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A
B
A
B
a
b
a
b
A
B
a
b
A
B
a
b
A
B
a
b
A
B
a
b
Parentals
(one per gamete)
Crossover
Synapsis
A
Meiosis I
Meiosis II
A
B
A
B
a
b
a
b
A
B
A
b
a
B
a
b
A
B
A
b
a
B
a
b
Parentals
(one per gamete)
Crossover
Synapsis
B
Meiosis I
Meiosis II
Recombinants
(one per gamete)
Linkage
High
Intermediate
Low
cM
Recombination Fraction (
θ
)
Recombination %
LOD Score
0.1 cM
1 cM
50 cM
0.001
θ
0.1%
1 gamete per 1000 gametes will contain a
recombinant chromosome
≥
+ 3
≤
–2
0.01
θ
1.0%
1 gamete per 100 gametes will contain a
recombinant chromosome
0.50
θ
50%
50 gametes per 100 gametes will contain a
recombinant chromosome
C
FIGURE 8-2. Linkage. (A) Absolute linkage (0cM, 0.00
, 0%, LOD score 3). During meiosis I in gamete formation , two
homologous chromosomes undergo synapsis (pairing of homologous chromosomes) and crossover. In this case, the loci
are so close together that a crossover point will never occur between the alleles (A, B, a, b). Therefore, only parental chro-
mosomes (parentals) will form. During meiosis II, the gametes that are formed will have the same allele pattern under con-
sideration as the parents. (B) Low linkage (50cM, 0.50
, 50%, LOD score 2). During meiosis I in gamete formation, two
homologous chromosomes undergo synapsis (pairing of homologous chromosomes) and crossover. In this case, the loci
are very far apart so that a crossover point will occur between the alleles (A, B, a ,b). Therefore, both parental chromo-
somes (parentals) and recombinant chromosomes (recombinants) will form. During meiosis II, the 50% of the gametes
that are formed will have the same allele pattern under consideration as the parents and 50% of the gametes that are
formed will have different allele pattern under consideration than the parents. (C) Measurements of linkage. Examples of
high, intermediate, and low linkage are shown.
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1.
In an autosomal recessive disease, the
people represented by the “2pq” figure in the
equation “p
2
2pq q
2
” are:
(A)
carriers of the disease
(B)
affected by the disease
(C)
those who have a new mutation
(D)
those without the mutant gene
2.
Duchenne muscular dystrophy is a lethal
X-linked recessive disease, which affects 1 in
3,500 boys. What is the carrier frequency of
this gene mutation in females?
(A)
1/3,500
(B)
1/1,750
(C)
1/59
(D)
3/50
(E)
1/25
3.
A study was undertaken of pregnant
women in North Carolina to determine the
ratios of women with sickle cell disease and
sickle cell trait. The study found that 60% of
the African American women in North
Carolina are homozygous for the normal
dominant hemoglobin allele. What percent-
age of the women would be carriers of sickle
cell trait?
(A)
5%
(B)
25%
(C)
35%
(D)
40%
4.
Huntington disease is an autosomal dom-
inant disease in which those who are
homozygous for the disease gene have a
clinical course that is no different from that
of heterozygotes. In a population where the
frequency of the Huntington gene is 0.08,
what is the frequency of those who are
homozygous for the gene?
(A)
0.064
(B)
0.016
(C)
0.08
(D)
0.0064
5.
In achondroplastic dwarfism, an autoso-
mal dominant disease, the gene is lethal in
homozygotes. If the frequency of the normal
allele is 0.99, what is the heterozygote
frequency?
(A)
0.98
(B)
0.01
(C)
0.02
(D)
0.99
6.
In South Africa, variegate porphyria is
found in white South Africans at a higher fre-
quency than would be expected if the popu-
lation was in Hardy-Weinberg equilibrium.
This population originated from a small
group of Dutch settlers. The most likely
explanation for the high frequency of varie-
gate porphyria in this population is:
(A)
selection for heterozygotes
(B)
the founder effect
(C)
immigration into the population
(D)
selection against heterozygotes
7.
It is believed that the cystic fibrosis (CF)
gene conferred protection to carriers during
the cholera epidemics of the Middle Ages.
The CF gene frequency would be expected to
do what in those populations?
(A)
increase
(B)
decrease
(C)
stay the same
(D)
become sex-linked recessive
8.
Early in the 20th century, eugenics propo-
nents thought that if those with genetic dis-
eases were prevented from having children,
then the diseases would disappear. They
were erroneous in their reasoning, however.
Why wouldn’t this strategy work for autoso-
mal recessive diseases?
(A)
because only females are carriers
(B)
because only males are carriers
(C)
because the genes would be “protected”
in carriers
(D)
because 1/4 of all children would be nor-
mal
9.
Hemochromatosis is an autosomal reces-
sive disease that is relatively common in the
population (1 in 500). The disease can be
treated successfully by periodic removal of
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BRS Genetics
blood (serial phlebotomy) but failure to rec-
ognize it can lead to a number of serious
conditions. Population screening for carriers
is being considered. What is the expected
carrier frequency in the population?
(A)
0.10%
(B)
0.20%
(C)
2.25%
(D)
8.5%
(E)
15%
10.
The recombinant % of two loci is 66%.
Which one of the following is the explana-
tion for this figure?
(A)
The two loci are far apart and there is low
linkage between them.
(B)
The two loci are close together and there
is high linkage between them.
(C)
The recombinant fraction between the
loci is low.
(D)
The LOD score is high.
11.
A couple has had four children, two boys
and two girls. The father is a carrier of a peri-
centric inversion of chromosome 8. Two of
the couple’s children were born with recom-
binant chromosomes and died shortly after
birth. What is the recombinant fraction for
the inversion?
(A)
0.10
(B)
0.25
(C)
0.50
(D)
0.75
12.
Which one of the following violates the
assumptions upon which the Hardy-
Weinberg Law is based?
(A)
There is no selection against any allele in
the population.
(B)
There is a constant mutation rate where
lethal genes are replaced by new muta-
tions.
(C)
There is a large population with assorta-
tive mating.
(D)
There is no migration into the popula-
tion.
13.
An autosomal recessive gene is lethal in
homozygotes. In a population at Hardy-
Weinberg equilibrium, how many genera-
tions will it take to eliminate the gene from
the population
(A)
1
(B)
25
(C)
100
(D)
It will not be eliminated from the
population.
14.
In West Africa, the sickle cell gene (S) fre-
quency is 0.15, while in the United States the
frequency of (S) is about 0.04. The most
likely explanation for this is which one of the
following?
(A)
selection for homozygotes
(B)
selection against homozygotes
(C)
selection for heterozygotes
(D)
selection against heterozygotes
15.
If a dominant gene is lethal (fitness
0),
why do you continue to see it in a popula-
tion?
(A)
some individuals are carriers
(B)
it is “protected” in homozygotes
(C)
it is “protected” in heterozygotes
(D)
it continues to arise as a new mutation
16.
The frequency of the autosomal recessive
disease phenylketonuria is 1 in 10,000. What
is the carrier frequency for this disease?
(A)
1/50
(B)
1/100
(C)
1/500
(D)
1/1,000
17.
Which one of the following best
describes the gene frequency for an X-linked
recessive disease?
(A)
it is equal to 1
(B)
it is equal to 2pq – 1
(C)
it is equal to its frequency in female
carriers
(D)
it is equal to the disease frequency
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1. The answer is (A).
In an autosomal recessive disease, those individuals who are repre-
sented by “2pq” are heterozygotes who carry a copy of a mutated disease-causing gene.
2. The answer is (B).
The gene frequency in X-linked recessive diseases is the same as the dis-
ease frequency. In this case, the gene frequency is 1/3,500. The carrier frequency is 2pq,
since p is almost equal to 1 the carrier frequency is 2
1 1/3,500 2/3,500 or 1/1,750.
3. The answer is (C).
Because 60% of the women are homozygous for the normal dominant
hemoglobin allele, that means that p
2
0.60 and the square root, or “p” is equal to 0.77.
That means that 1 – p, or “q”, the frequency of sickle cell trait, is equal to 0.23. The percent-
age of heterozygotes, or carriers of the sickle cell trait, is thus 2pq, or 2
0.77 0.23, which
is equal to 0.35 or 35%.
4. The answer is (D).
Because the frequency p of the disease gene is 0.08, then the homozy-
gote frequency would be p
2
or 0.0064.
5. The answer is (C).
Because the frequency of the normal allele “q” is 0.99, the frequency of
the mutated allele “p” is 1 – q or 1 – 0.99, which is 0.01. The frequency of “p” is almost equal
to 1, so the heterozygote frequency 2pq is thus 2
1 0.01 or 0.02.
6. The answer is (B).
In a small population, a greater proportion of “founding” individuals
may carry a gene than in the larger population. The small group of Dutch settlers in South
Africa reproduced only with members of the group. If the autosomal dominant mutant
gene that causes variegate porphyria was overrepresented in that small population, then a
larger proportion of individuals with the gene would be born into the group, and the fre-
quency of the gene would increase.
7. The answer is (A).
If heterozygosity for the CF gene conferred protection against cholera,
then heterozygotes would probably survive to reproduce. Those without the CF gene
would most likely die before producing any more children or die before they reached
reproductive age. Those who were homozygous for the gene had CF and died without
reproducing. Carriers would survive to reproduce, the CF gene would be passed on, and
the gene frequency would increase due to the positive selective pressure on the gene.
8. The answer is (C).
In autosomal recessive diseases, carriers of the disease gene are gener-
ally asymptomatic, so it is not obvious who does and who does not carry the gene. Since
those who are carrying the gene cannot be identified, the gene is “protected” from removal
from the population. Even if that could be accomplished, new mutations would occur that
would keep the gene in the population.
9. The answer is (D).
The frequency of those with the disease, 1 in 500, is equal to 0.02, which
is q
2
. The frequency of the disease gene “q” is the square root of 0.02, or 0.0447. The carrier
frequency is 2pq, and “p” is close to 1, but with the disease being so common, a more accu-
rate carrier frequency can be obtained by using the true value of “p”, which is 1 – q
1 –
0.0447
0.9553. So 2pq 2(0.9553)(0.0447) 0.0849 which is about 8.5%.
10. The answer is (A).
The recombinant percent of the two loci is very large, meaning that there
is a lot of recombination between them. The further apart two loci are, the higher the
recombination percent, so the two loci are far apart and not very tightly linked.
11. The answer is (C).
Of the couple’s four children, two of the children had a recombinant
chromosome resulting from crossing over in the inversion loop during paternal meiosis.
The recombinant fraction is the number of recombinant chromosomes divided by the
number of children. There were 2 recombinant chromosomes in the 4 children, which is
2/4 or 1/2, which is 0.50.
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12. The answer is (C).
Assortative mating, where “like mates with like” violates the assumption
that there is random mating in the population. Assortative mating would cause an increase
in the frequency of the gene (or genes) responsible for whatever characteristic was associ-
ated with the assortative mating, like tall with tall, short with short, etc.
13. The answer is (D).
The gene will not be eliminated from the population because it is
“protected” in heterozygotes who are not affected and can pass the gene on to succeeding
generations.
14. The answer is (C).
Heterozygotes for the sickle cell gene (S) have resistance to malaria. West
Africa is a malaria prone area and having resistance to malaria would enhance an individ-
ual’s chances of living to reproduce. Thus, the sickle cell gene would be “selected for” in the
population and its frequency would gradually increase because those with the gene would
have more reproductive success.
15. The answer is (D).
A lethal dominant gene appears in the population as the result of new
mutations. There are no “carriers” who are unaffected; it is lethal to all who inherit it.
16. The answer is (A).
The frequency of phenylketonuria, 1/10,000 is q
2
. So “q” is the square
root of 1/10,000 or 1/100. Because “p” is close to 1, the carrier frequency, 2pq is
2(1)(1/100)
2/100 1/50, so 1 in 50 is the carrier frequency.
17. The answer is (D).
The gene frequency for X-linked recessive diseases is the same as the fre-
quency of affected males, which is the disease frequency because females are generally not
affected.
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