Question 1

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 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.

Accepted Answer

The female is sn sn and ct+ ct+ while the male is sn+ sn+ and ct ct+. The F1 generation is all sn sn ct ct+. After interbreeding of the F1 generation, the F2 generation has the following genotype distribution:
sn+ ct+ : 225
sn+ ct : 31
sn ct+ : 36
sn ct : 208
Total: 500
The sum of the recombination frequencies is (31+36)/500 = 0.134. Therefore, the map distance is 0.134 x 100 = 13.4 cM or approximately 25 m.u. (centiMorgans and map units are interchangeable). Hence, the answer is option C.

Question 2

Which of the three genes, if any, is unlinked to the others?
A)T/t
B)R/r
C)W/w
D)All three genes are linked.

Accepted Answer

The question is asking which gene(s) are unlinked to the others. In genetics, linked genes are those located on the same chromosome and tend to be inherited together. Unlinked genes, on the other hand, are those located on different chromosomes or far apart on the same chromosome, and are more likely to assort independently during meiosis.

To determine whether the genes in the question are linked or unlinked, we need to look at the results of the test cross. A test cross involves crossing an individual with an unknown genotype (but which may be heterozygous or homozygous) with a homozygous recessive individual, and observing the phenotypic ratios of the offspring.

From the given data, we can see that the expected phenotypic ratio for each gene is 1:1, indicating that the genes are located on different homologous chromosomes and are therefore unlinked. This means that each gene is inherited independently of the others.

Therefore, the answer is C)W/w, as this gene is unlinked to the others.

Question 3

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 LM/lm * lm/lm would be Lm/lm?
A)10%
B)25%
C)50%
D)75%
E)100%

Accepted Answer

Since the L and M genes are 100 map units apart, we can assume they are on different arms of the chromosome and undergo independent assortment. Therefore, the F1 gametes will be LM and lm in equal proportions. When we cross an LM/lm F1 with an lm/lm individual, we will get four different possible gametes: LM, Lm, lm, and m. Half of the gametes will have an L allele and half will have an M allele, so the probability of getting an Lm genotype in the progeny is (1/2) x (1/2) = 1/4 or 25%.

Question 4

Which statement(s) below apply to the concept of gene linkage? 
A.The different alleles of two or more genes that are on a chromosome are inherited in an manner inconsistent with Mendel's Second law
B.Recombination between homologous chromosomes during meiosis results in different combinations of alleles for different genes
C.Recombination between sister chromatids during meiosis results in different combinations of alleles for different genes
A)A only
B)B only
C)C only
D)A and B
E)A, B, and C

Accepted Answer

Gene linkage refers to the tendency of genes on the same chromosome to be inherited together because they are physically connected. This can result in the inheritance of different combinations of alleles than expected based on Mendel's Second law. Recombination, or the swapping of genetic material between homologous chromosomes during meiosis, can disrupt this linkage and result in different combinations of alleles for different genes. Sister chromatid exchange, however, does not result in recombination between different chromosomes and therefore does not affect gene linkage. Therefore, both statements A and B apply to the concept of gene linkage, making choice D the correct answer.

Question 5

If only 100 progeny had been counted and the same proportion observed, how would the p value and the conclusion drawn about linkage change?
A)The p value would increase, and the likelihood of linkage increases.
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.

Accepted Answer

With a smaller sample size (100 progeny), the p value would increase due to reduced statistical power, making it harder to detect linkage, thus increasing the likelihood of concluding linkage.

Question 6

Which of the following linkage maps correctly shows the order and distance between the px, sp, and cn genes?
A)sp--0.21 mu--px--30.01 mu--cn
B)sp--30.01 mu--px--0.21 mu--cn
C)sp--0.2 mu--px--30 mu--cn
D)px--0.2 mu--sp--30.2 mu--cn
E)px--30.2 mu--sp--0.2 mu--cn

Accepted Answer

In genetic linkage maps, the distance between genes is measured in centiMorgans (cM) or map units (mu). The order of the genes is always mentioned from left to right. In option A, the distance between sp and px is 0.21 mu, and the distance between px and cn is 30.01 mu. This correctly shows the order and distance between the sp, px, and cn genes.

Question 7

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

Accepted Answer

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

Question 8

Given this data, which of the following is the most accurate range within which the p value falls, and what can be concluded?
A)0 < p < 0.01, genes C and D are most likely unlinked.
B)0 < p < 0.01, genes C and D are most likely linked.
C)0.01 < p < 0.05, genes C and D are most likely linked.
D)0.01 < p < 0.05, genes C and D are most likely unlinked.

Accepted Answer

A p-value in the range 0 < p < 0.01 suggests strong evidence against the null hypothesis, indicating that genes C and D are most likely unlinked.

Question 9

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

Accepted Answer

The answer of What is the genotype of the females...

Question 10

If two or more of the genes are linked, what map distance separates them?
A)4 mu
B)12 mu
C)24 mu
D)50 mu
E)None of the genes are linked to each other.

Accepted Answer

The answer of If two or more of the genes...

Question 11

What is the genotype of the tall, red, wide-leaved parent?
A)Tt Rr Ww
B)Tt Rr WW
C)TT RR WW
D)TT Rr Ww
E)TT RR Ww

Accepted Answer

The answer of What is the genotype of the tall,...

Question 12

How many degrees of freedom are there?
A)1
B)2
C)3
D)4

Accepted Answer

The number of degrees of freedom (DOF) corresponds to the number of independent variables that can be varied in a system without violating any constraints. In a 3-dimensional space, there are three axes (x, y, and z), each of which can be varied independently, leading to a total of 3 DOF. Therefore, the correct answer is choice C)3.

Question 13

What is the coefficient of coincidence in this region?
A)0
B)0.2
C)0.4
D)0.6
E)0.8

Accepted Answer

The coefficient of coincidence is calculated as the observed double crossovers divided by the expected double crossovers. In this case, we have observed 4 double crossovers (crossing over of both pairs of genes) out of a total of 10 gametes produced. The expected number of double crossovers can be calculated by multiplying the individual crossover frequencies, which is 0.2 x 0.3 = 0.06. Then, the coefficient of coincidence would be 4/0.06 = 0.67. However, this value is greater than expected, indicating some interference. The coefficient of coincidence is therefore 1 - interference, which is 1 - 0.6 = 0.4. Thus, the best choice is C)0.4.

Question 14

If the map distance between genes A and B is 10 map units and the map distance between genes B and C is 25 map units, what is the map distance between genes A and C?
A)15 map units
B)35 map units
C)either 15 map units or 35 map units, depending on the order of the genes
D)The map distance between A and C can not be predicted from these data.

Accepted Answer

The map distance between A and C can be either 15 or 35 map units, depending on the order of the genes on the chromosome. If A-B-C, then it's 35 map units; if C-B-A, then it's 15 map units.

Question 15

Which of the three genes is in the middle?
A)px
B)sp
C)cn
D)Insufficient data

Accepted Answer

The answer of Which of the three genes is in...

Question 16

The R/r and S/s genes are linked and 10 map units apart.In the cross Rs/rS * rs/rs what fraction of the progeny will be RS/rs?
A)5%
B)10%
C)25%
D)40%
E)45%

Accepted Answer

The genes are 10 map units apart, indicating a 10% recombination frequency. In the cross Rs/rS * rs/rs, the recombinant gametes (RS and rs) will each occur 5% of the time, leading to 5% RS/rs progeny.

Question 17

Calculate the $ \chi$² value used test the hypothesis that the C and D genes are unlinked. A)0 B)0.0576 C)10.8 D)14.4 E)Cannot be determined

Accepted Answer

The answer of Calculate the $ \chi$² value used test...

Question 18

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 Of these 4 phenotypic classes of offspring, which of the following arose from a parental gamete produced by the F1 females? A)sn⁺ ct⁺ and sn⁺ ct flies B)sn⁺ ct⁺ flies only C)sn ct and sn⁺ ct⁺ flies D)sn⁺ ct and sn ct⁺ flies E)sn ct and sn ct⁺ flies

Accepted Answer

The female parent is sn sn ct+ ct+. The male parent is sn+ sn+ ct ct. The F1 offspring will be all sn sn+ ct+ ct because of the reciprocal crosses. When these flies are intercrossed, we can set up a Punnett square to determine the offspring. Possible gametes from the female: sn ct+: sn ct: sn+ ct+: sn+ ct Possible gametes from the male: sn+ ct: sn+ ct+: sn ct: sn ct+ Crossing these gametes gives the following offspring: sn sn+ ct ct: 1/4 (from sn ct x sn+ ct) sn+ sn ct ct: 1/4 (from sn+ ct x sn ct+) sn sn ct ct+: 1/4 (from sn ct x sn ct+) sn+ sn+ ct+ ct+: 1/4 (from sn+ ct+ x sn ct+) So, the offspring include sn⁺ ct and sn ct⁺ flies (choice D).

Question 19

Which of the following linkage maps correctly shows the order and distance between the y, f, and v genes?
A)f--35 mu--y--15 mu--v
B)f--22 mu--y--15 mu--v
C)y--35 mu--f--22 mu--v
D)y--22 mu--v--15 mu--f
E)y--15 mu--v--22 mu--f

Accepted Answer

The answer of Which of the following linkage maps correctly...

Question 20

Calculate the coefficient of coincidence in this region.
A)0
B)0.16
C)0.33
D)0.5
E)0.66

Accepted Answer

The coefficient of coincidence is a measure of the randomness of a distribution. In this case, the distribution is the distribution of the number of events in each region. The coefficient of coincidence is calculated by dividing the number of pairs of events that occur in the same region by the total number of pairs of events. In this case, the number of pairs of events that occur in the same region is 16, and the total number of pairs of events is 100. Therefore, the coefficient of coincidence is 16/100 = 0.16.

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

Which of the three genes, if any, is unlinked to the others?

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 LM/lm * lm/lm would be Lm/lm?

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

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

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?

Given this data, which of the following is the most accurate range within which the p value falls, and what can be concluded?

What is the genotype of the females that gave rise to these progeny?

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

What is the genotype of the tall, red, wide-leaved parent?

How many degrees of freedom are there?

What is the coefficient of coincidence in this region?

If the map distance between genes A and B is 10 map units and the map distance between genes B and C is 25 map units, what is the map distance between genes A and C?

Which of the three genes is in the middle?

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

Calculate the $\chi$ ² value used test the hypothesis that the C and D genes are unlinked.

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

Calculate the coefficient of coincidence in this region.