Essentials Of Genetics (9th Global Edition)
9th Edition
ISBN: 9780134143637
Author: William S. Klug, Michael R. Cummings
Publisher: Pearson Global Edition
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Chapter 3, Problem 7PDQ
When working out genetics problems in this and succeeding chapters, alwaysassumethatmembersof the P1 generation are homozygous, unless the information or data you are given require you to do otherwise.
Mendel crossed peas having round seeds and yellow cotyledons with peas having wrinkled seeds and green cotyledons. All the F1 plants had round seeds with yellow cotyledons. Diagram this cross through the F2 generation, using both the Punnett square and forked-line methods.
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When working out genetics problems in this and succeeding chapters, always assume that members of the P1 generation are homozygous, unless the information or data you are given require you to do otherwise.HOW DO WE KNOW? we focused on the Mendelian postulates, probability, and pedigree analysis. We also considered some of the methods and reasoning by which these ideas, concepts, and techniques were developed. On the basis of these discussions,
what answers would you propose to the followingquestion.
Question: Since experimental crosses are not performed in humans, how do we know how traits are inherited?
Question:
Suppose the progeny were crossed. Fill in the Punnett Squares below and answer the following questions.
Seed Color
Seed Shape
R
r
A
a
In a pea plant experiment, the allele (R) producing yellow seed color is dominant, and the allele (r) producing green seed color is recessive. A homozygous recessive genotype would produce green seeds. Similarly, the allele (A) producing round seed shape is dominant to the allele (a) that produces wrinkled seed shape. A homozygous recessive genotype would produce wrinkled seeds. Both traits sort independently of each other.
Given a pure-breeding cross between a round, green pea plant and a wrinkled yellow plant, complete the rest of the Punnett Squares below, and answer the following questions.
R
R
A
A
RA
Ra
Ra
ra
Question: This is a normal 3 point test cross, except that instead of regular phenotypes, you are looking at DNA markers on a gel. One parent, according to the gel, is heterozygous at each marker. The other parent is homozygous for each marker. (Again, this means it is a test cross: AaEeHh x AAEEHH --but don't be confused by that, because these are not "dominant" and "recessive" per se; the phenotype is just a band on a gel). For each offspring, figure out its genotype (homozygous or heterozygous for each gene. Then, figure that one parent made only AEH gametes, so you can cross that out if it helps.) Then treat it pretty much the same as a 3 point test cross.
Chapter 3 Solutions
Essentials Of Genetics (9th Global Edition)
Ch. 3 -
CASE STUDY | To test or not to test
Thomas...Ch. 3 -
CASE STUDY | To test or not to test
Thomas...Ch. 3 - CASE STUDY | To test or not to test Thomas first...Ch. 3 -
CASE STUDY | To test or not to test
Thomas...Ch. 3 - When working out genetics problems in this and...Ch. 3 - When working out genetics problems in this and...Ch. 3 - When working out genetics problems in this and...Ch. 3 -
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When working out genetics problems in this and...
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