Question 1

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?

Accepted Answer

The degrees of freedom for a chi-square test for goodness of fit is calculated as the number of categories minus 1. If there are 4 categories, the degrees of freedom would be 4 - 1 = 3.

Question 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.
 $\begin{array}{|l|r|}
\hline \text { tall, red, wide } & 381 \
\hline \text { tall, white, wide } & 122 \
\hline \text { short, red, wide } & 118 \
\hline \text { short, white, wide } & 379 \
\hline & 1000 \
\hline
\end{array}$
-What is the genotype of the tall plant that has red flowers and wide leaves?

Accepted Answer

The progeny ratios suggest that the tall plant is heterozygous for all traits, with the genotype T R W / t r W.

Question 3

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)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?

Accepted Answer

D

Question 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:
 $\begin{array}{|l|r|}
\hline p x  s p  c n & 1461 \
\hline p x  s p  c n^{+} & 3497 \
\hline p x  s p{+ c n} & 1\
\hline p x  s p^{+}  c n+ & 11 \
\hline p x^{+}  s p  c n & 9 \
\hline p x^{+}  s p  c n+ & 0 \
\hline p x^{+}  s p+ c n & 3483 \
\hline p x^{+}  s p+ c n+ & 1539 \
\hline & 10,000 \
\hline
\end{array}$
-Which of the three genes is in the middle?

Accepted Answer

The answer of Females heterozygous for the recessive second chromosome...

Question 5

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?

Accepted Answer

The answer of A dihybrid testcross is made to determine...

Question 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.
 $\begin{array}{|l|r|}
\hline \text { tall, red, wide } & 381 \
\hline \text { tall, white, wide } & 122 \
\hline \text { short, red, wide } & 118 \
\hline \text { short, white, wide } & 379 \
\hline & 1000 \
\hline
\end{array}$
-If two or more of the genes are linked, what map distance separates them?

Accepted Answer

The recombinant phenotypes (tall, white, wide and short, red, wide) total 240 (122 + 118). The map distance is calculated as (number of recombinant offspring / total offspring) x 100 = (240/1000) x 100 = 24 m.u.

Question 7

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.

Accepted Answer

The p-value is less than 0.01, so the most accurate range within which it falls is 0.001 < p < 0.01.

Question 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:

$$\begin{array} { | l | r | } 
\hline y f v & 3210 \
\hline y f v ^ { + } & 72 \
\hline y f ^ { + } v & 1024 \
\hline y f ^ { + } v^ + & 678 \
\hline y ^ { + } f v & 690 \
\hline y ^ { + } f v ^+ & 1044 \
\hline y ^ { + } f ^+ v & 60 \
\hline y ^ { + } f ^+ v^ + & 3222 \
\hline & 10,000 \
\hline
\end{array}$$

-What is the coefficient of coincidence in this region?

Accepted Answer

The coefficient of coincidence is calculated as the ratio of observed double crossovers to expected double crossovers. Here, the observed double crossovers are 72 + 60 = 132. The expected double crossovers can be calculated using the recombination frequencies between the genes. The coefficient of coincidence is then 132 divided by the expected number, which results in approximately 0.4.

Question 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:
 $\begin{array}{|l|r|}
\hline p x  s p  c n & 1461 \
\hline p x  s p  c n^{+} & 3497 \
\hline p x  s p{+ c n} & 1\
\hline p x  s p^{+}  c n+ & 11 \
\hline p x^{+}  s p  c n & 9 \
\hline p x^{+}  s p  c n+ & 0 \
\hline p x^{+}  s p+ c n & 3483 \
\hline p x^{+}  s p+ c n+ & 1539 \
\hline & 10,000 \
\hline
\end{array}$
-What is the coefficient of coincidence in this region?

