Essentials Of Genetics (9th Global Edition)
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 24PDQ

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.

Draw all possible conclusions concerning the mode of inheritance of the trait expressed in each of the following limited pedigrees. (Each case is based on a different trait.)

Chapter 3, Problem 24PDQ, 

When working out genetics problems in this and succeeding chapters, always assume that members of

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Unpacking the Problem 44John and Martha are contemplating having children, butJohn’s brother has galactosemia (an autosomal recessivedisease) and Martha’s great-grandmother also had galactosemia. Martha has a sister who has three children, none ofwhom have galactosemia. What is the probability that Johnand Martha’s first child will have galactosemia?1. Can the problem be restated as a pedigree? If so, writeone.2. Can parts of the problem be restated by using Punnettsquares?3. Can parts of the problem be restated by using branchdiagrams?4. In the pedigree, identify a mating that illustratesMendel’s first law.5. Define all the scientific terms in the problem, and lookup any other terms about which you are uncertain.6. What assumptions need to be made in answering thisproblem?7. Which unmentioned family members must be considered? Why?8. What statistical rules might be relevant, and in whatsituations can they be applied? Do such situations existin this problem?9. What are two generalities…
Instruction: Solve for the genetic problems. a. Genotype of the parentsb. Phenotype of the parentsc. Punnett squared. Genotypic ratioe. Phenotypic ratio  Question: Two individuals with widow’s peaks want to have a child with a continuous hairline. Is this possible?
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.

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Essentials Of Genetics (9th Global Edition)

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