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 3, Problem 31.10P
Summary Introduction
To explain: The differences between sex-linked inheritance and autosomal inheritance.
Introduction: The transmission of a particular genetic trait from one generation to another is determined by the two major types of inheritance patterns, which are sex-linked inheritance and autosomal inheritance.
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Another cross in Drosophila involved the recessive, X-linked genes yellow (y), white (w) and cut (ct). A yellow-bodied, white-eyed female with normal wings was crossed to a male whose eyes and body were normal but whose wings were cut. The F1 females were wild type for all three traits, while the F1 males expressed the yellow-body and white eyes traits. The cross was carried to an F2 progeny and only male offspring were tallied. Based on the data shown here, a genetic map was constructed. a) Diagram the genotypes of the F1 parents. b) Construct a map, assuming the white is at locus 1.5 on the X-chromosome
*******ANSWER PART B NOT PART A!!!!
Phenotype
Male offspring
y + ct
9
+ w +
6
y w ct
90
+ + +
95
+ + ct
424
y w +
376
y + +
0
+ w ct
0
In fruit flies, red eyes (pr+_) are dominant to purple eyes (prpr) and normal wings (vg+_) are dominant to vestigial wings (vgvg). The genes are located on the same chromosome. A pure-breeding red-eyed fly with vestigial wings was crossed with a pure-breeding purple-eyed fly with normal wings. All of the F1 progeny had a WT phenotype. The recombination frequency between the two genes is 15%. If an F1 individual were test crossed, what percentage of the progeny would you expect to have the WT phenotype?
In fruit flies, you are mapping three genes in a three point cross. The mutants are hairy
body (h), sepia colored eyes (se) and female sterility (g). You cross a heterozygous parent
with a homozygous recessive parent and obtain the following results:
Type
Number
h se g.
5
+ se +
450
+ se g
27
++g_
h se +
+ + +
h + g.
h + +
TOTAL
is the gene in the middle and the distance in map units between se and g is
Oh; 16.4
se; 7.1
Oh; 7.1
70
82
7
327
32
1000
se; 16.4
Chapter 3 Solutions
Introduction to Genetic Analysis
Ch. 3 - Prob. 1PCh. 3 - Prob. 2PCh. 3 - Prob. 4PCh. 3 - Prob. 5PCh. 3 - Prob. 6PCh. 3 - Prob. 7PCh. 3 - Prob. 10PCh. 3 - Prob. 11PCh. 3 - Prob. 12PCh. 3 - Prob. 13P
Ch. 3 - Prob. 14PCh. 3 - Prob. 15PCh. 3 - Prob. 16PCh. 3 - Prob. 17PCh. 3 - Prob. 18PCh. 3 - Prob. 19PCh. 3 - Prob. 20PCh. 3 - Prob. 21PCh. 3 - Prob. 22PCh. 3 - Prob. 23PCh. 3 - Prob. 24PCh. 3 - Prob. 25PCh. 3 - Prob. 26PCh. 3 - Prob. 27PCh. 3 - Prob. 28PCh. 3 - Prob. 29PCh. 3 - Prob. 30PCh. 3 - Prob. 31PCh. 3 - Prob. 31.1PCh. 3 - Prob. 31.2PCh. 3 - Prob. 31.3PCh. 3 - Prob. 31.4PCh. 3 - Prob. 31.5PCh. 3 - Prob. 31.6PCh. 3 - Prob. 31.7PCh. 3 - Prob. 31.8PCh. 3 - Prob. 31.9PCh. 3 - Prob. 31.10PCh. 3 - Prob. 31.11PCh. 3 - Prob. 31.12PCh. 3 - Prob. 31.13PCh. 3 - Prob. 31.14PCh. 3 - Prob. 31.15PCh. 3 - Prob. 32PCh. 3 - Prob. 33PCh. 3 - Prob. 34PCh. 3 - Prob. 35PCh. 3 - Prob. 36PCh. 3 - Prob. 37PCh. 3 - Prob. 38PCh. 3 - Prob. 39PCh. 3 - Prob. 40PCh. 3 - Prob. 41PCh. 3 - Prob. 42PCh. 3 - Prob. 43PCh. 3 - Prob. 44PCh. 3 - Prob. 45PCh. 3 - Prob. 46PCh. 3 - Prob. 48PCh. 3 - Prob. 49PCh. 3 - Prob. 50PCh. 3 - Prob. 51PCh. 3 - Prob. 52PCh. 3 - Prob. 53PCh. 3 - Prob. 54PCh. 3 - Prob. 57P
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