Introduction to Genetic Analysis
11th Edition
ISBN: 9781464109485
Author: Anthony J.F. Griffiths, Susan R. Wessler, Sean B. Carroll, John Doebley
Publisher: W. H. Freeman
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Chapter 4, Problem 28P
Summary Introduction
To determine: The use of chi square value to determine the presence of linked genes.
Introduction. The chi square value x2 is calculated as the summation of observed values subtracted by the expected values and dividing the resultant by the number of expected values.
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From parents of genotypes A/A ⋅ B/B and a/a ⋅ b/b, adihybrid was produced. In a testcross of the dihybrid,the following seven progeny were obtained:A/a ⋅ B/b, a/a ⋅ b/b, A/a ⋅ B/b, A/a ⋅ b/b,a/a ⋅ b/b, A/a ⋅ B/b, and a/a ⋅ B/bDo these results provide convincing evidence of linkage?
Consider the following dihybrid testcross: B/b • E/e × b/b • e/e
For the progeny from this testcross, determine the relative proportions (from 0% to 100%) of each genotype if the two genes:
a) are linked (dominant alleles in cis conformation) with no crossing over: Be/be:
be/be:
BE/be:
bE/be:
b) assort independently.
B/b; E/e:
B/b; e/e:
b/b; E/e:
b/b; e/e:
c) are linked (dominant alleles in cis conformation) and 20 map units apart. Be/be:
be/be:
BE/be:
bE/be:
What will be the PR and GR of the following crosses? Give also their phenotypes. (Designate your own phenotype for each character)
AaBb x aabb
AAbb x AaBB
AaBb x aaBB
AABb x aabb
Chapter 4 Solutions
Introduction to Genetic Analysis
Ch. 4 - Prob. 1PCh. 4 - Prob. 5PCh. 4 - Prob. 12PCh. 4 - Prob. 13PCh. 4 - Prob. 14PCh. 4 - Prob. 15PCh. 4 - Prob. 16PCh. 4 - Prob. 17PCh. 4 - Prob. 18PCh. 4 - Prob. 19P
Ch. 4 - Prob. 20PCh. 4 - Prob. 21PCh. 4 - Prob. 21.1PCh. 4 - Prob. 21.2PCh. 4 - Prob. 21.3PCh. 4 - Prob. 21.4PCh. 4 - Prob. 21.5PCh. 4 - Prob. 21.6PCh. 4 - Prob. 21.7PCh. 4 - Prob. 21.8PCh. 4 - Prob. 21.9PCh. 4 - Prob. 21.10PCh. 4 - Prob. 21.11PCh. 4 - Prob. 21.12PCh. 4 - Prob. 21.13PCh. 4 - Prob. 21.14PCh. 4 - Prob. 21.15PCh. 4 - Prob. 21.16PCh. 4 - Prob. 21.17PCh. 4 - Prob. 21.18PCh. 4 - Prob. 21.19PCh. 4 - Prob. 21.20PCh. 4 - Prob. 21.21PCh. 4 - Prob. 21.22PCh. 4 - Prob. 21.23PCh. 4 - Prob. 21.24PCh. 4 - Prob. 21.25PCh. 4 - Prob. 21.26PCh. 4 - Prob. 22PCh. 4 - Prob. 23PCh. 4 - Prob. 24PCh. 4 - Prob. 25PCh. 4 - Prob. 26PCh. 4 - Prob. 27PCh. 4 - Prob. 28PCh. 4 - Prob. 29PCh. 4 - Prob. 30PCh. 4 - Prob. 31PCh. 4 - Prob. 32PCh. 4 - Prob. 33PCh. 4 - Prob. 34PCh. 4 - Prob. 35PCh. 4 - Prob. 36PCh. 4 - Prob. 37PCh. 4 - Prob. 38PCh. 4 - Prob. 38.1PCh. 4 - Prob. 38.2PCh. 4 - Prob. 38.3PCh. 4 - Prob. 38.4PCh. 4 - Prob. 38.5PCh. 4 - Prob. 38.6PCh. 4 - Prob. 38.7PCh. 4 - Prob. 38.8PCh. 4 - Prob. 38.9PCh. 4 - Prob. 38.10PCh. 4 - Prob. 38.11PCh. 4 - Prob. 38.12PCh. 4 - Prob. 38.13PCh. 4 - Prob. 38.14PCh. 4 - Prob. 38.15PCh. 4 - Prob. 38.16PCh. 4 - Prob. 38.17PCh. 4 - Prob. 38.18PCh. 4 - Prob. 38.19PCh. 4 - Prob. 38.20PCh. 4 - Prob. 38.21PCh. 4 - Prob. 38.22PCh. 4 - Prob. 38.23PCh. 4 - Prob. 38.24PCh. 4 - Prob. 39PCh. 4 - Prob. 40PCh. 4 - Prob. 41PCh. 4 - Prob. 42PCh. 4 - Prob. 43PCh. 4 - Prob. 44PCh. 4 - Prob. 45PCh. 4 - Prob. 46PCh. 4 - Prob. 47PCh. 4 - Prob. 48PCh. 4 - Prob. 49PCh. 4 - Prob. 50PCh. 4 - Prob. 51PCh. 4 - Prob. 52PCh. 4 - Prob. 53PCh. 4 - Prob. 54PCh. 4 - Prob. 55PCh. 4 - Prob. 56PCh. 4 - Prob. 57PCh. 4 - Prob. 58PCh. 4 - Prob. 59PCh. 4 - Prob. 60PCh. 4 - Prob. 62PCh. 4 - Prob. 63PCh. 4 - Prob. 64PCh. 4 - Prob. 65PCh. 4 - Prob. 66PCh. 4 - Prob. 67PCh. 4 - Prob. 68PCh. 4 - Prob. 69P
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- The trihybrid e+f+g+/efg is test crossed to the triple recessive efg/efg and the following genotypes were obtained in the progeny: efg = 2; efg = 28; e'fg* = 9; e*rg 32; efg 73; e*fg* = 10; ef g* =10; e'fg 2; %3D "The odd gene is The gene order is The gene distance between e-f is The gene distance between f-g is The gene distance between e-g isarrow_forwardWhat are the expected phenotypic ratios from the following cross:Tt Rr yy Aa × Tt rr YY Aa, where T = tall, t = dwarf, R = round,r = wrinkled, Y = yellow, y = green, A = axial, a = terminal; T, R,Y, and A are dominant alleles. Note: Consider using the multiplication method in answering this problemarrow_forwardThe pedigree below shows the phenotypes of the ABO blood groups and Rhesus factors [positive (+) and negative (-)] for several members of a family. I (B+ AB- 1 2 3 4 II O- A+ В- B- AB+ A+ 1 2 4 5 6 a. What are the ABO blood group genotypes of individuals I-1 and I-2? b. Which child/ren of individual I-4 can donate blood to him? c. Which individual in the pedigree can donate blood to all the other individuals in the pedigree?arrow_forward
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