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Mitotic recombination can occasionally produce a twin spot. Let’s suppose an animal species is heterozygous for two genes that govern fur color and length: one gene affects pigmentation, with dark pigmentation (A) dominant to albino (a); the other gene affects hair length, with long hair (L) dominant to short hair (l). The two genes are linked on the same chromosome. Let’s assume an animal of this species is
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Genetics: Analysis and Principles
- Consider two loci A and B that are 30 cM apart on chromosome 1 in the fruit fly. A paracentric inversion is found to span about half of the distance between these loci, but not include either A or B. What is the approximate recombination frequency expected for these loci in … flies that are heterozygous for this inversion? RF(A-B) for heterozygous flies is … flies that are homozygous for this inversion? RF(A-B) for homozygous flies is …arrow_forwardIn a breed of cattle, rough coat is dominant over smooth coat and black coat is codominant to red. What do the following statements mean a) The loci of the two pairs of gene concerned are on the same chromosome b) The loci of the two pairs of gene concerned are on different chromosomesarrow_forwardGenes a and b are 20 cM apart. An a+ b+/a+ b+ individual was mated with an a b/a b individual.(a) Diagram the cross and show the gametes produced by each parent and the genotype of the F1.(b) What gametes can the F1 produce, and in what proportions?(c) If the F1 was crossed to a b/a b individuals, what off-spring would be expected, and in what proportions?(d) Is this an example of the coupling or repulsion link-age phase?(e) If the F1 were intercrossed, what offspring would be expected, and in what proportions?arrow_forward
- Two recessive traits in mice—droopy ears and flaky tail—arecaused by genes that are located 6 mu apart on the same chromosome.A true-breeding mouse with normal ears (De) and a flaky tail ( ft)was crossed to a true-breeding mouse with droopy ears (de) and anormal tail (Ft). The F1 offspring were then crossed to mice withdroopy ears and flaky tails. If this testcross produced 100 offspring,what is the expected outcome of phenotypes?arrow_forwardTwo recessive traits in mice—droopy ears and flaky tail—arecaused by genes that are located 6 mu apart on the same chromosome.A true-breeding mouse with normal ears (De) and a flaky tail ( ft)was crossed to a true-breeding mouse with droopy ears (de) and anormal tail (Ft). The F1 offspring were then crossed to mice withdroopy ears and flaky tails. If this testcross produced 100 offspring,What topic in genetics does this question address?arrow_forwardTwo recessive traits in mice—droopy ears and flaky tail—arecaused by genes that are located 6 mu apart on the same chromosome.A true-breeding mouse with normal ears (De) and a flaky tail ( ft)was crossed to a true-breeding mouse with droopy ears (de) and anormal tail (Ft). The F1 offspring were then crossed to mice withdroopy ears and flaky tails. If this testcross produced 100 offspring,Make a drawing. Predict the outcome. Make a calculation.arrow_forward
- You are working with a hypothetical fly and have found color and wing mutants. Preliminary work indicates that the mutant traits are recessive and the associated genes are not sex-linked, but beyond that, you have no information. You first look at 2 genes, each with 2 alleles. "B" or “b" for body color and "W" or "w" for wing surface. The red-body phenotype is dominant to the yellow-body phenotype and smooth wings are dominant to crinkled wings.arrow_forwardWild-type mice have brown fur and short tails. Loss of function of a particular gene produces white fur, while loss of function of another gene produces long tails, and loss of function at a third locus produces agitated behavior. Each of these loss of function alleles is recessive. If a wild-type mouse is crossed with a triple mutant, and their F1 progeny is test-crossed, the following recombination frequencies are observed among their progeny. Produce a genetic map for these loci. Brown, short tailed, normal: 955 White, short tailed, normal: 16 Brown, short tailed, agitated: 0 White, short tailed, agitated: 36 Brown, long tailed, normal: White, long tailed, normal: Brown, long tailed, agitated: 46 0 14 White, long tailed, agitated: 933arrow_forwardThe genes S and T show complete linkage. If a heterozygous individual (whose parents were SSTT x sstt) were to produce gametes, which distribution below is most likely? a) S,s,T and t gametes in a 1:1:1:1 ratio b) ST and st gametes in a 1:1 RATIO c) ST, St, sT and st gametes in a 1:1:1:1 ratio d) SsTt gamete is the only type producedarrow_forward
- Part A You start your experiments with the eyeless mutation on chromosome IV. You cross the reciprocal translocation strain to the eyeless pure line to generate F, flies that are both translocation heterozygotes and Ee heterozygotes. You decide to testcross F, males and females in two separate experiments to take advantage of the fact that crossing over does not occur in male Drosophila. F, males x pure eyeless females (Note: There is no crossing over F, females x pure eyeless males (Note: Crossing over in the F, males.) Cross may occur in the F, females.) 4 eyeless, fully fertile 4 wild-type, semi-sterile 4 eyeless, semi-sterile 4 wild-type, fully fertile 4 eyeless, fully fertile 4 wild-type, semi-sterile 4 eyeless, semi-sterile 4 wild-type, fully fertile F, Progeny What can you conclude from these results? Select the two correct statements. > View Available Hint(s) O The F2 progeny in both experiments contain recombinants. O The F2 progeny contain recombinants only when F, females…arrow_forwardA new gene is being investigated in fruit flies. The recessive allele of this gene (b) causes the wings to develop a blue color, while the dominant allele (b+) permits wild-type colorless wings to develop. Preliminary studies indicate that this new gene is located on the X-chromosome. You decided to perform a two-point testcross to determine its positionrelative to the well-established garnet eyes gene (g). You cross a female heterozygous for both genes with a testcross male fly and obtain the male offspring results shown in table 1, below. Is the original female a coupling or repulsion heterozygote? What is the map distance between genes b and g?. Based on these results, gene b must be located between what two genes on the map? You perform another two-point testcross between gene b and gene v and obtained the results in table 2, below. Now, you can localize gene b to be specifically between which two genes?arrow_forwardA cross in Drosophila melanogaster involved the recessive X-linked genes for white eye (w), yellow body (y), and cut wings (c). A wild-type tri-hybrid female was crossed with wild-type males and only the male offspring were tallied. On the basis of the results shown below, which of the choices shown best represents the genetic map of the three loci on the X-chromosome? Phenotype Male Offspring + у t 494 394 + + ct 28 w y + 35 + у + 105 w + ct 101 w y ct 5 5.4 mu 17.6 mu ct 6.1 mu 18.3 mu w ct 17.6 mu 5.4 mu ct 18.3 mu 6.1 mu ct 6.1 mu 18.3 mu ct Submit Request Answer de Feedback Next >arrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning