Concept explainers
A
a. Determine the gene order and identify the alleles on thehomologous X chromosomes in the trihybrid females.
b. Calculate the recombination frequencies between eachof the gene pairs.
c. Compare the recombination frequencies and speculateabout the source of any apparent discrepancies in therecombination data.
d. Use
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Genetic Analysis: An Integrated Approach (2nd Edition)
- The following X-linked recessive traits are found in fruit flies:vermilion eyes are recessive to red eyes, miniature wings are recessiveto long wings, and sable body is recessive to gray body. A cross wasmade between wild-type males with red eyes, long wings, and graybodies and females with vermilion eyes, miniature wings, and sablebodies. The heterozygous female offspring from this cross, whichhad red eyes, long wings, and gray bodies, were then crossed tomales with vermilion eyes, miniature wings, and sable bodies. Thefollowing data were obtained for the F2 generation (including bothmales and females):1320 vermilion eyes, miniature wings, sable body1346 red eyes, long wings, gray body102 vermilion eyes, miniature wings, gray body90 red eyes, long wings, sable body42 vermilion eyes, long wings, gray body48 red eyes, miniature wings, sable body2 vermilion eyes, long wings, sable body1 red eyes, miniature wings, gray bodyAnalyze data. Make a drawing. Make a calculation.arrow_forwardBow-legs is hypothesized to be X-linked recessive trait in Drosophila melanogaster. The P1 virgin females were, once again, homozygous wild type but the males were bow-legged. There were 52 wild type males and 67 wild type females in the F1 generation. The F2 generation contained 30 wild type males, 75 wild type females, 40 bow-legged males and no bow-legged females. Is this what is to be expected? Use chi-square to prove your position.arrow_forwardAnother recessive mutation in Drosophila, ebony (e), is on anautosome (chromosome 3) and causes darkening of the bodycompared with wild-type flies. What phenotypic F1 and F2 maleand female ratios will result if a scalloped-winged female withnormal body color is crossed with a normal-winged ebony male?Work this problem by both the Punnett square method and theforked-line method.arrow_forward
- Let’s suppose that two different X-linked genes exist in mice,designated with the letters N and L. Gene N exists in a dominant,normal allele and in a recessive allele, n, that is lethal. Similarly,gene L exists in a dominant, normal allele and in a recessive allele,l, that is lethal. Heterozygous females are normal, but males thatcarry either recessive allele are born dead. Explain whether or notit would be possible to map the distance between these two genesby making crosses and analyzing the number of living and deadoffspring. You may assume that you have strains of mice in whichfemales are heterozygous for one or both genes.arrow_forwardOne of the X chromosomes in a particular Drosophila female had a normal order of genes but carried recessive alleles of the genes for yellow body color (y), vermilion eye color (v), and forked bristles (f), as well as the dominant X-linked Bar eye mutation (B). Her other X chromosome carried the wild-type alleles of all four genes, but the region including y+, v+, and f+ (but not B+) was inverted with respect to the normal order of genes. This female was crossed to a wild- type male in the cross diagrammed her. The cross produced the following male offspring:* table in figure a. Why are there no male offspring with the allele combinations y v f+, v+ v+ f, y v+ f+, or y+ v f (regardless of the allele of the Bar eye gene)?b. What kinds of crossovers produced the y v f b+ and v+ y+ f+ B offspring? Can you determine any genetic distances from these classed of progeny?c. What kinds of crossovers produced the y+ v f+ B+ and y v+ f B offspring?arrow_forwardOne of the X chromosomes in a particular Drosophila female had a normal order of genes but carried recessive alleles of the genes for yellow body color (y), vermilion eye color (v), and forked bristles (f), as well as the dominant X-linked Bar eye mutation (B). Her other X chromosome carried the wild-type alleles of all four genes, but the region including y+, v+, and f+ (but not B+) was inverted with respect to the normal order of genes. This female was crossed to a wild-type male in the cross diagrammed her. The cross produced the following male offspring: Y v f B 48 y+ v+ f+ B+ 45 y v f B+ 11 y+ v+ f+ B 8 y v f B 1 y+ v+ f+ B+ 1 a. Why are there no male offspring with the allele combinations y v f+, v+ v+ f, y v+ f+, or y+ v f (regardless of the allele of the Bar eye gene)? b.What kinds of crossovers produced the y v f b+ and v+ y+ f+ B offspring? Can you determine any genetic distances from these classed of progeny? c. What kinds of crossovers produced the…arrow_forward
- In Drosophila, a cross was made between females expressing thethree X-linked recessive traits, scute bristles (sc), sable body (s),and vermilion eyes (v), and wild-type males. All females were wildtype in the F1, while all males expressed all three mutant traits.The cross was carried to the F2 generation and 1000 offspringwere counted, with the results shown in the following table. Nodetermination of sex was made in the F2 data. Question:Calculate the coefficient of coincidence; does this represent positive or negative interference? Phenotype Offspringsc s v 314+ + + 280+ s v 150sc + + 156sc + v 46+ s + 30sc s + 10+ + v 14arrow_forwardIn Drosophila, the vermilion eye color is determined by a recessive allele, v, of an X-linked gene. The wildtype color is determined by the v+ allele and causes a brick red eye color. In a cross of a heterozygous female with a wild type male you observe 340 red eye females, 136 red eye males, and 90 vermillion males. Do these results follow your expectations?arrow_forwardIn drosophila, a recessive mutation (m-) of a maternal effect gene results in an abnormal phenotype wherein homozygous (m-m-) females produce eggs that cannot support embryonic development. Homozygous (m-m-) males, however, can still produce viable sperm. Using m+ to denote a normal gene, determine the genotypes and phenotypes of the F1s produce by a cross between a heterozygous female and a recessive male. From the offspring, backcross the recessive female with the paternal strain. What are the genotypes and phenotypes of the F2s? Show COMPLETE cross for both cases. If m-m- females produce useless eggs, then how are m-m- produced?arrow_forward
- You discover a new X-linked mutation for antenna shape in Drosophila (fruit flies) which produces long antennae. You call this new allele long (XL2), and discover it is incompletely dominant with the short antennae allele (XL1), so the heterozygotes (XL1XL2) have antennae that are intermediate in length (and XL2 homozygote females have long antennae). Males only have either long or short antennae. You cross a female with long antennae with a male with short antennae. What proportion of their male offspring do you expect to have short antennae? Group of answer choices 0.75 0.50 0 1.00 0.25arrow_forwardIn Drosophila, vermilion eye color is due to a recessive allele (v) located on the X chromosome. Curved wings are due to a recessive allele (cu) located on one autosome, and ebony body is due to a recessive allele (e) located on another autosome. A vermilion male is mated to a curved, ebony female, and the F1 males are phenotypically wild-type. If these males were backcrossed to curved, ebony females, what proportion of the F2 offspring will be wild-type males?arrow_forwardMiniature wings (Xm) in Drosophila result from an X-linked allele that is recessive to the allele for long wings (X +). Give the genotypes of the parents in each of the following crosses. Male parent Female parent Male offspring Female offspring a. long long 231 long, 250 miniature 560 long 250 miniature b. miniature long 610 long 632 long c. miniature long 410 long, 417 miniature 412 long, 415 miniature 417 miniature 415 miniature d. long miniature 753 miniature 761 long e. long long 625 long 630 longarrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning