On the
a. Using any genotypes you choose, design two separatecrosses, one to test recombination between genes Y and
b. Can any cross reveal genetic linkage between gene Y and gene F? Why or why not?
c. Why is “independent assortment” the genetic term thatbest describes the observations of a genetic cross between gene Y and gene F?
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Genetic Analysis: An Integrated Approach (3rd Edition)
- 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_forwardIn 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_forwardBlack body (b) and purple eye (pr) are recessive autosomal mutations in Drosophila. Bridges are crossed b/b females with pr/pr males and in the F2 observed 684 wild type, 371 black-bodied, and 300 purple-eyed flies. Do these results indicate that the b and pr genes are closely linked? Explain. (Remember that there is no crossing-over in male Drosophila)..arrow_forward
- 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 sn ct 36 sn ct+ 13 sn+ ct 12 sn+ ct+ 39 What is the map distance between sn and ct?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 the following cross: Male parent Female parent Male offspring Female offspring Long Miniature 750 miniature 761 long O male: X* / X* and female X™ /x+ O male: X*/Xt and female Xm /xm O male: X*/ Y and female Xm /xm O male: Xm/ Y and female Xm /xmarrow_forwardIn Drosophila, the gene loci for curved wings and purple eyes are 20 centimorgans (map units) apart from each other on an autosomal chromosome. Wild-type flies were mated with double mutant flies with cv cv p p genotypes and curved wing/purple eye phenotypes. AlIl of the F1 flies had wild type phenotypes. The F, flies were crossed with double mutant flies with cv cv p'p genotypes and curved wing/purple eye phenotypes. What are the predicted percentages of flies in the F2 generation with wild type wings and eyes, curved wings and wild type eyes, wild type wings and purple eyes, and curved wings and purple eyes?arrow_forward
- In Drosophila, the sepia mutation (se, chromosome 3, position 26) results in dark brown eyes, while cinnabar (cn, chromosome 2, position 57.5) results in bright orange-red eyes. True breeding, wild type females are mated with true breeding males homozygous recessive for both traits. Using Drosophila notation, diagram the P1 and F1 crosses. P1 F1 Fill in the chart with phenotypic ratios that would be expected in the F2 generation. Use the space provided to show your work. Phenotype Females Males Overall (♀and ♂) =1 =1arrow_forwardGiven the following alleles that control the traits of Drosophila melanogaster:R = red eyes L = long wings B = brown bodyr = sepia eyes l = vestigial wings b = black body A testcross between a triple heterozygous parent and a recessive parent revealed the followingexpressed alleles: 55 sepia long brown 8 sepial long black187 sepia vestigial black 1 sepia vestigial brown43 red vestigial black 196 red long brown2 red long black 9 red vestigial brown Determine the order of the loci and the distance between them in centimorgan (cM). Construct agene map to show your results. Show COMPLETE solution.arrow_forwardIn Drosophila, the autosomal recessive brown eyecolor mutation displays interactions with both theX-linked recessive vermilion mutation and the autosomal recessive scarlet mutation. Flies homozygous for brown and simultaneously hemizygous orhomozygous for vermilion have white eyes. Fliessimultaneously homozygous for both the brown andscarlet mutations also have white eyes. Predict theF1 and F2 progeny of crossing the following truebreeding parents:a. vermilion females × brown malesb. brown females × vermilion malesc. scarlet females × brown malesd. brown females × scarlet malesarrow_forward
- 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 progeny phenotypes are missing? Why?arrow_forwardProduce a Punnett square to illustrate the dihybrid cross described below: There are two common alleles for the TAS2R38 gene on Chromosome 7. This gene encodes a seven-transmembrane G-protein coupled receptor. This receptor controls the ability to taste glucosinolates. Phenylthiocarbamide (PTC) is a synthetic glucosinolate. The recessive TAS2R38 allele produces a non-functional receptor. The father in this dihybrid cross is heterozygous for these alleles, meaning that he can taste PTC. The mother is homozygous recessive, meaning that she cannot taste PTC The father has X-Linked Protoporphyria which means that he is very sensitive to sunlight exposure, he is hemizygous for the dominant causative mutation. The mother is homozygous wild type at the same locus. Add a file here showing your diagram.arrow_forwardIN DROSOPHILA, AN X-LINKED RECESSIVE MUTATION, Xm CAUSES MINIATURE WINGS. LIST THE F₂ PHENOTYPIC RATIOS IF: A MINIATURE-WINGED FEMALE IS CROSSED WITH A NORMAL MALE AND A MINIATURE-WINGED MALE IS ● ● CROSSED WITH A NORMAL FEMALE. WHAT WOULD THE PHENOTYPIC RATIO FROM (A) BE IF THE MINIATURE- WINGED GENE WERE AUTOSOMAL? ASSUME IN ALL CASES THAT THE P1 INDIVIDUALS ARE TRUE-BREEDING.arrow_forward
- Biology: The Dynamic Science (MindTap Course List)BiologyISBN:9781305389892Author:Peter J. Russell, Paul E. Hertz, Beverly McMillanPublisher:Cengage Learning