of offspring are produced: H I, H i, h I, and h i.How many degrees of freedom would there be in a chi-square test for goodness of fit of the null hypothesis that the H and I genes are unlinked?

A)1
B)2
C)3
D)4
E)0
Question
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.
Question
Sturtevant's detailed mapping studies of the X chromosome of Drosophila supported what genetic principle?

A)That genes are arranged in a linear order on the chromosomes.
B)That genes are carried on chromosomes.
C)That sex determination is controlled by the X and Y chromosomes.
D)That segregation of an allelic gene pair is accompanied by disjunction of homologous chromosomes.
E)That different pairs of chromosomes assort independently.
Question
The Q gene locus is 10 map units from the R gene locus which is 40 map units from the S gene locus: Q--10 m.u.--R--40 m.u.--S
Which interval would likely show the higher ratio of double to single chiasmata?

A)Q-R
B)R-S
C)The ratios would be the same in the two intervals.
D)Two chiasmata never occur in the same interval.
Question
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
Question
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
Question
In tetrad analysis, NPD asci can result from (Select all that apply. )

A)independent assortment of unlinked genes.
B)double crossovers between linked genes.
C)single crossovers between linked genes.
D)single crossovers between a gene and a centromere.
Question
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
Question
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}

-Which phenotypes are parental?

A)wild-type and orange
B)brown and bright-red
C)wild-type and brown
D)bright-red and orange
E)There is no way to determine this.
Question
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
Question
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}  <strong>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. )</strong> A)p > 0.5 B)0.1 < p < 0.5 C)p < 0.1 D)p < 0.05 E)p < 0.01 <div style=padding-top: 35px>

-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
Question
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
Question
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
Question
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
Question
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
Question
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
Question
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
Question
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
Question
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}  <strong>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 ?<sup>2</sup> value for a chi-square test for goodness of fit of the null hypothesis?</strong> A)0.5 B)1.0 C)2.0 D)0.4 E)20 <div style=padding-top: 35px>

-What is the ?2 value for a chi-square test for goodness of fit of the null hypothesis?

A)0.5
B)1.0
C)2.0
D)0.4
E)20
Question
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
Question
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
Question
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}  <strong>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}     -If you want to analyze this data for evidence of linkage between sn and car, what is the null hypothesis?</strong> A)Genes sn and car are linked. B)Genes sn and car are not linked. C)Genes sn and car are located close together on the same chromosome. D)Crossing-over sometimes occurs between sn and car. <div style=padding-top: 35px>

-If you want to analyze this data for evidence of linkage between sn and car, what is the null hypothesis?

A)Genes sn and car are linked.
B)Genes sn and car are not linked.
C)Genes sn and car are located close together on the same chromosome.
D)Crossing-over sometimes occurs between sn and car.
Question
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.
Question
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.
Question
In humans, the genes for red-green color blindness (R = normal, r = color blind) and hemophilia A (H = normal, h = hemophilia) are both X-linked and only 3 map units apart.

-Suppose a woman has four sons, and two are color blind but have normal blood clotting and two have hemophilia but normal color vision.What is the probable genotype of the woman?

A)H R / h r
B)H r / h r
C)h r / h R
D)H r / h R
E)H R / H r
Question
In humans, the genes for red-green color blindness (R = normal, r = color blind) and hemophilia A (H = normal, h = hemophilia) are both X-linked and only 3 map units apart.

-A woman whose mother is color blind and whose father has hemophilia A is pregnant with a boy.If the alleles for color blindness and hemophilia A are rare in the population, what is the probability that the baby will have normal vision and normal blood clotting?

A)0
B)0.03
C)0.485
D)0.47
E)0.015
Question
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}

-The mother of these progeny resulted from a cross between two flies from true-breeding lines.What are the genotypes of these two lines?

A)pr cn+ / pr cn+ and pr+ cn / pr+ cn
B)pr+ cn+ / pr+ cn+ and pr cn / pr cn
C)pr+ cn+ / pr cn and pr cn / pr cn
D)pr+ cn / pr cn and pr cn+ / pr cn
E)More than one of these could be true.
Question
Two genes are considered linked when there are more F2 progeny with recombinant genotypes than parental genotypes in the offspring of a dihybrid testcross.
Question
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
Question
Twin spotting provides evidence of what genetic event?

A)meiotic recombination
B)mitotic recombination
C)linkage
D)mutation
E)biological evolution
Question
A coefficient of coincidence of 0.5 in a region of three genes means that half as many double crossovers were observed as would have been expected if crossovers in the two intervals were independent.
Question
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.
Question
Mitotic recombination occurs between homologous chromosomes.In which of the following would you not expect to encounter mitotic recombination?

A)e.coli
B)tobacco plants
C)homo sapiens
D)drosophila melanogaster
Question
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.
Question
The hypothesis that predicts no linkage between genes is known as the null hypothesis.
Question
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
Question
Chiasmata are structures that show where recombination occurred between sister chromatids.
Question
Chiasmata can be seen through a light microscope and are sites of recombination.
Question
Another name for a chromosome is a ________, because it contains alleles that are often inherited together.

A)linkage group
B)crossing over group
C)genetic recombinant
D)bivalent
Question
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.
Question
Genes that are not syntenic are not linked.
Question
Large sectors suggest a mitotic recombination event late in the growth of a yeast colony.
Question
The genetic distance from one end of a linkage group and the other may exceed 50 m.u.because the distances between many gene pairs are added together to make the map.
Question
Sectors with a different phenotype in an otherwise uniform yeast colony may be evidence of mitotic recombination.
Question
In some fungi, such as the bread mold Neurospora crassa, the arrangement of spores in the ascus directly reflects the order in which they were produced during meiosis.What is the collection of spores produced by meiosis in Neurospora crassa called?

A)unordered tetrad
B)unordered octad
C)ordered tetrad or ordered octad
D)ordered pentad
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Deck 5: Linkage, Recombination, and the Mapping of Genes on Chromosomes
1
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}  <strong>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}     -Using the chi-square test for goodness of fit, how many degrees of freedom are in this data set?</strong> A)1 B)2 C)3 D)4

-Using the chi-square test for goodness of fit, how many degrees of freedom are in this data set?

A)1
B)2
C)3
D)4
3
2
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?
T R W / t r W?
3
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}  <strong>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?</strong> 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
14.4
4
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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5
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}  <strong>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}     -What is a reasonable conclusion based on the chi-square analysis?</strong> A)There is a high probability that the deviation from the expected results is due chance. B)One can say with a high degree of confidence that genes C and D are linked. C)The analysis supports the null hypothesis. D)Genes C and D are most likely unlinked.

-What is a reasonable conclusion based on the chi-square analysis?

A)There is a high probability that the deviation from the expected results is due chance.
B)One can say with a high degree of confidence that genes C and D are linked.
C)The analysis supports the null hypothesis.
D)Genes C and D are most likely unlinked.
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6
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}

-If two or more of the genes are linked, what map distance separates them?

A)4 m.u.
B)12 m.u.
C)24 m.u.
D)50 m.u.
E)None of the genes are linked to each other.
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7
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}  <strong>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.</strong> 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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8
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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9
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}

-What is the coefficient of coincidence in this region?

A)0
B)0.16
C)0.33
D)0.5
E)0.66
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10
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}

-This cross is not useful to determine if one of the genes is linked to the others.Which gene?

A)gene T
B)gene R
C)gene W
D)This cross shows that all three genes are linked.
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11
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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12
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}

-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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13
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 following linkage maps correctly shows the order and distance between the px, sp, and cn genes?

A)sp--0.21 m.u.--px--30.01 m.u.--cn
B)sp--30.01 m.u.--px--0.21 m.u.--cn
C)sp--0.2 m.u.--px--30 m.u.--cn
D)px--0.2 m.u.--sp--30.2 m.u.--cn
E)px--30.2 m.u.--sp--0.2 m.u.--cn
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14
Suppose the L and M genes are on the same chromosome but separated by 100 map units.What fraction of the progeny from the cross L M / l m × l m / l m would be L m / l m?

A)10%
B)25%
C)50%
D)75%
E)100%
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15
The pairwise map distances for four linked genes are as follows: A-B = 22 m.u. , B-C = 7 m.u. , C-D = 9 m.u. , B-D = 2 m.u. , A-D = 20 m.u. , A-C = 29 m.u.What is the order of these four genes?

A)ABCD
B)ADBC
C)ABDC
D)BADC
E)CADB
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16
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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17
In Drosophila, singed bristles (sn)and cut wings (ct)are both caused by recessive, X-linked alleles.The wild-type alleles (sn+ and ct+)are responsible for straight bristles and intact wings, respectively.A female homozygous for sn and ct+ is crossed to a sn+ct male.The F1 flies are interbred.The F2 males are distributed as follows: snct13snct+36sn+ct39sn+ct+12\begin{array} { | l | l | } \hline \operatorname { sn } c t & 13 \\\hline \operatorname { sn } c t ^ { + } & 36 \\\hline s n ^ { + } c t & 39 \\\hline s n ^ { + } c t ^ { + } & 12 \\\hline\end{array} What is the map distance between sn and ct?

A)12 m.u.
B)13 m.u.
C)25 m.u.
D)50 m.u.
E)75 m.u.
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18
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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19
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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20
If the map distance between genes A and B is 20 map units and the map distance between genes B and C is 35 map units, what is the map distance between genes A and C?

A)15 map units
B)55 map units
C)More information is needed to distinguish between 15 and 55 map units.
D)Gene C must be located on a different nonhomologous chromosome.
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21
Suppose an individual is heterozygous for alternate alleles of gene A (Aa).Under what conditions would a crossover in a somatic cell of this individual lead to a clone of cells that are homozygous for a? (Pick the most precise answer. )

A)The crossover would have to occur between the A locus and the centromere and involve two homologous (nonsister)chromatids.
B)The crossover would have to occur between the A locus and the end of the chromosome and involve two homologous (nonsister)chromatids.
C)The crossover would have to occur between the A locus and the end of the chromosome and involve two nonhomologous chromosomes.
D)The crossover would have to occur between the A locus and the centromere and involve two sister chromatids.
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22
The measured distance between genes D and E in a two-point testcross is 50 map units.Where are genes D and E in relation to each other? (Select all that apply. )

A)D and E are on different homologous chromosomes.
B)D and E are on the same chromosome, at least 50 map units apart.
C)D and E are on the same chromosome, exactly 50 map units apart.
D)D and E are on the same chromosome, less than 50 map units apart.
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23
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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24
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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25
If the recombination frequency between two genes is close to 50%, what could be true about the location of the two genes? (Select all that apply. )

A)They are on nonhomologous chromosomes.
B)They are far apart on the same chromosome.
C)They are very close together on the same chromosome.
D)None of the choices could be true.
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26
When analyzing octads (ordered tetrads), second-division (MII)segregations result from

A)single crossovers between linked genes.
B)double crossovers between linked genes.
C)single crossovers between a gene and a centromere.
D)independent assortment of unlinked genes.
E)nondisjunction of homologs.
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27
In Drosophila, the genes y (yellow body)and car (carnation eyes)are located at opposite ends of the X chromosome.In doubly heterozygous females (y+ car+ / y car), a single chiasma is observed somewhere along the X chromosome in 90% of the examined oocytes.No X chromosomes with multiple chiasmata are observed.What percentage of the male progeny from such a female would be recombinant for y and car?

A)5%
B)10%
C)45%
D)55%
E)90%
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28
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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29
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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30
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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31
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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32
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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33
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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34
A dihybrid testcross is made between genes H and I.Four categories of offspring are produced: H I, H i, h I, and h i.How many degrees of freedom would there be in a chi-square test for goodness of fit of the null hypothesis that the H and I genes are unlinked?

A)1
B)2
C)3
D)4
E)0
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35
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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36
Sturtevant's detailed mapping studies of the X chromosome of Drosophila supported what genetic principle?

A)That genes are arranged in a linear order on the chromosomes.
B)That genes are carried on chromosomes.
C)That sex determination is controlled by the X and Y chromosomes.
D)That segregation of an allelic gene pair is accompanied by disjunction of homologous chromosomes.
E)That different pairs of chromosomes assort independently.
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37
The Q gene locus is 10 map units from the R gene locus which is 40 map units from the S gene locus: Q--10 m.u.--R--40 m.u.--S
Which interval would likely show the higher ratio of double to single chiasmata?

A)Q-R
B)R-S
C)The ratios would be the same in the two intervals.
D)Two chiasmata never occur in the same interval.
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38
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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39
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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40
In tetrad analysis, NPD asci can result from (Select all that apply. )

A)independent assortment of unlinked genes.
B)double crossovers between linked genes.
C)single crossovers between linked genes.
D)single crossovers between a gene and a centromere.
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41
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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42
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}

-Which phenotypes are parental?

