Genetics Homework 1
Due: February 5, 2024 at 11:59 PM
1. Determine if the following events occur during Mitosis, Meiosis I, Meiosis II, Mitosis and Meiosis,
Mitosis and Meiosis I, or Mitosis and Meiosis II. (Answers may be used once, more than once or not at
all. Be as accurate as possible with your answer.)
__f_ 1. The chromosome number is reduced by half.
__a_ 2. The amount of DNA in the cell is twice the amount of a normal diploid cell in G1.
_b__ 3. DNA is replicated during S phase.
__f_ 4. Homologous chromosomes separate.
_d__ 5. Sister chromatids separate.
__b_ 6. Chromosomes condense and align along the metaphase plate.
__f_ 7. Homologous chromosomes align and undergo crossing over.
___e 8. Has one round of division per round of DNA replication.
__f_ 9. The number of chromatids is equal to the number of chromosomes of a diploid cell in
G1.
_f__ 10. Tetrads align on the metaphase plate.
a. Mitosis and Meiosis I only
b. Both Mitosis and Meiosis
c. Meiosis II only
d. Mitosis and Meiosis II only
e. Mitosis only
f. Meiosis I only
b. Both Mitosis and Meiosis
Explanation: The statement "3. DNA is replicated during S phase" is true for both Mitosis and Meiosis.
In both processes, DNA replication occurs during the S phase of the cell cycle, which is part of
interphase preceding both Mitosis and Meiosis. This is when the cell duplicates its DNA to prepare for
division.
f. Meiosis I only
Explanation: The statement "4. Homologous chromosomes separate" is true only for Meiosis I. During
Anaphase I of Meiosis I, homologous chromosomes separate and move to opposite poles of the cell.
This does not occur in Mitosis or Meiosis II.
d. Mitosis and Meiosis II only
Explanation: The statement "5. Sister chromatids separate" is true for both Mitosis and Meiosis II.
During Anaphase of Mitosis and Anaphase II of Meiosis II, sister chromatids separate and move to
opposite poles of the cell.
b. Both Mitosis and Meiosis
Explanation: The statement "6. Chromosomes condense and align along the metaphase plate" is true for
both Mitosis and Meiosis. During Metaphase of Mitosis and Metaphase I and II of Meiosis,
chromosomes condense and align along the metaphase plate in preparation for separation.
f. Meiosis I only
Explanation: The statement "7. Homologous chromosomes align and undergo crossing over" is true only
for Meiosis I. During Prophase I of Meiosis I, homologous chromosomes pair up and exchange genetic
material in a process known as crossing over.
e. Mitosis only
Explanation: The statement "8. Has one round of division per round of DNA replication" is true only for
Mitosis. Mitosis involves a single round of division following DNA replication, resulting in two
identical daughter cells. In contrast, Meiosis involves two rounds of division following DNA replication,
resulting in four genetically unique daughter cells.
f. Meiosis I only
Explanation: The statement "9. The number of chromatids is equal to the number of chromosomes of a
diploid cell in G1" is true only for Meiosis I. After DNA replication in the S phase, each chromosome
consists of two sister chromatids. Therefore, at the start of Meiosis I, the number of chromatids is indeed
equal to the number of chromosomes of a diploid cell in G1.
f. Meiosis I only
Explanation: The statement "10. T
2. You are studying the inheritance of 8 unlinked genes (A, B, C, D, E, F, G and H). Assuming that the
dominant allele for each gene shows complete dominance, how many phenotypic classes would you
expect to see if you testcrossed an individual that is heterozygous for all eight genes.
256
3. In continuing your study of the heritance of the 8 unlinked genes from the previous question (Genes
A, B, C, D, E, F, G, and H), you make the following cross: AAbbCcDdEeFFGghh x
AabbCcDDeeFfGgHh. What is the probability you will get an individual that is either homozygous
dominant at all eight loci or homozygous recessive at all eight loci? (Be sure to completely reduce your
answer.)
0
4. Continuing on with your study of the 8 unlinked genes from question 2 above (Genes A, B, C, D, E,
F, G and H), you make the following cross: AAbbCcDdEeFFGgHh x AabbCcDDeeFfGgHh. What is the
probability you will get an individual that is AAbbCcDDEeFfggHH or AabbCCDdeeFFGGhh? (Be sure
to completely reduce your answer.)
3/1024
The probability of getting an individual that is AAbbCcDDEeFfggHH or AabbCCDdeeFFGGhh is
1/256.
Explanation:
To calculate the probability, we need to consider each gene separately. For each gene, we calculate the
probability of getting the desired genotype and then multiply all these probabilities together.
