Question 1

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?

Accepted Answer

The map distance between genes depends on their order on the chromosome. If the order is A-B-C, the distance between A and C is 35 map units (10 + 25). If the order is A-C-B or B-C-A, the distance could be different, potentially 15 map units if A and C are closer together than either is to B. Thus, without knowing the order of the genes, both 15 and 35 map units are possible distances.

Question 2

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?

Accepted Answer

The fraction of progeny that will be RS/rs is 5% because the R/r and S/s genes are 10 map units apart, indicating a 10% recombination frequency. Since RS/rs progeny result from a single crossover event between the R/r and S/s genes, only half of the recombinant progeny (i.e., 5% of the total) will be RS/rs, with the other half being rS/Rs.

Question 3

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

B

Question 4

Which of the following processes can generate recombinant gametes?

Accepted Answer

Crossing over between two linked heterozygous loci and independent assortment of two unlinked heterozygous loci both contribute to the generation of recombinant gametes by mixing alleles in new combinations. Segregation of alleles in a heterozygote does not create recombinant gametes; it simply separates alleles into different gametes.

Question 5

The map of a chromosome interval is Q-10 m.u.-R-40 m.u.-S.Which interval would likely show the higher ratio of double to single chiasmata?

Accepted Answer

The interval with the highest ratio of double to single chiasmata is typically the middle of the chromosome, which in this case is interval R-S. Therefore, choice B is the best answer.

Question 6

Suppose a three-point testcross was conducted involving the genes X,Y,and Z.If the most abundant classes are XYz and xyZ and the rarest classes are xYZ and Xyz,which gene is in the middle?

Accepted Answer

The answer of Suppose a three-point testcross was conducted involving...

Question 7

In an individual with genotype Lpq/lPQ the L gene is in the middle.What would be the genotypes of the gametes produced by a double crossover surrounding the L locus?

Accepted Answer

Double crossover surrounding the L locus will result in the exchange of genetic material between the two homologous chromosomes, resulting in the formation of gametes with the following genotypes: LPQ and lpq.

Question 8

In humans,the genes for red-green colour blindness (R = normal,r = colour-blind)and haemophilia A (H = normal,h = haemophilia)are both X-linked and only 3 map units apart.Suppose a woman has four sons,and two are colour-blind but have normal blood clotting and two have haemophilia but normal colour vision.What is the probable genotype of the woman?

Accepted Answer

Since the woman has four sons, each son has a 50% chance of inheriting either X chromosome from the mother. Therefore, the two colour-blind sons inherited the mother's X chromosome with the R allele, and the two sons with haemophilia inherited the mother's X chromosome with the h allele. 
Since the two genes are only 3 map units apart, they are likely to be inherited together most of the time, unless a crossover event occurred between them during meiosis. 
To produce the observed pattern of inheritance, the woman must be a carrier of both R and H alleles on one of her X chromosomes and h and normal allele on the other X chromosome. 
Therefore, the probable genotype of the woman is Hr/hR.

Question 9

Crossing over takes place in paired bivalents consisting of ______ chromatids,and involves _______ of the chromatids.

Accepted Answer

The answer of Crossing over takes place in paired bivalents...

Question 10

Recombination frequencies near 50 percent suggest that

Accepted Answer

Recombination frequencies near 50% indicate that two genes assort independently, which can happen if they are on different chromosomes or if they are on the same chromosome but are so far apart that crossing over between them is very likely, effectively mimicking independent assortment.

Question 11

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

Accepted Answer

The synaptonemal complex is the zipper-like connection that holds paired homologs together during early prophase of meiosis. This complex plays an important role in crossing over and genetic recombination between homologous chromosomes. Spindle fibers are responsible for separating chromosomes during cell division, while a chiasma is a point of crossing over between homologous chromosomes.

Question 12

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

Accepted Answer

The answer of The map of a chromosome interval is...

Question 13

The measured distance between genes D and E in a two-point test cross is 50 map units.What does this mean in physical terms?

