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Q1: Assuming these two genes sort independently, how many progeny would you expect to show the green striped phenotype? (Enter your answers as a whole number, e.g. 1)
Q2:
Which of the following are recombinant
A. Green and Spotted
B. Green and Striped
C. Yellow and Spotted:
D. Yellow and Striped
Step by step
Solved in 2 steps
- In Drosophila flies, the allele b gives a black body, and the allele b+ gives brown, the wild-typephenotype. The allele wx of a separate gene gives waxy wings, and wx+ gives non-waxy. The allele cn ofa third gene gives cinnabar eyes, and cn+ gives red. A female heterozygous for these three genes istestcrossed, and 745 progenies are produced which are phenotypically classified as follows:5 b+ wx+ cn+4 b wx cn53 b+ wx cn 49 b wx+ cn+287 b+ wx+ cn279 b wx cn+ 33 b+ wx cn+35 b wx+ cn Make a linkage map of the three genes. Compute for interference and explain what the derived valuemeans. Show complete solutions to support your answers.When Calvin Bridges observed a large number of offspring from a cross of white-eyed female Drosophila tored-eyed males, he found very rare white-eyed femalesand red-eyed males among the offspring. He was ableto show that these exceptions resulted from nondisjunction, such that the white-eyed females had received twoXs from the egg and a Y from the sperm, while thered-eyed males had received no sex chromosome fromthe egg and an X from the sperm. What progeny wouldhave arisen from these same kinds of nondisjunctionalevents if they had occurred in the male parent? Whatwould their eye colors have been?The a, b, and c loci are all on different chromosomesin yeast. When a b+ yeast were crossed to a+ b yeastand the resultant tetrads analyzed, it was found thatthe number of nonparental ditype tetrads was equal tothe number of parental ditypes, but there were no tetratype asci at all. On the other hand, many tetratypeasci were seen in the tetrads formed after a c+ wascrossed with a+ c, and after b c+ was crossed withb+ c. Explain these results.
- Mitotic recombination can occasionally produce a twin spot.Let’s suppose an animal species is heterozygous for two genesthat govern fur color and length: one gene affects pigmentation,with dark pigmentation (A) dominant to albino (a); the other geneaffects hair length, with long hair (L) dominant to short hair (l).The two genes are linked on the same chromosome. Let’s assume ananimal of this species is AaLl; A is linked to l, and a is linked to L.Draw the chromosomes labeled with these alleles, and explainhow mitotic recombination could produce a twin spot with onespot having albino pigmentation and long fur and the other havingdark pigmentation and short fur.A wild-type fruit fly (heterozygous for gray body color andred eyes) is mated with a black fruit fly with purple eyes. Theoffspring are wild-type, 721; black purple, 751; gray purple, 49;black red, 45. What is the recombination frequency betweenthese genes for body color and eye color? Using informationfrom problem 3, what fruit flies (genotypes and phenotypes)would you mate to determine the order of the body color, wingsize, and eye color genes on the chromosome?DRAW IT Suppose, as in the questionat the bottom of Figure 15.9, the parental(P generation) flies were true-breedingfor gray body with vestigial wings andblack body with normal wings. Draw thechromosomes in each of the four possiblekinds of eggs from an F1 female, andlabel each chromosome as “parental”or “recombinant.”
- . In nature, the plant Plectritis congesta is dimorphic forfruit shape; that is, individual plants bear either winglessor winged fruits, as shown in the illustration.Wingless fruit Winged fruitPlants were collected from nature before floweringand were crossed or selfed with the following results:Number of progenyPollination Winged WinglessWinged (selfed) 91 1*Winged (selfed) 90 30Wingless (selfed) 4* 80Winged × wingless 161 0Winged × wingless 29 31Winged × wingless 46 0Winged × winged 44 0*Phenotype probably has a nongenetic explanation.Interpret these results, and derive the mode ofinheritance of these fruit-shaped phenotypes. Usesymbols. What do you think is the nongeneticexplanation for the phenotypes marked by asterisks inthe table?You design Drosophila crosses to provide recombinationdata for gene a, which is located on the chromosome shownin Figure 15.12. Gene a has recombination frequencies of14% with the vestigial wing locus and 26% with the browneye locus. Approximately where is a located along thechromosome?Hemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To calculate the recombination frequency between the two genes, you draw a large number of pedigrees that include grandfathers with both hemophilia and color blindness, their daughters (who presumably have one chromosome with two normal alleles and one chromosome with two mutant alleles), and the daughters sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia, 24 have neither of the traits, 1 has color blindness only, and 1 has hemophilia only. How many centimorgans (map units) separate the hemophilia locus from the locus for color blindness?
- The following corn loci are on one arm of chromosome9 in the order indicated (the distances between themare shown in map units):c-bz-wx-sh-d-centromere12 8 10 20 10C gives colored aleurone; c, white aleurone.Bz gives green leaves; bz, bronze leaves.Wx gives starchy seeds; wx, waxy seeds.Sh gives smooth seeds; sh, shrunken seeds.D gives tall plants; d, dwarf.A plant from a standard stock that is homozygous for allfive recessive alleles is crossed with a wild-type plantfrom Mexico that is homozygous for all five dominantalleles. The F1 plants express all the dominant allelesand, when backcrossed to the recessive parent, give thefollowing progeny phenotypes:colored, green, starchy, smooth, tall 360white, bronze, waxy, shrunk, dwarf 355colored, bronze, waxy, shrunk, dwarf 40white, green, starchy, smooth, tall 46colored, green, starchy, smooth, dwarf 85white, bronze, waxy, shrunk, tall 84colored, bronze, waxy, shrunk, tall 8white, green, starchy, smooth, dwarf 9colored, green, waxy, smooth,…In the video game Animal Crossing: New Horizons, flowering breeding is based in genetics. Each flower's color is determined by the genotype at three or four unlinked genes: R, Y, W, and S. The genotype of the elusive blue rose is RR YY ww ss. In the game, one way to get a blue rose is to cross two roses with the Rr Yy Ww ss genotype. A) What types of gametes and in what proportions will a Rr Yy Ww ss rose produce? B) In a cross Rr Yy Ww ss x Rr Yy Ww ss what are the possible offspring genotypes and at what frequency will they each appear? Show your work. C) What proportion of the offspring of the cross will be blue roses?On a camping trip, you find one male snail on a desertedisland that coils to the right. However, in this same area, youfind several shells (not containing living snails) that coil tothe left. Therefore, you conclude that you are not certain ofthe genotype of this male snail. On a different island, you finda large colony of snails of the same species. All of these snailscoil to the right, and every snail shell that you find on thissecond island coils to the right. With regard to the maternaleffect gene that determines coiling pattern, how would youdetermine the genotype of the male snail that you found onthe deserted island? In your answer, describe your expectedresults