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

ملء الشاشة (f)
exit full mode
سؤال
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
استخدم زر المسافة أو
up arrow
down arrow
لقلب البطاقة.
سؤال
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?
سؤال
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
سؤال
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
سؤال
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.
سؤال
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.
سؤال
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
سؤال
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
سؤال
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
سؤال
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.
سؤال
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.
سؤال
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
سؤال
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
سؤال
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%
سؤال
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
سؤال
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
سؤال
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.
سؤال
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.
سؤال
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%
سؤال
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.
سؤال
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.
سؤال
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.
سؤال
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
سؤال
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.
سؤال
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.
سؤال
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.
سؤال
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%
سؤال
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
سؤال
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
سؤال
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%
سؤال
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
سؤال
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+.
سؤال
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.
سؤال
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
سؤال
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.
سؤال
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.
سؤال
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.
سؤال
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
سؤال
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
سؤال
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.
سؤال
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
سؤال
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.
سؤال
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
سؤال
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
سؤال
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
سؤال
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
سؤال
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
سؤال
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
سؤال
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
سؤال
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
سؤال
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
سؤال
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
سؤال
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
سؤال
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
سؤال
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.
سؤال
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.
سؤال
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.
سؤال
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
سؤال
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
سؤال
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.
سؤال
Two genes are considered linked when there are more F2 progeny with recombinant genotypes than parental genotypes in the offspring of a dihybrid testcross.
سؤال
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
سؤال
Twin spotting provides evidence of what genetic event?

A)meiotic recombination
B)mitotic recombination
C)linkage
D)mutation
E)biological evolution
سؤال
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.
سؤال
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.
سؤال
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
سؤال
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.
سؤال
The hypothesis that predicts no linkage between genes is known as the null hypothesis.
سؤال
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
سؤال
Chiasmata are structures that show where recombination occurred between sister chromatids.
سؤال
Chiasmata can be seen through a light microscope and are sites of recombination.
سؤال
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
سؤال
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.
سؤال
Genes that are not syntenic are not linked.
سؤال
Large sectors suggest a mitotic recombination event late in the growth of a yeast colony.
سؤال
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.
سؤال
Sectors with a different phenotype in an otherwise uniform yeast colony may be evidence of mitotic recombination.
سؤال
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
فتح الحزمة
قم بالتسجيل لفتح البطاقات في هذه المجموعة!
Unlock Deck
Unlock Deck
1/78
auto play flashcards
العب
simple tutorial
ملء الشاشة (f)
exit full mode
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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%
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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%
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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%
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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%
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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+.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
63
Twin spotting provides evidence of what genetic event?

A)meiotic recombination
B)mitotic recombination
C)linkage
D)mutation
E)biological evolution
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
68
The hypothesis that predicts no linkage between genes is known as the null hypothesis.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
70
Chiasmata are structures that show where recombination occurred between sister chromatids.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
71
Chiasmata can be seen through a light microscope and are sites of recombination.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
74
Genes that are not syntenic are not linked.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
75
Large sectors suggest a mitotic recombination event late in the growth of a yeast colony.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
77
Sectors with a different phenotype in an otherwise uniform yeast colony may be evidence of mitotic recombination.
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
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
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.
فتح الحزمة
k this deck
locked card icon
فتح الحزمة
افتح القفل للوصول البطاقات البالغ عددها 78 في هذه المجموعة.