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4. A woman has a Widow's peak (dominant), but she does not know her genotype. She marries a man who has a straight hairline and they have 13 children. Nine have widow's peaks and four have straight hairlines a. What are the genotypes of the parents? b. What are the genotypes of the children? C. Explain how you arrived at your answers. G host 5. In humans, Phenylketonuria (PKU) and Lactose intolerance are both autosomal recessive traits. If two individuals who are heterozygous for both traits marry and have children, what is the chance of having a child with PKU and Lactose intolerance? 6. If a person who is heterozygous for both PKU and Lactose intolerance marries someone who has both traits (has PKU and is lactose intolerant), what is the chance of having a child with both traits? 7. A woman who is heterozygous for astigmatism and cannot roll her tongue marries a man who has normal vision and is homozygous dominant for tongue rolling. What are the genotypes and phenotypes of their children for these traits? 8. If a carrier for Hemophilia (a sex-linked, recessive trait) marries a man without the trait: a. What is the chance that a child will be a carrier? b. What is the chance of having an afflicted son (a son with hemophilia)? c. What is the chance of having a daughter with hemophilia?

Answer 1

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4. Let the dominant be WW(widow peak) and recessive be ww(straight hair line)

a)genotype of mother-Ww

Genotype of father-ww.

b)Genotype of children :-with widow peak-Ww, straight hair line-ww.

c)If the father has a straight hair line then he will be ww.If the mother is WW(dominant),then there is no chance of children having straight hair line. As when we cross them all children will bear Ww(widow peak).But we have four children who have straight hair so that is why mother has to be heterozygous dominant i.e. Ww.

5.There is a 1/16th chance of child having PKU and Lactose intolerance.

Let Pp be gene for PKU and Ll be the gene for Lactose intolerance.

So the resulting genes in the gametes can be PL,Pl,pL,pl.

So doing a cross using punnett square only 1 individual will be there among 16 with ppll which has both PKU and Lactose intolerance.

6.There is a 1/4 chance of child having both traits.because the person who is heterozygous will have the gene PL,Pl,pL,pl in the gametes whereas the person who has both PKU and Lactose intolerance will have only gene pl in the gametes.

So by crossing them we get 4 type of offspring PpLl,Ppll,ppLl,ppll. ppll is the one with both traits.

7.Woman who can't roll her tongue is homozygous recessive. Let the genes be tt(TT for dominant). If she is heterozygous for astigmatism let it be Aa.

Her husband is dominant for tongue rolling i.e. TT. And has normal vision AA.

The children will have AATt and AaTt. That means 50% of children will be heterozygous for astigmatism and and 100% will be heterozygous for tongue rolling.

8.Let the mother be XX' and father be XY. Where X is normal X chromosome and X' chromosome has trait for haemophilia.

So gametes from the mother will be either X or X'.

From father it will be X and Y.

On crossing we will get, 4 possibilities.

XX',XX,XY,X'Y.

a)possibility of child being a carrier i.e. XX'=25%

b)chance of having hemophilic son i.e. X'Y=25%

c)chamce of javing a daughter with hemophilia ime.X'X'=0%.