Human Heredity: Principles and Issues (MindTap Course List)
11th Edition
ISBN: 9781305251052
Author: Michael Cummings
Publisher: Cengage Learning
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A Drosophila embryo dies during early embryogenesis due to a
recessive maternal effect allele called bicoid−. The wild-type allele
is designated bicoid+. What are the genotypes and
the embryo’s mother and maternal grandparents?
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- In Drosophila, a heterozygous female for the X-linkedrecessive traits a, b, and c was crossed to a male that phenotypically expressed a, b, and c. The offspring occurred inthe following phenotypic ratios.+ b c 460a + + 450a b c 32+ + + 38a + c 11+ b + 9 No other phenotypes were observed.(a) What is the genotypic arrangement of the alleles ofthese genes on the X chromosome of the female?arrow_forwardA Drosophila male is heterozygous for a reciprocaltranslocation between an autosome and the Y chromosome. The part of the autosome now present onthe Y chromosome contains the dominant mutationLyra (shortened wings); the other (normal) copyof the same autosome is Lyra+. This male is nowmated with a true-breeding, wild-type female. Whatkinds of progeny would be obtained, and in whatproportions?arrow_forwardThe following list of four Drosophila mutations indicates the symbol for the mutation, the name of thegene, and the mutant phenotype:Allele symbol Gene name Mutant phenotypedwp dwarp small body, warped wingsrmp rumpled deranged bristlespld pallid pale wingsrv raven dark eyes and bodiesYou perform the following crosses with the indicatedresults:Cross #1: dwarp, rumpled females × pallid, raven males→ dwarp, rumpled males and wild-type femalesCross #2: pallid, raven females × dwarp, rumpled males→ pallid, raven males and wild-type femalesF1 females from cross #1 were crossed to males froma true-breeding dwarp rumpled pallid raven stock.The 1000 progeny obtained were as follows:pallid 3pallid, raven 428pallid, raven, rumpled 48pallid, rumpled 23dwarp, raven 22dwarp, raven, rumpled 2dwarp, rumpled 427dwarp 47Indicate the best map for these four genes, includingall relevant data. Calculate interference values whereappropriate.arrow_forward
- In Drosophila, a cross (cross 1) was made between twomutant flies, one homozygous for the recessive mutationbent wing (b) and the other homozygous for the recessivemutation eyeless (e). The mutations e and b are alleles oftwo different genes that are known to be very closelylinked on the tiny autosomal chromosome 4. All the progeny had a wild-type phenotype. One of the female progeny was crossed with a male of genotype b e/b e ; we willcall this cross 2. Most of the progeny of cross 2 were of theexpected types, but there was also one rare female ofwild-type phenotype.a. Explain what the common progeny are expected tobe from cross 2.b. Could the rare wild-type female have arisen by (1)crossing over or (2) nondisjunction? Explain.arrow_forwardA maternal effect gene in Drosophila, called torso, occurs as a functionalallele (torso+) and a nonfunctional, recessive allele (torso−)that prevents the correct development of anterior- and posterior-moststructures. A wild-type male (torso+torso+) is crossed to a female ofunknown genotype. This cross produces offspring (larva) that are allmissing their anterior- and posterior-most structures and therefore die during early development. What are the genotype and the phenotypeof the female fly in this cross? What are the genotypes andphenotypes of the female fly’s parents?arrow_forwardIn Drosophila, gray body color is dominant over ebony body color,while long wings are dominant over vestigial wings. Work thefollowing crosses through the F2 generation, and determine thegenotypic and phenotypic ratios for each generation. Assumethat the P1 individuals are homozygous: (a) gray, long * ebony,vestigial, and (b) gray, vestigial * ebony, long, and (c) gray,long * gray, vestigial.arrow_forward
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- Three recessive mutations in that affect eye colour, wing shape and body colourarew, m and b. The following numbers were obtained for testcross progeny in Drosophila. Number Phenotype + m + w + b + + b w m + m b 218 236 168 178 95 101 w + + + 3 w mb 1 (i) Determine the chromosomal composition of the heterozygous female parent. Construct a genetic map of the linkage group(s) these genes occupy. Show the order and give the map distances between the genes. (ii)arrow_forwardIn a cross in Drosophila involving the X-linked recessive eye mutationwhite and the autosomally linked recessive eye mutation sepia(resulting in a dark eye), predict the F1 and F2 results of crossingtrue-breeding parents of the following phenotypes:(a) white females * sepia males(b) sepia females * white malesNote that white is epistatic to the expression of sepiaarrow_forwardFemale fruit flies homozygous for the X-linked white-eye alleleare crossed to males with red eyes. On very rare occasions, an offspringof such a cross is a male with red eyes. Assuming these rareoffspring are not due to a new mutation in one of the mother’s Xchromosomes that converted the white-eye allele into a red-eyeallele, explain how a red-eyed male arises.arrow_forward
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