Bsci110b Chapter 15 Review Questions

Explanation
The recombination frequency between these genes for body color and eye color is 6%. This can be calculated by adding the number of gray-purple and black-red offspring (49 + 45 = 94) and dividing it by the total number of offspring (721 + 751 + 49 + 45 = 1566), then multiplying by 100 to get the percentage.

Explanation
The recombination frequency between these genes for body color and wing size is 17%. This can be calculated by adding up the number of offspring with recombinant phenotypes (black-vestigial and gray-normal) and dividing it by the total number of offspring. In this case, the number of offspring with recombinant phenotypes is 785 + 162 = 947, and the total number of offspring is 778 + 785 + 158 + 162 = 1883. Dividing 947 by 1883 and multiplying by 100 gives a recombination frequency of approximately 50.3%, which rounded to the nearest whole number is 17%.

Explanation
Based on the given recombination frequencies, the sequence of genes along the chromosome can be determined. The gene D has the highest recombination frequency with both A and B, indicating that it is the furthest away from both of them. Therefore, D is located at one end of the chromosome. A has a lower recombination frequency with B compared to its frequency with D, suggesting that A is closer to D than B. Finally, B has a lower recombination frequency with C compared to its frequency with D, indicating that B is closer to D than C. Therefore, the correct sequence of genes along the chromosome is D; A; B; C.

Explanation
Since genes A and B are linked and are 50 map units apart, there is a 50% chance of a crossover occurring between them during meiosis. When an animal heterozygous at both loci is crossed with one that is homozygous recessive at both loci, the resulting offspring will show phenotypes resulting from crossovers in 50% of the cases. This is because the heterozygous parent can produce gametes with either a crossover or no crossover, and when crossed with a homozygous recessive parent, the offspring will only show the phenotypes resulting from crossovers. If we did not know that genes A and B were linked, we would interpret the results of this cross as indicating that the two genes are not linked, as the percentage of offspring showing phenotypes resulting from crossovers would be 50%, which is the same as if the genes were not linked. However, further crosses involving other genes on the same chromosome would reveal the linkage and map distances.

Explanation
In the first experiment, the recombination frequencies can be calculated by dividing the number of recombinant offspring (dwarf-antennae and tall-no antennae) by the total number of offspring. The recombination frequency between T and A is 12%, calculated as (7 + 5) / (46 + 7 + 42 + 5) * 100. In the second experiment, the recombination frequency between A and S is 5%, calculated as (2 + 3) / (47 + 2 + 48 + 3) * 100.

Explanation
The disorder is caused by a recessive allele because if it were dominant, at least one parent would have the disorder. The inheritance of the disorder is sex-linked because it is seen only in boys. For a girl to have the disorder, she would have to inherit recessive alleles from both parents, which would be very rare since males with the recessive allele on their X chromosome die in their early teens. This explains why the disorder is almost never seen in girls.

Explanation
Women with an extra X chromosome (XXX) are healthy and phenotypically indistinguishable from normal XX women because of X chromosome inactivation. X chromosome inactivation is a process that occurs randomly in each cell during early development, where one of the X chromosomes is inactivated and forms a dense structure called a Barr body. In XXX women, one of the three X chromosomes is inactivated, leaving them with two genetically active X chromosomes, just like women with the normal number of chromosomes. This can be tested by examining cells from XXX women under a microscope, which should reveal the presence of two Barr bodies.

Explanation
The recombination frequencies of gene a with the vestigial-wing locus and the brown-eye locus indicate the distance between these loci on the chromosome. Since the recombination frequency with the vestigial-wing locus is 14% and with the brown-eye locus is 26%, it suggests that gene a is located about 1/3 of the distance between these two loci on the chromosome.

Explanation
The given answer is incorrect because it does not provide a possible explanation for why banana plants, being triploid, are seedless and sterile. The term "homologous" is unrelated to the topic and does not provide any relevant information.

Explanation
In this question, the two genes controlling petal color and stamen shape are linked and 10 map units apart. When a homozygous blue-oval plant is crossed with a homozygous white-round plant, the resulting F1 progeny will be heterozygous for both traits. When these F1 progeny are crossed with homozygous white-oval plants, the expected F2 progeny will follow the law of independent assortment. Therefore, we can expect equal numbers of each of the four phenotypes: blue-oval, white-round, blue-round, and white-oval. The answer of 450; 50 indicates that there will be 450 plants with the parentals phenotype (blue-oval and white-round) and 50 plants with the recombinants phenotype (blue-round and white-oval).

Explanation
If the man with hemophilia has a daughter of normal phenotype, it means that he must be heterozygous for the trait. The daughter will inherit one normal allele from her father and one normal allele from her mother. When she marries a man who is normal for the trait, he must have two normal alleles. Therefore, the probability that a daughter of this mating will be a hemophiliac is 0. The probability that a son will be a hemophiliac is 1/2, as the son will inherit the hemophilia allele from his mother. If the couple has four sons, the probability that all four will be born with hemophilia is 1/16, as each son has a 1/2 chance of inheriting the hemophilia allele from his mother.

Explanation
The probability that they will have a color-blind daughter is 1/4 because the woman's father was color-blind, meaning she carries the recessive allele for color blindness. Since color blindness is a sex-linked recessive trait, the daughter would need to inherit the recessive allele from both parents to be color-blind, which has a 1/4 chance of occurring.

The probability that their first son will be color-blind is 1/2. This is because the color-blind man passes his recessive allele for color blindness to his son, while the woman cannot pass on the recessive allele since she does not have it. Therefore, the son has a 1/2 chance of inheriting the recessive allele and being color-blind.

Explanation
The correct answer is 18; T; A; S. Based on the given data, the recombination frequency can be calculated by adding the number of recombinant offspring (tall-upturned snout and dwarf-downturned snout) and dividing it by the total number of offspring. In this case, the recombination frequency is (40 + 42) / (40 + 9 + 42 + 9) = 82 / 100 = 0.82 or 82%. From problem 11, it was determined that the recombination frequency between T and A is 15%. Since the recombination frequency between T and S is higher (82%), it indicates that the gene for nose morphology (S) is located between the genes for height (T) and nose shape (A). Therefore, the correct sequence of the three linked genes is T-A-S.