Accepted Answer

In genetic mapping, a distance of 50 map units suggests that the genes are either on different chromosomes (and thus assort independently) or they are on the same chromosome but far enough apart that recombination between them occurs with a frequency that approaches the maximum possible (50%), indicating they could be linked but appear to assort independently due to the high recombination rate. This does not necessarily mean they are exactly 50 map units apart if they are on the same chromosome; they could be further apart, leading to the same observed recombination frequency.

Question 14

A dihybrid test cross is made between genes H and I.You wish to use the χ² test to test the hypothesis that the H and I genes are unlinked.How many degrees of freedom would there be in this test?

Accepted Answer

In a dihybrid test cross, there are four possible phenotypic classes. The degrees of freedom for a chi-square test is calculated as the number of phenotypic classes minus one, which is 4 - 1 = 3. However, since we are testing for linkage (independence), we subtract one more degree of freedom, resulting in 2 - 1 = 1 degree of freedom.

Question 15

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

Accepted Answer

The map distance between the sn and ct genes can be calculated using the number of recombinant offspring (those with a combination of traits different from the parents) divided by the total number of offspring, multiplied by 100 to get the percentage, which is equivalent to map units (m.u.). In this case, the recombinant offspring are those with sn ct+ and sn+ ct phenotypes, which total 36 + 39 = 75. The total number of offspring is 13 + 36 + 39 + 12 = 100. Thus, 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 given answer choices suggest a misunderstanding of the calculation, the correct approach to calculate map distance in this context would be to recognize the error in interpreting the data for this specific question. The correct calculation for map distance should consider only the recombinant phenotypes relative to the total, which typically does not exceed 50% due to the nature of genetic linkage. The error in the initial explanation points to a misinterpretation of the data provided, suggesting a need for clarification in understanding genetic linkage and recombination calculations.

Question 16

In Drosophila,the genes b,c,and sp are linked and in the order given,the distances being b-c = 30 m.u.and c-sp = 20 m.u.These intervals exhibit 90% interference.How many double crossovers would be recovered in a three-point cross involving b,c,and sp out of 1000 progeny?

Accepted Answer

The answer of In Drosophila,the genes b,c,and sp are linked...

Question 17

The T/t and S/s genes are linked and 10 map units apart.In the cross Ts/tS × ts/ts what fraction of the progeny will be Ts/ts?

Accepted Answer

The answer of The T/t and S/s genes are linked...

Question 18

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 males are distributed as follows.What is the value of ?² for a test of the hypothesis that the sn and car genes are unlinked? $$\begin{array} { | l | l | } \hline s n \text { car } ^ { + } & 55 \ \hline s n \text { car } ^ { + } & 45 \ \hline s n ^ { + } \mathrm { car } ^ { + } & 45 \ \hline \mathrm { sn } ^ { + } \mathrm { car } ^ { + } & 55 \ \hline \end{array}$$

Accepted Answer

The hypothesis of unlinked genes suggests that the offspring should exhibit a 1:1:1:1 ratio for the four phenotypes. Given that there are 200 total offspring (55+45+45+55), we expect 50 of each phenotype under the hypothesis. The chi-square test value ($\chi^2$) is calculated as $\sum \frac{(O-E)^2}{E}$, where O is the observed frequency and E is the expected frequency. For each phenotype, the calculation is $\frac{(55-50)^2}{50} + \frac{(45-50)^2}{50} + \frac{(45-50)^2}{50} + \frac{(55-50)^2}{50} = \frac{25}{50} + \frac{25}{50} + \frac{25}{50} + \frac{25}{50} = 2.0$.

Question 19

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?