For the first genotype (AAbbCcDDEeFfggHH):
- The probability of getting AA is 1/2 (since one parent is AA and the other is Aa).
- The probability of getting bb, cc, dd, ee, gg, and HH is 1 (since both parents are homozygous for these
genes).
- The probability of getting DD is 1/4 (since both parents are Dd).
- The probability of getting Ff is 1/2 (since one parent is FF and the other is Ff).
Multiplying these probabilities together gives us 1/2 * 1 * 1 * 1 * 1 * 1 * 1/4 * 1/2 = 1/16.
For the second genotype (AabbCCDdeeFFGGhh):
- The probability of getting aa, bb, dd, ee, and hh is 1 (since both parents are homozygous for these
genes).
- The probability of getting CC, FF, and GG is 1/4 (since both parents are heterozygous for these genes).
- The probability of getting Aa is 1/2 (since one parent is AA and the other is Aa).
Multiplying these probabilities together gives us 1 * 1 * 1/4 * 1 * 1 * 1/4 * 1/4 * 1/2 = 1/64.
Since these are two independent events, we add the probabilities together to get the total probability:
1/16 + 1/64 = 1/256.
5. Evaluate the following pedigree:
What is the most likely mode of inheritance?
autosomal recessive
6. Again, consider the following pedigree:
What is the probability that Individuals III-1 and III-2 have an affected child assuming individual III-1 is
heterozygous?
1/4
7. The urofacial syndrome has two symptoms -- a smile that appears to be a frown and blockage of the
urinary system, which can be fatal if untreated. It is an autosomal recessive condition. Using the
pedigree below, determine all those individuals that must be carriers (i.e. heterozygous). (To get full
credit you must select all that apply.)
1 1 and 2
2 4 and 5 2
3 3
A 1 and 4 5
8. John and Sarah are contemplating having children, but both have a family history of Tay Sach's
Disease, which is lethal usually by the age of 2 or 3 years. John had a brother die of Tay Sach's and
Sarah's great grandparents (on her father's side) had a daughter die of the disease. All other members of
the family are healthy. What is the probability that John and Sarah's first child will be affected with Tay
Sachs? (Be sure to completely reduce your answer.)
autosomal recessive
1) The probability that their child will have alkaptonuria is 1/4 or 25%.
Explanation:
1) Alkaptonuria is an autosomal recessive disease. This means that both parents must be carriers
of the disease for their child to have it. If both parents are carriers, there is a 25% chance that their
child will inherit the disease. This is because there is a 1/2 chance that the child will inherit the
disease allele from each parent, and these probabilities multiply together (1/2 * 1/2 = 1/4).
9. Consider the following pedigree:
What is the most likely mode of inheritance?
autosomal dominant
10. A recently married man and woman discover they each had an uncle with alkaptonuria, also known
as "black urine disease", a rare autosomal disease caused by a recessive allele of a single gene. They are
about to have their first child and they come to you to ask what the probability is that their child will
have alkaptonuria. What do you tell them? (Be sure to completely reduce your answer.)
1/16
The probability that their child will have alkaptonuria is 1/16 or 6.25%. The chance that Jack and Jill
will have a child with Alpers Syndrome is approximately 1 in 4,000,000 or 0.000025%.
Explanation:
For alkaptonuria, since both parents had an uncle with the disease, it suggests that each of their
parents are carriers (heterozygous). The chance that each parent is a carrier is 1/2. If both parents
are carriers, the Punnett square method shows that there is a 1/4 chance that a child will inherit the
disease. Therefore, the overall probability is 1/2 (mother carrier) * 1/2 (father carrier) * 1/4 (child
affected) = 1/16 or 6.25%.
For Alpers Syndrome, Jack's grandfather's brother had the disease, which suggests that Jack's
grandfather was a carrier. However, this does not necessarily mean that Jack is a carrier. Given
that the frequency of heterozygotes in the general population is 1 in 2000, the chance that Jack is a
carrier is 1/2000. Jill has no family history of the disease, so we assume her chance of being a
carrier is also 1/2000. If both are carriers, there is a 1/4 chance their child will have the disease.
Therefore, the overall probability is 1/2000 (Jack carrier) * 1/2000 (Jill carrier) * 1/4 (child affected)
= 1 in 4,000,000 or 0.000025%.
11. Jack and Jill are thinking about starting a family and they come to you with a question about a rare
recessive genetic disease called Alpers Syndrome. Alpers causes progressive neurodegeneration in
infants and children and is usually lethal within the first 10 years of life. Jack's grandfather had a brother
die of Alpers Syndrome, while Jill has no family history of the disease. If the frequency of heterozygotes
for Alpers is 1 in 2000 in the general population, what is the chance that Jack and Jill will have a child
with Alpers Syndrome? (Be sure to completely reduce your answer.)