Accepted Answer

The coefficient of coincidence (COC) is calculated as the observed double recombinants divided by the expected double recombinants. The observed double recombinants are the offspring that show recombination for both traits, which are the ones with low counts (1 and 11), totaling 12. The expected double recombinants can be calculated by multiplying the recombinant frequencies of each gene pair. The recombinant frequency for px-sp can be found by adding the recombinants for these genes (11+1)/(10,000) = 0.0012, and for sp-cn by adding their recombinants (3483+1539)/(10,000) = 0.5022. Multiplying these gives an expected frequency of 0.0012 * 0.5022 = 0.0006. The total expected double recombinants would be 0.0006 * 10,000 = 6. The COC is then 12 observed / 6 expected = 2. However, this calculation seems to misunderstand the question's intent, which is to calculate the COC as the observed double recombinants over the expected based on independent assortment frequencies. The correct approach involves calculating the expected double recombinants based on the single recombination frequencies between each gene pair, not directly from the observed counts. Given the misunderstanding in the calculation, let's correct the approach:To calculate the coefficient of coincidence (COC), we first need to determine the expected number of double recombinants if crossing over events were independent. The COC is defined as the ratio of the observed number of double recombinants to the expected number of double recombinants.The expected number of double recombinants can be calculated by multiplying the frequencies of the single crossovers and then by the total number of offspring. However, without clear single crossover frequencies provided directly, and given the misunderstanding in the initial explanation, we cannot accurately calculate the COC without the correct formula and data interpretation.The correct calculation involves understanding the genetic distances between the genes and using them to estimate the expected number of double recombinants. Since the initial explanation led to a confusion rather than a clear answer, and without re-calculating based on the correct genetic distances and frequencies, the selection of "B) 0.16" as the correct answer was premature and not based on a correctly outlined calculation process here. For a precise calculation, one would typically use the formula: COC = (Number of observed double recombinants) / (Number of expected double recombinants based on independent assortment). The expected number is often calculated from the product of the distances between genes (as probabilities), but without these distances or a clear method to calculate them from the provided data, the correct COC calculation process was not accurately described. Given the data and the typical approach to calculating COC, a reevaluation of the calculation method is necessary for an accurate answer.

Question 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.
 $\begin{array}{|l|r|}
\hline \text { tall, red, wide } & 381 \
\hline \text { tall, white, wide } & 122 \
\hline \text { short, red, wide } & 118 \
\hline \text { short, white, wide } & 379 \
\hline & 1000 \
\hline
\end{array}$
-This cross is not useful to determine if one of the genes is linked to the others.Which gene?

Accepted Answer

The distribution of phenotypes suggests that the gene for leaf width (W) segregates independently from the genes for height (T) and flower color (R), as the wide leaf trait appears in roughly equal proportions across tall/short and red/white categories, indicating no linkage with T or R.

Question 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?

Accepted Answer

With fewer progeny counted, the statistical power decreases, making it harder to detect a significant deviation from the expected ratios for independent assortment. This would result in a higher p value, suggesting a lower likelihood of rejecting the null hypothesis (no linkage), thus implying a decreased likelihood of linkage.

Question 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:
 $\begin{array}{|l|r|}
\hline p x  s p  c n & 1461 \
\hline p x  s p  c n^{+} & 3497 \
\hline p x  s p{+ c n} & 1\
\hline p x  s p^{+}  c n+ & 11 \
\hline p x^{+}  s p  c n & 9 \
\hline p x^{+}  s p  c n+ & 0 \
\hline p x^{+}  s p+ c n & 3483 \
\hline p x^{+}  s p+ c n+ & 1539 \
\hline & 10,000 \
\hline
\end{array}$
-What is the genotype of the females that gave rise to these progeny?

Accepted Answer

The answer of Females heterozygous for the recessive second chromosome...

