PCB 3063: Study Guide

Topics: Allele, Dominance, Zygosity Pages: 7 (1647 words) Published: November 17, 2012
PCB 3063
Spring 2012

Problem Set 1


1. Determine the types of gametes produced by each of the following individuals: a. Aa1/2 A, 1/2 a
b. AaBb1/4 AB, 1/4 Ab, 1/4 aB, 1/4 ab
c. AABb1/2 AB, 1/ Ab
d. AaBBCc1/4 ABC, 1/4 aBC, 1/4 ABc, 1/4 aBc

2. Use the Punnett square to determine the genotypes in the progeny of each of the
following crosses:
a. Dd x Dd
b. AaBB x AaBB
c. CcEE x CCEe
Notice: in every case, each parent produces only two types of gametes.


3. In guinea pigs, rough coat (R) is dominant over smooth coat (r). A rough coated guinea pig is bred to a smooth one, giving eight rough and seven smooth progeny in the F1 generation. a. What are the genotypes of the parents and their offspring? The recessive trait is observed in the progeny, so the rough-coated parent must be heterozygous. P: Rr (rough) x rr (smooth)

F1: 1/2 Rr, 1/2 rr
b. If one of the rough F1 animals is mated to its rough parent, what progeny would you expect? This would be a monohybrid cross: Rr x Rr => 1/4 RR, 1/2 Rr, 1/4 rr. 4. In maize, a dominant allele A is necessary for seed color, as opposed to colorless (a). Another gene has a recessive allele w that results in waxy starch, as opposed to normal starch (W). The two genes segregate independently. What are the phenotypes and relative frequencies of offspring from each of the following crosses? Notice: The question specifies phenotypic ratios.

a. AaWw x AaWw
This is a dihybrid cross:
9/16 A_W_ (normal)
3/16 A_ww (waxy)
3/16 aaW_ (colorless)
1/16 aaww (waxy, colorless)
b. AaWW x AaWW
This works like a monohybrid cross because both parents are homozygotic for WW. 3/4 A_WW (normal), 1/4 aaWW (colorless)

5. In humans, alkaptonuria is a metabolic disorder in which affected persons produce black urine. Alkaptonuria results from an autosomal allele (a) that is recessive to the allele for normal metabolism (A). Sally has normal metabolism, but her brother has alkaptonuria. Sally’s father has alkaptonuria, and her mother has normal metabolism. a. Construct a pedigree of this family and indicate the genotypes of Sally, her mother, her father, and her brother. Sally’s mother must be a carrier. Sally is also a carrier because she received one alkaptonuria allele from her father. [pic]

Caution: this is not X-linked inheritance.
b. If Sally’s parents have another child, what is the probability that this child will have alkaptonuria? aa (father) x Aa (mother)
1/2 Aa (normal metabolism)
1/2 aa (alkaptonuria)

6. Both John and Cathy have normal vision. After 10 years of marriage to John, Cathy gave birth to a color-blind daughter (color blindness is an X-linked recessive trait). John filed for divorce, claiming that he is not the father of the child. Is John justified in his claim of non-paternity? Explain your answer. Give the genotypes of John, Cathy and the child. Since color blindness is an X-linked recessive trait, the color-blind daughter must be homozygous for the color blindness allele. That means that she inherited a color blindness allele from each parent. John can not be the father, because he has no color blindness alleles (he has normal vision, so he is hemizygous for the normal vision allele). Cathy is a carrier. She is also a big cheater! John: CY

Cathy: Cc
Daughter: cc
Whoever the father of the girl is must be cY and color blind.


7. If the pedigree above illustrates an autosomal dominant trait, then individual I-1: (Note: the “carrier” symbol was not used in the above pedigree.) a. must be homozygous dominant
b. must be heterozygous
c. must be homozygous recessive
d. could be either homozygous dominant or heterozygous
e. could be either homozygous recessive or heterozygous
This is a dominant trait and...
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