Problem 1
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Consider the following pedigree for a rare human muscle disease
A. What unusual feature distinguishes this pedigree?
B. Where in the cell do you think the mutant DNA resides that is responsible for the phenotype
Answer Problem 1
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A. What unusual feature distinguishes this pedigree?
The pattern of clearly shows maternal inheritance. However males and females are affected equally.
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Where in the cell do you think the mutant DNA resides that is responsible for the phenotype
Mostly likely, the mutant DNA is mitochondrial.
Problem 2
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The following pedigree shows the recurrence of a rare neurological disease (large black symbols) and
spontaneous fetal abortion (small black symbols) in one family. (Slashes mean that the individual is deceased.)
Provide and explanation for this pedigree in regard to cytoplasmic segregation of defective mitochondria.
Answer Problem 2
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Provide and explanation for this pedigree in regard to
cytoplasmic segregation of defective mitochondria.
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Each cell has many mitochondria, each with numerous
genomes. Cytoplasmic segregation of mitochondria mixtures
is routinely found within the same cell.
The best explanation for this pedigree is that the mother in
generation I experienced a mutation in a single cell that was a
progenitor of her egg cells (primordial germ cell). By chance
alone, the two males with the disorder in the second
generation were from egg cells that had experienced a great
deal of cytoplasmic segregation prior to fertilization, whereas
the two females received mixture.
The spontaneous abortions that occurred for the first woman in
generation II were the result of extensive cytoplasmic
segregation in her primordial germ cells: aberrant
mitochondria were retained. The spontaneous abortions of the
second woman in generation II also came from such cells.
The normal children of this woman were the result of extensive
segregation in the opposite direction: normal mitochondria
were retained. The affected children of this woman were from
egg cells that had undergone less cytoplasmic segregation by
the time of fertilization so that they developed to term but still
suffered from the disease.
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Problem 3
• A study made in 1958 in the
mining town of Ashibetsu in the
Hokkaido province of Japan
revealed the frequencies of
MN blood type genotypes (for
individuals and for married
couples) shown in the
following table to the right
• A. Is the population in HardyWeinberg equilibrium with
respect to the MN blood types.
• B. Is mating random with
respect to MN blood types
Genotype
Number of individuals or
couples
Individuals
LM/LM
406
LM/LN
744
LN/LN
332
Total
1482
Couples
LM/LMXLM/LM
58
LM/LMX LM/LN
202
LM/LNX LM/LN
190
LM/LMXLN/LN
88
LM/LNXLN/LN
162
LN/LNXLN/LN
41
Total
741
Answer Problem 3
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A. Is the population in Hardy-Weinberg equilibrium
with respect to the MN blood types.
If the population is in equilibrium then the HardyWeinberg equation is true and p2+2pq+q2=1
First you need to calculate the frequency of the
genotypes which are
– p=fA/A+1/2fA/a=frequency of A
– q=fa/a+1/2fA/a=frequency of a
M
L =p=406/1482+1/2(744/1482)=0.52
LN=q=(332+1/2(744))/1482)=0.48
Genotype
Two be in equilibrium the genotypes should be
distributed
Couples
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LM/LM=p2(1482)=0.52x0.52(1482)=401
LM/LN=2pq(1482)=2x0.52X0.48X1482=740
LN/LN=q2(1482)=0.48X0.48X1482=341
This compares well with the data, so the population
is in equilibrium
Number of individuals or
couples
Individuals
LM/LM
406
LM/LN
744
LN/LN
332
Total
1482
LM/LMXLM/LM
58
LM/LMX LM/LN
202
LM/LNX LM/LN
190
LM/LMXLN/LN
88
LM/LNXLN/LN
162
LN/LNXLN/LN
41
Total
741
Answer Problem 3
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B. Is mating random with respect to MN blood types
If the mating is random with respect to blood type,
then the following frequency of mating should occur
–
LM/LMXLM/LM=p2xp2x741=0.524x741=54
– LM/LMX LM/LN or LM/LNX LM/LM
=(2)xp2x2pqx741=2x0.522X2x0.52X0.48=200
– LM/LNX
LM/LN=2pqx2pq741=4x0.522x0.482x741=185
– LM/LMXLN/LN or LN/LNXLM/LM
=(2)p2xq2x741=2x0.52x0.52x0.48x0.48x741=
92
– LM/LNXLN/LN or LN/LNXLM/LN
=(2)2pqxq2x741=2x2x0.52x0.48x0.48x0.48x7
41=170
– LN/LNXLN/LN=q2xq2x741=0.484x741=39
Genotype
These number compare nicely with the actual data
therefore the matings are random
Number of individuals or
couples
Individuals
LM/LM
406
LM/LN
744
LN/LN
332
Total
1482
Couples
LM/LMXLM/LM
58
LM/LMX LM/LN
202
LM/LNX LM/LN
190
LM/LMXLN/LN
88
LM/LNXLN/LN
162
LN/LNXLN/LN
41
Total
741
Problem 4
• Colorblindness results from a sex-linked recessive allele. One in
every 10 males is colorblind.
• a. What proportion of all women are colorblind
• b. How many colorblind men are there for each colorblind woman
• c. In what proportion of marriages would colorblindness affect half
the children of each sex?
• d. In what proportion of marriages would all children be normal?
• e. In a population that is not in equilibrium, the frequency of the
allele for colorblindness is 0.2 in woman and 0.6 in men. After one
generation of random mating, what proportion of the female
progeny will be color blind? What proportion of the male progeny?
