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 (2nd Edition)
- The recessive, X-linked z1mutation of the Drosophilagene zeste (z) can produce a yellow (zeste) eye coloronly in flies that have two or more copies of the wildtype white (w) gene. Using this property, tandem duplications of the w+ gene called w+Rwere identified.Males with the genotype y+ z1w+R spl+ / Y thus havezeste eyes. These males were crossed to females withthe genotype y z1 w+R spl / y+ z1 w+R spl+. (These fourgenes are closely linked on the X chromosome, in theorder given in the genotype, with the centromere tothe right of all these genes: y = yellow bodies; y+ =tan bodies; spl = split bristles; spl+ = normal bristles.) Out of 81,540 male progeny of these females,the following exceptions were found:Class A 2430 yellow bodies, zeste eyes, wild-type bristlesClass B 2394 tan bodies, zeste eyes, split bristlesClass C 23 yellow bodies, wild-type eyes, wild-type bristlesClass D 22 tan bodies, wild-type eyes, split bristlesa. What were the phenotypes of the remainder of the81,540 males…arrow_forwardThe phenotype of vestigial (short) wings (vg) in Drosophila melanogaster is caused by a recessive mutant gene that independently assorts with a recessive gene for hairy (h) body. Assume that a cross is made between a fly that is homozygous for normal wings and has a hairy body and a fly with vestigial wings that is homozygous for normal body. The wild-type F1 flies were crossed among each other to produce 1024 F2 offspring. Which phenotypes would you expect among the F2 offspring, and how many of each phenotype would you expect? Group of answer choices 192 wild type, 256 vestigial, 64 hairy, and 192 vestigial and hairy All vestigial and hairy. 576 wild type, 192 vestigial, 192 hairy, and 64 vestigial and hairy All wild type 256 wild type; 256 vestigial, 256 hairy, and 256 vestigial and hairyarrow_forwardIn 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: genotype number sn ct 15 sn ct+ 34 sn+ ct 33 sn+ct+ 18 What is the map distance between sn and ct?arrow_forward
- In Drosophila, a cross was made between a yellowbodied male with vestigial (not fully developed)wings and a wild-type female (brown body). The F1generation consisted of wild-type males and wild-typefemales. F1 males and females were crossed, and theF2 progeny consisted of 16 yellow-bodied males withvestigial wings, 48 yellow-bodied males with normalwings, 15 males with brown bodies and vestigialwings, 49 wild-type males, 31 brown-bodied femaleswith vestigial wings, and 97 wild-type females.Explain the inheritance of the two genes in questionbased on these results.arrow_forwardIn the fruit fly Drosophila melanogaster, the following genes and mutations are known:Wing size: recessive allele for tiny wings t; dominantallele for normal wings T.Eye shape: recessive allele for narrow eyes n;dominant allele for normal (oval) eyes N.For each of the four following crosses, give thegenotypes of each of the parents.Male FemaleWings Eyes Wings Eyes Offspring1 tiny oval × tiny oval 78 tiny wings, oval eyes24 tiny wings, narrow eyes2 normal narrow × tiny oval 45 normal wings, oval eyes40 normal wings, narrow eyes38 tiny wings, oval eyes44 tiny wings, narrow eyes3 normal narrow × normal oval 35 normal wings, oval eyes29 normal wings, narrow eyes10 tiny wings, oval eyes11 tiny wings, narrow eyes4 normal narrow × normal oval 62 normal wings, oval eyes19 tiny wings, oval eyesarrow_forwardIn Drosophila, ebony body colour is produced by a recessive gene a and wild-type (gray) body colour by its dominant allele a+. Vestigial wings are governed by a recessive gene vg, and normal wing size (wild type) by its dominant allele vg+. If wild-type dihybrid flies are crossed and produce 256 progeny, how many of these progeny flies are expected in each phenotypic class?arrow_forward
- In 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_forwardIn Drosophila melanogaster, ebony body color is determined by the e allele. The e+ allele produces the wild-typehoney-colored body. In heterozygotes, the body is honey-colored except for a dark marking called the trident seen on the thorax. The e+ allele is thus considered to be incompletely dominant to the e allele.a. When female e+ e+ flies are crossed to male e+ e flies, what is the probability that progeny will have the tridentmarking? Animals with the marking mate among themselves. Of 100 progeny, how many would be expected to have a trident, how many would have ebony bodies, and how many would have honey-colored bodies?arrow_forwardConsider the following three autosomal recessive mutations in Drosophila:vestigial wings (v); wild type is long (v+)black body color (b); wildtype is gray (b+)plum eyes (p); wildtype is red (p+)A vestigal, gray, red female (homozygous for all three genes) is crossed with a long wing, black, plum male (homozygous for all three genes). The F1 female progeny are mated with triple homozygous recessive males. Here is the phenotypic data for the F2 progeny:vestigal; gray; red 580long wings; black; plum 592vestigal; black; red 45long; gray; plum 40vestigal; black; plum 89long; gray; red 94vestigal; gray; plum 3long; black; red 5A total of 1448 progeny were counted.Which one of the following values is the approximate distance between the plum eye color and black body color loci?arrow_forward
- In Drosophila melanogaster white (w) and miniature (m) wings are controlled by X-linked recessive genes with a recombination fruequency between them of approximately 38%. Show the sexes, phenotypes and proportions of offspring expected from the following mating:a. ++/wm female X wm maleb. +m/w+ female X w+ malec. w+/+m female X ++ maleIf we assume that white eyes and miniature wings are not x-linked but are linkedto the autosomal genes, what phenotypic frequencies would you expect from this cross: ++/wm female X ++/wm male?arrow_forwardIn 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_forwardIn Drosophila, raspberry eyes (ras) is a sex-linked trait (X chromosome) and recessive. Polished eyes (pol) is autosomal and recessive. A raspberry, polished male is crossed with a homozygous wild-type female, producing an F1 in which both males and females are phenotypically wild-type. The F1 are inbred to form an F2. At what frequency (the number of total flies) would you expect to find a phenotypically raspberry (but not polished) male in the F2? 1. 1/16 2. 3/16 3. 1/4 4. 6/16 5.9/16arrow_forward
- Biology: The Dynamic Science (MindTap Course List)BiologyISBN:9781305389892Author:Peter J. Russell, Paul E. Hertz, Beverly McMillanPublisher:Cengage Learning