chapter 19
- a: AB
b: AB and aB
c: Ab and ab
d: AB, aB, Ab, and ab
- a: AaBB will occur in all the offspring.
b: 25% AABB; 25% AaBB; 25% AABb; 25% AaBb
c: 25% AaBb; 25% Aabb; 25% aaBb; 25% aabb
d: 1/16 AABB (6.25%)
1/8 AaBB (12.5%)
1/16 aaBB (6.25%)
1/8 AABb (12.5%)
1/4 AaBb (25%)
1/8 aaBb (12.5%)
1/16 AAbb (6.25%)
1/8 Aabb (12.5%)
1/16 aabb (6.25%)
- a: ABC
b: ABc, aBc
c: ABC, ABc, aBC, aBc
d: ABC, ABc, AbC, Abc, aBC, aBc, abC, abc
- a: Both parents must be heterozygous for the trait,
having one dominant allele and one recessive
allele.
b: Both parents are homozygotes for the albinism
trait—both have two copies of the recessive form of
the allele.
c: The parent with typical pigmentation is heterozy-
gous for the albinism allele. The probability that
any one child will have the albinism phenotype
is 50 percent. However, with a small sample of
only four offspring, there is a high probability of a
deviation from a 1:1 ratio due to the random mixes
of alleles that occur during meiosis and fertilization
(discussed in Chapter 18).- Because Molly does not exhibit the recessive hip
disorder, she must be either homozygous dominant
(HH) for this trait, or heterozygous (Hh). If the father
is homozygous dominant (HH), then he and Molly
cannot pro duce offspring that are homozygous
recessive (hh), and so none of their offspring will have
the undesirable phenotype. How ever, if Molly is a
heterozygote for the trait, notice that the probability
is 1/2 (50 percent) that a puppy will be heterozygous
(Hh) and so carry the trait. - a: The mother must be heterozygous (IAi). The man
having type B blood could have fathered the child
if he were also heterozygous (IBi).
b: If the man is heterozygous, then he could be the
father. However, because any other type B het-
erozygous male could also be the father, one
cannot say that this particular man absolutely
must be. Actually, any male who could contribute
an O allele (i) could have fathered the child. This
would include males with type O blood (ii) or
type A blood who are heterozygous. - The probability is 1/2 (50 percent) that a child of this
couple will be a heterozygote and have sickle-cell
trait. The probability is 1/4 (25 percent) that a child
will be homozygous for the sickling allele and so will
have sickle-cell anemia. - For these ten traits, all the man’s sperm will carry
identical genes. He cannot produce genotypically
different sperm. The woman can produce eggs with
four genotypes. This example underscores the fact that
the more heterozygous gene pairs that are present, the
more genetically different gametes are possible.
Answers to Critical Thinking Genetics Problems
Appendix V
A-12 Appendix V
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