Accepted Answer

A cross between LM/lm and lm/lm produces progeny with the genotype LM/lm and lm/lm. Since the L and M genes are separated by 100 map units, they are far enough apart that crossover events can occur between them during meiosis. Crossover events can result in the exchange of alleles between homologous chromosomes, leading to the production of recombinant gametes. The frequency of recombination between two genes is proportional to their physical distance on the chromosome. In this case, the distance between L and M is 100 map units, so the frequency of recombination between them is 100/2 = 50%. This means that 50% of the progeny from the LM/lm × lm/lm cross will be recombinant (either Lm/lm or LM/LM). The other 50% will be non-recombinant (either LM/lm or lm/lm). Since we are interested in the frequency of Lm/lm progeny specifically, we need to consider the genotypes of the recombinant progeny. Of the recombinant progeny, half will be Lm/lm and half will be LM/LM. Therefore, the fraction of progeny with the Lm/lm genotype is 50% × 1/2 = 25%.

Question 20

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?

Accepted Answer

The answer of In Drosophila,the genes y (yellow body)and car...

Quiz 3: The Chromosome Theory of Inheritance

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?

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?

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?

Which of the following processes can generate recombinant gametes?

The map of a chromosome interval is Q-10 m.u.-R-40 m.u.-S.Which interval would likely show the higher ratio of double to single chiasmata?

Suppose a three-point testcross was conducted involving the genes X,Y,and Z.If the most abundant classes are XYz and xyZ and the rarest classes are xYZ and Xyz,which gene is in the middle?

In an individual with genotype Lpq/lPQ the L gene is in the middle.What would be the genotypes of the gametes produced by a double crossover surrounding the L locus?

Crossing over takes place in paired bivalents consisting of chromatids,and involves _ of the chromatids.

Recombination frequencies near 50 percent suggest that

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

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

The measured distance between genes D and E in a two-point test cross is 50 map units.What does this mean in physical terms?

A dihybrid test cross is made between genes H and I.You wish to use the χ² test to test the hypothesis that the H and I genes are unlinked.How many degrees of freedom would there be in this test?

In Drosophila,the genes b,c,and sp are linked and in the order given,the distances being b-c = 30 m.u.and c-sp = 20 m.u.These intervals exhibit 90% interference.How many double crossovers would be recovered in a three-point cross involving b,c,and sp out of 1000 progeny?

The T/t and S/s genes are linked and 10 map units apart.In the cross Ts/tS × ts/ts what fraction of the progeny will be Ts/ts?

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?

Quiz 3: The Chromosome Theory of Inheritance

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?

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?

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?

Which of the following processes can generate recombinant gametes?

The map of a chromosome interval is Q-10 m.u.-R-40 m.u.-S.Which interval would likely show the higher ratio of double to single chiasmata?

Suppose a three-point testcross was conducted involving the genes X,Y,and Z.If the most abundant classes are XYz and xyZ and the rarest classes are xYZ and Xyz,which gene is in the middle?

In an individual with genotype Lpq/lPQ the L gene is in the middle.What would be the genotypes of the gametes produced by a double crossover surrounding the L locus?

Crossing over takes place in paired bivalents consisting of ______ chromatids,and involves _______ of the chromatids.

Recombination frequencies near 50 percent suggest that

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

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

The measured distance between genes D and E in a two-point test cross is 50 map units.What does this mean in physical terms?

A dihybrid test cross is made between genes H and I.You wish to use the χ2 test to test the hypothesis that the H and I genes are unlinked.How many degrees of freedom would there be in this test?

In Drosophila,the genes b,c,and sp are linked and in the order given,the distances being b-c = 30 m.u.and c-sp = 20 m.u.These intervals exhibit 90% interference.How many double crossovers would be recovered in a three-point cross involving b,c,and sp out of 1000 progeny?

The T/t and S/s genes are linked and 10 map units apart.In the cross Ts/tS × ts/ts what fraction of the progeny will be Ts/ts?

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?

Crossing over takes place in paired bivalents consisting of chromatids,and involves _ of the chromatids.

A dihybrid test cross is made between genes H and I.You wish to use the χ² test to test the hypothesis that the H and I genes are unlinked.How many degrees of freedom would there be in this test?