Assignment 3Linkage and Recombination1. In corn, the genes an (anther ear), br (brachytic), and f (fine stripe)are linked. Testeross data are as follows:NumberNumber3552ProgenyProgeny+++88an ++++f+ br ++ br f21an +fan br +an br f21739955Determine the linkage map and the genotype of the homozygous parentsused to obtain the heterozygote for testcross.

In corn , there are three types of linked genes :- A) Anther ear is represented as an B )…

1. In corn, the genes an (anther ear), br (brachytic), and f (fine stripe) are linked. Testeross data are as follows: Progeny Number Progeny Number +++ 88 an ++ 355 ++f + br + + br f 21 2 an +f an br + 17 55 399 an br f Determine the linkage map and the genotype of the homozygous parents used to obtain the heterozygote for testeross.

1. In corn, the genes an (anther ear), br (brachytic), and f (fine stripe) are linked. Testeross data are as follows: Progeny Number Progeny Number +++ 88 an ++ 355 ++f + br + + br f 21 2 an +f an br + 17 55 399 an br f Determine the linkage map and the genotype of the homozygous parents used to obtain the heterozygote for testeross.

Human Anatomy & Physiology (11th Edition)

11th Edition

ISBN:9780134580999

Author:Elaine N. Marieb, Katja N. Hoehn

Publisher:Elaine N. Marieb, Katja N. Hoehn

Chapter1: The Human Body: An Orientation

Section: Chapter Questions

Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...

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Consider section J, the following is a karyotype of: 75 X » » N 5 1 2 3 SeXlY 6 13 19 MEK NYET 14 # 20 8 X 15 00 ** 21 9 4 16 22 10 X 11 17 12 16 18 Y O an abnormal karyotype of a human AMAB missing a Y chromosome an abnormal karyotype of a human AFAB showing trisomy 13 a normal karyotype of a human AMAB a normal karyotype of a human AFAB
9. Make a pedigree for each of the following situations. For each individual, write the individual's genotype (when possible) next to the individual's symbol (e.g. O xty, I Gg): a. Two parents do not have cystic fibrosis and they have a daughter with cystic fibrosis and a son who does not have cystic fibrosis. The daughter grows up and she mates with a male who does not have cystic fibrosis. Their only child is a boy and he has cystic fibrosis. b. A man with hemophilia mates with a female without hemophilia. They have one son and one daughter. The daughter has hemophilia and the son does not have hemophilia. The son grows up, and he marries and mates with a female. Their only child is a boy, and he has hemophilia.
#9 help
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7-8. Assume that white color is dominant over yellow color in squash. If pollen from the anthers of a heterozygous white-fruited plant is placed on the pistil of a yellow- fruited plant, show using ratios the genotypes and phenotypes you would expect the seeds from this cross to produce. Genotypes = 1/2 Ww 1/2 Ww = 1:1 ratio | Phenotypes = All white = 1:0 ratio Genotypes = 1/2 WW 1/2 ww = 1:1 ratio | Phenotypes = 1/2 white 1/2 yellow = 1:1 ratio Genotypes = 3/4 Ww 1/4 ww = 1:1 ratio | Phenotypes = 3/4 white 1/4 yellow = 1:1 ratio Genotypes = 1/2 Ww 1/2 ww = 1:1 ratio | Phenotypes = 1/2 white 1/2 yellow = 1:1 ratio
8. Answer the following questions for the two following pedigrees In Pedigree A, is the red highlighted trait dominant or recessive? Why? а. b. What is the genotype of individual C? In Pedigree B, is the red highlighted trait dominant or recessive? Why? а. d. What is the genotype of individual D? Pedigree A Pedigree B mele male : foncle I femole teffected D offected
Calculate the map distances between the genes.
In a diploid plant species, an F1 with the genotype Mm Rr Ss is test crossed to a pure breeding recessive plant with the genotype mm rr ss. The offspring genotypes are as follows: Genotype Number Mm Rr Ss 687 Mm Rr ss 5 Mm rr Ss 68 Mm rr ss 196 mm Rr Ss 185 mm Rr ss 72 mm rr Ss 8 mm rr ss 679 Total 1900 1. What is the interference value for this data set?
7. Instead of one big square it is much easier to make two small Punnett squares, one for each gene. In the space at the bottom of page, draw the two squares (from 4d and 5b on previous page). Indicate the phenotypic ratios for each square and then multiply to get the combined probabilities. For example, the probability of having both diseases requires multiplying the probability of each event. a. Is the inheritance of an "A" or "a" allele dependent on (or affected by) the inheritance of a "B" or b" allele? Why or why not? (hint: Which of Mendel's Laws is involved?) b. What is the chance that their child will be phenotypically normal (for both genes)? Show your work: c. What is the chance that their child will be genotype aabb (affected with both diseases)? Show your work: d. What is the chance that they will have a boy who has genotype AaBb (carrier for both diseases like his parents)? Show your work:
4. In Drosophila melanogaster, a yellow-bodied (dark coloured) male with vestigial (underformed) wings is crossed to a wildtype female (brown body, red eyes). The F1 generation are all wildtypes. An F1 male and female are crossed and the F2 progeny consists of 16 vestigial-winged males with yellow bodies, 48 yellow males with normal wings, 15 males with vestigial wings and yellow bodies, 49 wildtype males, 31 vestigial winged females with brown bodies and 97 wildtype females. Explain the inheritance pattern of the 2 genes.
7. In the guinea pig, one locus involved in the control of coat color may be occupied by any of four alleles: C (full color), ck (sepia), cd (cream), or c (albino), with an order of dominance of: C>ck > cd > c. In the following crosses, determine the parental genotypes and predict the phenotypic ratios that would result. a) Sepia x cream, where both guinea pigs had an albino parent b) Sepia x cream, where the sepia guinea pig had an albino parent and the cream guinea pig had two sepia parents c) Sepia x cream, where the sepia guinea pig had two full color parents and the cream guinea pig had two sepia parents d) Sepia x cream, where the sepia guinea pig had a full color parent and an albino parent and the cream guinea pig had two full color parents.