Heredity ❮ 113
❯ Answers and Explanations
- A—The crossover frequencies are an indication
of the distance between the different genes on a
chromosome. The farther apart they are, the
greater chance there is that they will cross over
during prophase I of meiosis. You are first told
that A and B cross over with a frequency of 35
percent, so imagine that they are 35 units apart
on a chromosome map.
A (35) B B (15) C A (20) C
We can then tell you that B and C have a frequency
of 15 percent. They are 15 units apart on the map,
recessive phenotype, you know the individual
has the genotype Gg.
- D—The Punnett square shown below shows all
the possible gamete combinations from this cross.
Two-sixteenths or one-eighth of the possible
gametes will be AaBbCc. A quick way to deter-
mine the number of possible gametes that an
individual can produce given a certain genotype is
to use the formula 2n. For example, an individual
who is AABbCc can have 2^2 =4 possible gametes
because Bb and Cc are heterozygous.
- A—Person A must have genotype Bb because he
has some children that have the recessive condi-
tion and some that do not. Because his wife is pure
recessive, she can contribute only a b. The father
must therefore be the one who contributes the B
to the child who does not have the condition,
and the second b to the one with the condition.
- B—Person B most likely has a genotype of BB.
Because he does not have the condition, we
know that his genotype is either BB or Bb. If it
were Bb, then when crossed with his wife who
has a genotype of bb, 50 percent of his children
would be expected to have the recessive condi-
tion. None of the children have the condition,
which leads you to believe that he is most likely
BB. (This test is, of course, not 100 percent
accurate. Answer choice B is not certain, but is
the most probable conclusion.)
- C—We know that neither parent in the question
has the recessive condition. We therefore need to
calculate the probability that each of them is Bb.
The probability that person C is Bb is 1. Because
his mother has the condition, she mustpass a b to
him during gamete formation. So the only possible
genotypes he can have are Bb and bb. Since he does
not have the condition, he must be Bb with a prob-
ability of 1. The probability that person D is Bb is
0.67. Neither of her parents has the condition, but
but you cannot yet be sure what side of gene A
that C is on. Gene A and C cross with 20 percent
frequency. This means that gene C must be in
between A and B.
A (20) C (15) B A (10) D D (25) B
Gene A crosses over with D 10 percent of the
time, and D crosses with B 25 percent of the
time; therefore, D must also be in between A
and B. It is closer to A than it is to B. You can
use this knowledge to eliminate answer choices
B and C.
A(10)D(10)C (15) B
Gene A crosses over with E with a frequency of 5
percent. You do not know which side of A gene
E is on until you know its crossover frequency
with B. Because the question tells you that it has
a 40 percent frequency with B, you know that it
must be on the leftof A. This completes your
map, leaving A as the correct answer.
- C—This is a test cross. To determine the geno-
type of an individual showing the dominant
phenotype, you cross that individual with a
homozygous recessive individual for the same
trait. If they have no offspring with the recessive
phenotype, then the individual displaying the
dominant phenotype is most likely GG. If
approximately one-half of the offspring have the
ABC AbC ABc Abc aBC abC aBc abc
ABC AABBCC AABbCC AABBCc AABbCc AaBBCC AaBbCC AaBBCc AaBbCc
AbC AABbCC AAbbCC AABbCc AAbbCc AaBbCC AabbCC AaBbCc AabbCc
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