Bio 304 Genetics Exam Page 4
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424. In domestic fowl (Gallus domesticus) the gene for plumage color is sex-linked and
females are the heterogametic sex (ZW) while females are homogametic (ZZ). The dominant
allele G determines gold coloration and its recessive allele g determines silver plumage. A
cross is made between a silver male and a gold female, then the progeny are allowed to
mate, producing an F2 generation. Give the genotypes, phenotypes and proportions you
expect for among the F1 and F2 animals (1 point each):
Genotype(s)
Phenotype(s)
Percentage
F1 males
___Gg_________
___gold___________
____100%____
F1 females
___g___________
____silver
______
____100%____
F2 males
__Gg, gg
__
__gold, silver
___
__50%, 50%
__
F2 females
____G,
g_______
__gold, silver
____
__50%,
50%__
23. Consider the following human pedigree of the rare condition microphthalmia (small,
nonfunctional eyes), which occurs equally frequently in males and females within the general
population. In responding to the questions below, assume that all people marrying into the
pedigree do not carry the abnormal allele. (3 pts. each)
A
B
a) What is the mode of inheritance of this trait (e.g., linkage, dominant/recessive)?
autosomal, recessive
b) If individuals A and B have a child, what is the probability that the child will be
microphthalmic?
probability that A is heterozygous = ½; probability that B is heterozygous = ½
If both are heterozygous, probability that child is homozygous recessive = ¼
Overall: ½ X ½ X ¼ = 1/16
c) If the first child of A × B is normal, what is the probability that their second child will be
microphthalmic?
Same as b, 1/16
d) If the first child of A × B has the disease, what is the probability that their second child
will be microphthalmic?
Now you know that A and B are heterozygous;
¼
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