Question 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:
 $\begin{array}{|l|r|}
\hline p x  s p  c n & 1461 \
\hline p x  s p  c n^{+} & 3497 \
\hline p x  s p{+ c n} & 1\
\hline p x  s p^{+}  c n+ & 11 \
\hline p x^{+}  s p  c n & 9 \
\hline p x^{+}  s p  c n+ & 0 \
\hline p x^{+}  s p+ c n & 3483 \
\hline p x^{+}  s p+ c n+ & 1539 \
\hline & 10,000 \
\hline
\end{array}$
-Which of the following linkage maps correctly shows the order and distance between the px, sp, and cn genes?

Accepted Answer

The correct order of the genes is sp-px-cn. The distance between sp and px is 0.21 m.u., and the distance between px and cn is 30.01 m.u.

Question 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?

Accepted Answer

The answer of Suppose the L and M genes are...

Question 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?

Accepted Answer

The answer of The pairwise map distances for four linked...

Question 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:

$$\begin{array} { | l | r | } 
\hline y f v & 3210 \
\hline y f v ^ { + } & 72 \
\hline y f ^ { + } v & 1024 \
\hline y f ^ { + } v^ + & 678 \
\hline y ^ { + } f v & 690 \
\hline y ^ { + } f v ^+ & 1044 \
\hline y ^ { + } f ^+ v & 60 \
\hline y ^ { + } f ^+ v^ + & 3222 \
\hline & 10,000 \
\hline
\end{array}$$

-Which of the following linkage maps correctly shows the order and distance between the y, f, and v genes?

Accepted Answer

The gene order is y-v-f. The recombination frequencies are calculated from the F2 data: y-v is 15 m.u. and v-f is 22 m.u.

Question 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: $$\begin{array} { | l | l | } \hline \operatorname { sn } c t & 13 \ \hline \operatorname { sn } c t ^ { + } & 36 \ \hline s n ^ { + } c t & 39 \ \hline s n ^ { + } c t ^ { + } & 12 \ \hline \end{array}$$ What is the map distance between sn and ct?

Accepted Answer

The map distance between the sn and ct genes is calculated by adding the number of recombinant offspring and dividing by the total number of offspring, then multiplying by 100 to get the percentage, which represents the map units (m.u.). Recombinant offspring are those that exhibit a combination of traits not seen in the parents, which in this case are "sn ct+" and "sn+ ct" phenotypes. The total number of recombinant offspring is 36 + 39 = 75. The total number of offspring is 13 + 36 + 39 + 12 = 100. Therefore, the map distance is (75/100) * 100 = 75%. However, since the maximum recombination frequency between two genes on the same chromosome is 50%, and the actual calculation seems to have been misunderstood, the correct interpretation of the data should reflect the actual method for calculating map distance, which involves recombinants and total offspring. The correct calculation should be (36 + 39) / 100 = 75 / 100 = 0.75 * 100 = 75 m.u., which seems incorrect based on standard genetic principles. The correct approach is to recognize that the calculation error led to an incorrect interpretation, and the actual map distance calculation formula should yield a result within the 0-50 m.u. range for linked genes. Given the numbers, the correct calculation for map distance should involve understanding that recombinants are a fraction of the total, and the error in the initial explanation suggests a misunderstanding of the genetic mapping principles or a typographical error in presenting the options or interpreting the data.

Question 18

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

Accepted Answer

The synaptonemal complex is the structure that forms between homologous chromosomes during meiosis and facilitates their alignment, pairing, and recombination.

Question 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?

Accepted Answer

Since the R and S genes are linked and 10 map units apart, the expected frequency of recombinant offspring is 10%. However, the question asks specifically for the fraction of progeny that will be R S / r s, which is one of the two possible recombinant types. Therefore, we divide the total recombinant percentage (10%) by 2, resulting in 5%.

Question 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?

Accepted Answer

In genetic mapping, the map distance between two genes reflects the frequency of recombination between them. If genes A and B are 20 map units apart, and genes B and C are 35 map units apart, the distance between A and C could be either the sum of the two distances (if B is between A and C) or the difference (if C is between A and B or vice versa). Without knowing the order of the genes, it's impossible to determine the exact distance between A and C.

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