114 ❯ STEP 4. Review the Knowledge You Need to Score High
she has a brother who is bb. This means that each
of her parents must be a carrier for the condition
(Bb). You know that this woman is not bb, because
she does not have the condition. As a result, there
are only three possible genotypes from the cross
remaining. Two of these three are Bb, giving her a
probability of 2 ⁄ 3, or 0.67, of being Bb. The prob-
ability that bothperson C and person D are Bb is
(1)×(0.67)=(0.67). Now it is necessary to calcu-
late the probability that two Bb parents will pro-
duce a kid who is bb. The Punnett square says that
there is a 0.25 chance of this result. To calculate the
probability that they will have a child with the
recessive condition, you multiply the probability
that they are both Bb (0.67) times the probability
that two individuals Bb will produce a bb child
(0.25). Thus, the probability of an affected child
being produced from these two parents is 1 ⁄ 6.- C—Hemophilia is an X-linked condition. An
XY male with hemophilia gets his Y chromo-
some from his father, and his X chromosome
from his mother. All that is needed for the
hemophilia condition to occur is a copy of the
defective recessive allele from his mother. - D—Types O and A would prove that he was not
the father of this particular child. If the mother
has type O blood, this means that her genotype
is ii and she mustpass along an i allele to her
child. The baby has type B blood, and her geno-
type could be IBi or IBIB. Since the mother must
give an i, then the baby’s genotype must be IBi.
It follows that the father must provide the IB
allele to the baby to complete the known geno-
type. If he is type O, he won’t have an IBto pass
along since his genotype would be ii. This would
also be the case if he were type A, because his
genotype would be either IAIAor IAi. Therefore,
those two blood types would prove that he is not
the father of this child. - C—To figure out this problem, you need to
know the genotype of the mother. The father is
black, meaning that his genotype is gg. The two
of them produced a squirrel that is also black,
which means that the gray mother gave a g to
the baby. The mother’s genotype is Gg. A cross
of Gg ×gg produces a phenotype ratio of 1:1
gray:black. They have a 0.5 chance of producing
another black baby.
10. B—According to this scenario, yellow and white
are the only colors possible. If white were domi-
nant, and both parents were Ww, you couldpro-
duce a yellow offspring if the two recessive w’s
combined. If it were intermediate inheritance,
you probably would not produce a straight yellow
tulip in the offspring because they would either
meet halfway (incomplete dominance), or both
express fully (codominance). If yellow were dom-
inant, then you could produce a yellow offspring
only if there were a Y allele in one of the parents.
A cross of yy ×yy would produce only white
tulips if white were recessive.
11. A—This problem involves incomplete domi-
nance. The genotype of the pink offspring from
the first generation is RW. When the two RW
snapdragons are mated together, they produce
the following results:
The offspring will be 25 percent red (RR), 50 percent
pink (RW), and 25 percent white (WW).- C—In a problem like this, you will save time by
thinking about the laws of probability. The geno-
type is RrBBCcDDEe. How many possible com-
binations of the R gene are there? There are two:
R and r. How many for B? Only one: B. Following
the same logic, C has two, D has one, and E has
two as well. Now you multiply the possibilities:
(2× 1 × 2 × 1 ×2)=8. There are 8 possible
gametes from this genotype. Another way to arrive
at this answer is by use of the expression 2n, where
nis the number of hybrid traits being examined. In
this case it would be 2^3 or 8 possible gametes. - C—Down syndrome is most often due to a tri-
somy of chromosome 21. Klinefelter syndrome
is a trisomy of the sex chromosomes (XXY).
Patau syndrome is a trisomy of chromosome 15.
Edwards syndrome is a trisomy of chromosome
18. Turner syndrome, the only nontrisomy
listed in this problem, is a monosomyof the sex
chromosomes (XO).