• f. What will the allelic frequencies be in the male and in the female
progeny in part e?
Answer Problem 4
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a. What proportion of all women are colorblind
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b. How many colorblind men are there for each colorblind
woman
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For the condition to be true the mothers must be heterozygous and
the fathers must be color blind. The frequency of heterozygous
woman is 2pq and the frequency of affected males if q. Therefore
the frequency of a random marriage would be
2pqxq=2X0.9x0.1x0.1=0.018
In what proportion of marriages would all children be normal?
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½ colorblind
X Xcb
q/q2=10 to 1=frequency in men/frequency in females
c. In what proportion of marriages would colorblindness affect
half the children of each sex?
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Assuming population in Hardy-Weinberg equilibrium and that the
allelic frequency is the same in both sexes, we can directly calculate
the frequency of colorblindness allele as q=0.1 (Because this trait is
sex linked, q is equal to the frequency of affected males.) For
females to be colorblind they must be homozygous for the allele, so
their frequency is q2=0.01
All children will be phenotypically normal only if the mother is
homozygous for the noncolorblind allele. The fathers genotype does
not matter. Therefore the frequency would be p2=.9x.9=0.81
Xcb
X/Xcb
Xcb/Xcb
Y
X/Y
Xcb/Y
X
X
Xcb
X/Xcb
X/Xcb
Y
X/Y
X/Y
no colorblind
children
Answer Problem 4Fathers genotype
0.4X
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e. In a population that is not in equilibrium, the frequency of
the allele for colorblindness is 0.2 in woman and 0.6 in men.
After one generation of random mating, what proportion of
the female progeny will be color blind? What proportion of
the male progeny?
– easiest to draw a diagram showing possiblities
– So the frequency in females will be 0.12 and the
frequency in males will be 0.2
f. What will the allelic frequencies be in the male and in the
female progeny in part e?
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From the analysis of the results in (e), the frequency of the
colorblind allele will be 0.2 in males (the same as in the females of
the previous generation) and 1/2(0.08+0.48)+0.12=0.4 in females
Mother
0.8X 0.32X/X
0.2Xcb 0.08X/Xcb
0.6Xcb
Y
0.48X/Xcb 0.8X/Y
0.12Xcb/Xcb 0.2Xcb/Y
Problem 5
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A woman (II2 in the pedigree) wishes to know the probability that she is a carrier of Duchenne muscular dystrophy.
a. What is the probability if she has another affected male child
b. What is the probability if she has another unaffected male child
c. What is the probability if she has another unaffected female child
d. If she has another unaffected male child what is the probability she will have another affected child
I
II
III
Answer Problem 5
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A woman (II2 in the pedigree) wishes to know the
probability that she is a carrier of Duchenne
muscular dystrophy.
This problem is best solved using Bayes’ Theorem
It is easiest to make a table
I
Ancestral information
Hypothesis 1
II-2 Is a carrier
Hypothesis 2
II-4 Is Not a
carrier
Prior probability
1/2
1/2
Conditional
probability
(½)3=1/8
1
Joint probability
1/8X1/2=1/16
1/2x1
Posterior
probability
(1/16)/(1/16+1/2)
=1/9
1/2/(1/16+1/2)
=8/9
II
Considers children
III
So the probability she is a carrier is 1/9
Answer Problem 5
a. What is the probability if she has another affected
male child
If she has an affected child then she is a carrier =1
I
II
III
Answer Problem 5
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b. What is the probability if she has another
unaffected male child
This problem is best solved using Bayes’ Theorem
It is easiest to make a table
I
II
Hypothesis 1
II-2 Is a carrier
Hypothesis 2
II-4 Is Not a
carrier
Prior probability
1/2
1/2
Conditional
probability
(½)4=1/16
1
Joint probability
1/8X1/2=1/32
1/2x1
Posterior
probability
(1/32)/(1/32+1/2)
=1/17
1/2/(1/32+1/2)
=16/17
If she has another
unaffected male child
then the conditional
probability will change
III
So the probability she is a carrier is 1/17
Answer Problem 5
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c. What is the probability if she has another
unaffected female child
This problem is best solved using Bayes’ Theorem
It is easiest to make a table
I
II
If she has a female
child no additional
information is gained
because the disease is
X-linked recessive
Hypothesis 1
II-2 Is a carrier
Hypothesis 2
II-4 Is Not a
carrier
Prior probability
1/2
1/2
Conditional
probability
(½)3=1/8
1
Joint probability
1/8X1/2=1/16
1/2x1
Posterior
probability
(1/16)/(1/16+1/2)
=1/9
1/2/(1/16+1/2)
=8/9
III
So the probability she is a carrier is 1/9
Answer Problem 5
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d. If she has another unaffected male child what is
the probability she will have another affected child
This problem is best solved using Bayes’ Theorem
It is easiest to make a table
I
II
Hypothesis 1
II-2 Is a carrier
Hypothesis 2
II-4 Is Not a
carrier
Prior probability
1/2
1/2
Conditional
probability
(½)4=1/16
1
Joint probability
1/8X1/2=1/32
1/2x1
Posterior
probability
(1/32)/(1/32+1/2)
=1/17
1/2/(1/32+1/2)
=16/17
If she has another
unaffected male child
then the conditional
probability will change
III
So the probability she is a
carrier is 1/17 if she had four
unaffected boys. To pass on
the recessive allele is ½ and
the chance the child is male
is 1/2. So the chance she
has an affected child is
1/17X1/2X1/2=1/68