A)wild-type and orange
B)brown and bright-red
C)wild-type and brown
D)bright-red and orange
E)There is no way to determine this.
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43
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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44
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}  <strong>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. )</strong> 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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45
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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46
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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47
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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48
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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49
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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50
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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51
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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52
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}  <strong>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 ?<sup>2</sup> value for a chi-square test for goodness of fit of the null hypothesis?</strong> A)0.5 B)1.0 C)2.0 D)0.4 E)20

-What is the ?2 value for a chi-square test for goodness of fit of the null hypothesis?

A)0.5
B)1.0
C)2.0
D)0.4
E)20
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53
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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54
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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55
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}  <strong>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}     -If you want to analyze this data for evidence of linkage between sn and car, what is the null hypothesis?</strong> A)Genes sn and car are linked. B)Genes sn and car are not linked. C)Genes sn and car are located close together on the same chromosome. D)Crossing-over sometimes occurs between sn and car.

-If you want to analyze this data for evidence of linkage between sn and car, what is the null hypothesis?

A)Genes sn and car are linked.
B)Genes sn and car are not linked.
C)Genes sn and car are located close together on the same chromosome.
D)Crossing-over sometimes occurs between sn and car.
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56
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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57
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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58
In humans, the genes for red-green color blindness (R = normal, r = color blind) and hemophilia A (H = normal, h = hemophilia) are both X-linked and only 3 map units apart.

-Suppose a woman has four sons, and two are color blind but have normal blood clotting and two have hemophilia but normal color vision.What is the probable genotype of the woman?

A)H R / h r
B)H r / h r
C)h r / h R
D)H r / h R
E)H R / H r
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59
In humans, the genes for red-green color blindness (R = normal, r = color blind) and hemophilia A (H = normal, h = hemophilia) are both X-linked and only 3 map units apart.

-A woman whose mother is color blind and whose father has hemophilia A is pregnant with a boy.If the alleles for color blindness and hemophilia A are rare in the population, what is the probability that the baby will have normal vision and normal blood clotting?

A)0
B)0.03
C)0.485
D)0.47
E)0.015
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60
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}

-The mother of these progeny resulted from a cross between two flies from true-breeding lines.What are the genotypes of these two lines?

A)pr cn+ / pr cn+ and pr+ cn / pr+ cn
B)pr+ cn+ / pr+ cn+ and pr cn / pr cn
C)pr+ cn+ / pr cn and pr cn / pr cn
D)pr+ cn / pr cn and pr cn+ / pr cn
E)More than one of these could be true.
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61
Two genes are considered linked when there are more F2 progeny with recombinant genotypes than parental genotypes in the offspring of a dihybrid testcross.
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62
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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63
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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64
A coefficient of coincidence of 0.5 in a region of three genes means that half as many double crossovers were observed as would have been expected if crossovers in the two intervals were independent.
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65
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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66
Mitotic recombination occurs between homologous chromosomes.In which of the following would you not expect to encounter mitotic recombination?

A)e.coli
B)tobacco plants
C)homo sapiens
D)drosophila melanogaster
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67
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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68
The hypothesis that predicts no linkage between genes is known as the null hypothesis.
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69
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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70
Chiasmata are structures that show where recombination occurred between sister chromatids.
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71
Chiasmata can be seen through a light microscope and are sites of recombination.
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72
Another name for a chromosome is a ________, because it contains alleles that are often inherited together.

A)linkage group
B)crossing over group
C)genetic recombinant
D)bivalent
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73
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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74
Genes that are not syntenic are not linked.
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75
Large sectors suggest a mitotic recombination event late in the growth of a yeast colony.
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76
The genetic distance from one end of a linkage group and the other may exceed 50 m.u.because the distances between many gene pairs are added together to make the map.
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77
Sectors with a different phenotype in an otherwise uniform yeast colony may be evidence of mitotic recombination.
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78
In some fungi, such as the bread mold Neurospora crassa, the arrangement of spores in the ascus directly reflects the order in which they were produced during meiosis.What is the collection of spores produced by meiosis in Neurospora crassa called?

A)unordered tetrad
B)unordered octad
C)ordered tetrad or ordered octad
D)ordered pentad
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