Study Guide

Topic: FamilyChildren
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Last updated: May 17, 2019

PCB 3063 Spring 2012 Problem Set 1 ANSWERS 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.

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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. [pic] 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.

[pic] 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 I-1 is affected, but he had unaffected children. Therefore he must be heterozygous. 8. The following pedigree is for an X-linked trait: (Note: the “carrier” symbol was not used in this pedigree. ) [pic] a. Is this trait dominant or recessive? Recessive: unaffected females III-4 and IV-1have affected sons.

b. Indicate the genotypes of all the individuals in the pedigree. Affected males are hemizygous for the recessive allele • Affected females are homozygous for the recessive allele • Unaffected males are hemizygous for the dominant allele • Heterozygotes are carrier females Pedigree with the carrier symbols: [pic] Notice: the exact genotype of individual IV-4 is uncertain.9. The following pedigree illustrates the inheritance of Nance-Horan syndrome, a rare X-linked genetic condition in which affected persons have cataracts and abnormally shaped teeth. (Note: the “carrier” symbol was not used in this pedigree. [pic] If III-2 and III-7 mated, what would be the expected genotypic and phenotypic ratios in their progeny? Draw a Punnett square.

The trait is recessive because some unaffected mothers have affected sons (and males get their X from their mothers). III-2 is an affected male, so he must be hemizygous for the condition. III-7 is a carrier female because her father is affected. [pic] 10.

If traits R1 and R2 exhibit incomplete dominance over each other, what will be the phenotypic ratio in the progeny of the cross R1R1 x R1R2 ? :1 (individual R1R1 produces only one type of gamete) 1/2 R1R1, 1/2 R1R2 11. In shorthorn cattle, coat color may be red, white, or roan. Roan is an intermediate phenotype. The following data were obtained from various crosses: red x red-> all red white x white-> all white red x white-> all roan roan x roan-> 1/4 red: 1/2 roan: 1/4 white a. How is coat color inherited? Incomplete dominance: heterozygotes have an intermediate color (roan) b. What are the genotypes of parents and offspring in each cross? RR x RR => all RR rr x rr => all rrRR x rr => all Rr Rr x Rr => 1/4 RR, 2/4 Rr, 1/4 rr [pic] 12. Bar is an X-linked mutation in Drosophila that exhibits incomplete dominance. Flies that are homozygous for Bar have bar-shaped eyes.

Heterozygous flies have bean-shaped eyes (an intermediate phenotype). a. What will be the outcome of a cross between a normal female and a bar-eyed male? Normal females must be homozygous for the normal allele because Bar is an incomplete dominant. Bar-eyed males are hemyzygous Bar. [pic] b. Use the Punnett square to determine the genotypes and phenotypes of the F2 generation [pic]Note: always indicate sex along with the phenotypes for X-linked inheritance.

13. If traits LM and LN exhibit codominance relative to each other, what will be the phenotypic ratio in the progeny of the cross LMLN x LMLN ? 1:2:1 (monohybrid ratio): 1/4 LMLM (M), 2/4 LMLN(MN), 1/4 LNLN (N) 14. In a rare species of frog, red color (Y) is dominant over yellow color (y, a null allele). The character is autosomal. The cross YY x Yy produced over 100 frogs: 96% red and 4% yellow.

Which complication of Mendelian genetics can explain this outcome? a. recessive lethality of the y allele . codominance c. incomplete dominance d. incomplete penetrance e. variable expressivity All the frogs in F1 are Y_ so they should all be red, but a small percentage is not. There are no yy individuals, so recessive lethality can not explain this. It is not incomplete dominance or codominance, because those would yield 1:1 ratios.

It is not variable expressivity because yellow is a recessive phenotype, not a variation of the dominant phenotype. 15. In foxes, two alleles of a single gene, P and p, may result in lethality (PP), platinum coat (Pp), or silver coat (pp).Notice: this is an autosomal trait because nothing is indicated otherwise. a. Is the P allele behaving dominantly or recessively in causing lethality? Recessive: heterozygotes survive. b.

Is the P allele behaving dominantly or recessively in causing platinum coat color? Dominant: heterozygotes are platinum, while the pp homozygotes are silver (two p alleles are necessary for the silver coat, therefore silver is recessive). c. What ratios are obtained when platinum foxes are interbred? Pp x Pp => 1/4 PP (dead), 1/2 Pp (platinum), 1/4 pp (silver) Apparent ratio: 2/3 platinum: 1/3 silver [pic] 16.

In a rare species of duck there is an X-linked allele (E1) that results in animals with only one eye (pictured above). The normal phenotype results from the wild-type allele (E), which is also necessary for survival. Use the Punnet square to determine the genotypic and phenotypic ratios in the progeny of a cross between a one-eyed female and a normal male. Since E is necessary for survival, the E’ allele is recessive in causing lethality (because homozygous E’E’ would die) but dominant in causing the one-eye condition (since only heterozygotes would be able to carry the E’ allele, and they would be one-eyed).Notice: the one-eyed male (pictured above) cannot occur! [pic] 17. The g allele in the Chupacabra is X-linked recessive lethal. Heterozygous individuals have gray hair instead of the normal black hair. Use the Punnett square to determine the outcome of a cross between a normal male and a gray female.

I made this up! The gray female must be heterozygous because she needs at least one copy of the normal allele to survive. Since she’s also gray, the g allele is recessive in causing lethality but dominant in causing color. [pic] Important: always indicate sex along with the phenotypes for X-linked inheritance.


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