380: Problem set #3:
problems 7, 18, 20, 29, 32, 33, 34, 37, 38.
Give the genotypic ratios that may appear among the progeny of simple
crosses and the genotypes of the parents that may give rise to each ratio.
Aa × Aa
Aa × aa
AA x Aa
AA × AA
aa × aa
AA × aa
Joe has a white cat named Sam. When Joe crosses Sam with a black
cat, he obtains one-half white kittens and one-half black kittens. When the black
kittens are interbred, they produce all black kittens. On the basis of these results,
would you conclude that white or black coat color in cats is a recessive trait?
Explain your reasoning.
The black coat color is likely recessive. When Sam was crossed with a black cat,
one- half the offspring were white and one-half were black. This ratio potentially
indicates that one of the parental cats is heterozygous dominant while the other
parental cat is homozygous recessive—a testcross. The interbreeding of the
black kittens produced only black kittens, indicating that the black kittens are
likely to be homozygous, and thus the black coat color is the recessive trait.
If the black allele was dominant, we would have
the black kittens to be
heterozygous, containing a black coat color allele and a white coat color allele.
Under this condition, we would expect one-fourth of the progeny from the
interbred black kittens to have white coats. Because this did not happen, we can
conclude that the black coat color is recessive.
In humans, alkaptonuria is a metabolic disorder in which affected
persons produce black urine Alkaptonuria results from an 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.
Give the genotypes of Sally, her mother, her father, and her brother.
father, who has alkaptonuria, must be aa. Her brother, who also has alkaptonuria,
must be aa as well. Because both parents must have contributed one a allele to
her brother, Sally’s mother, who is phenotypically normal, must be heterozygous
(Aa). Sally, who is normal, received the A allele from her mother but must have
received an a allele from her father. The genotypes of the individuals are: Sally
(Aa), Sally’s mother (Aa), Sally’s father (aa), and Sally’s brother (aa).