2) For Alpers Syndrome, we know that Jack is likely a carrier because his grandfather's brother
had the disease. However, Jill has no family history of the disease, so we assume she is not a
carrier. The chance that she is a carrier is 1 in 2000. If Jack is a carrier and Jill is not, their child
has a 0% chance of having the disease. If both are carriers, their child has a 25% chance of having
the disease. Therefore, the overall chance is the product of the chance that Jill is a carrier and the
chance that a child of two carriers has the disease (1/2000 * 1/4 = 1/8000). However, this is the
chance for any given child. The chance that any of their children will have the disease is
approximately 1 in 8,000,000 (1/8000 * 1/1000).
2) The chance that Jack and Jill will have a child with Alpers Syndrome is approximately 1 in
8,000,000.
12. In a sample of 3840 families with 7 children, how many families would be expected to have 5 or
more boys?
870
13. You are studying a variety of pea plant that has either red flowers or white flowers. When you cross
true-breeding red-flowering plants with true-breeding white-flowering plants, you get all red-flowering
plants in the F1 generation. If you then allow the F1 plants to self-fertilize, you observe that 75% of the
F2 plants are red-flowering plants, while 25% of the F2 plants are white-flowering plants. Next you
make 640 different monohybrid crosses and decide to analyze only the first 8 plants produced from each
of these crosses (i.e. a total of 5120 progeny plants total). How many of these crosses would be expected
to produce 6 red plants and 2 white plants?
199 or 200
Explanation:
The computation of the number of these crosses is shown below:
P (red) = 75% = a
P (white) = 25% = b
Now determine the probability of 6 red + 2 white and then multiplied it by 640
Now use the binomial theorem
now the required probability is
= 28 × (0.75)^6 × (0.25)^2 × 640
= 199.25
= 199 or 200
199 or 200 Explanation: The computation of the number of these crosses is shown below: P (red) = 75% = a P
(white) = 25% = b Now determine the probability of 6 red + 2 white and then multiplied it by 640 Now use the
binomial theorem
The question is asking how many of the 640 monohybrid crosses would be expected to produce 6 red plants and
2 white plants. To determine this, we need to calculate the probability of getting 6 red plants and 2 white plants
in each cross.
In the F2 generation, we observed that 75% of the plants were red-flowering and 25% were white-flowering.
This means that the probability of getting a red plant (a) is 0.75 and the probability of getting a white plant (b) is
0.25.
To calculate the probability of getting 6 red plants and 2 white plants, we can use the binomial theorem. The
binomial theorem allows us to calculate the probability of a specific combination of outcomes in a series of
independent events.
The formula for the binomial theorem is:
P(x) = (nCx) * (a^x) * (b^(n-x))
Where:
P(x) is the probability of getting x red plants and (n-x) white plants
n is the total number of plants (in this case, 8)
x is the number of red plants (in this case, 6)
a is the probability of getting a red plant (0.75)
b is the probability of getting a white plant (0.25)
nCx is the number of combinations of n items taken x at a time
By plugging in the values, we can calculate the probability of getting 6 red plants and 2 white plants in each
cross. Then, we multiply this probability by the total number of crosses (640) to find the expected number of
crosses that would produce this outcome.
Therefore, the correct answer is 199 or 200 crosses.
199 or 200 Explanation: The computation of the number of these crosses is shown below: P (red) = 75% = a P
(white) = 25% = b Now determine the probability of 6 red + 2 white and then multiplied it by 640 Now use the
binomial theorem
The question is asking how many of the 640 monohybrid crosses would be expected to produce 6 red plants and
2 white plants. To determine this, we need to calculate the probability of getting 6 red plants and 2 white plants
in each cross.
In the F2 generation, we observed that 75% of the plants were red-flowering and 25% were white-flowering.
This means that the probability of getting a red plant (a) is 0.75 and the probability of getting a white plant (b) is
0.25.
To calculate the probability of getting 6 red plants and 2 white plants, we can use the binomial theorem. The
binomial theorem allows us to calculate the probability of a specific combination of outcomes in a series of
independent events.
The formula for the binomial theorem is:
P(x) = (nCx) * (a^x) * (b^(n-x))
Where:
P(x) is the probability of getting x red plants and (n-x) white plants
n is the total number of plants (in this case, 8)
x is the number of red plants (in this case, 6)
a is the probability of getting a red plant (0.75)
b is the probability of getting a white plant (0.25)
nCx is the number of combinations of n items taken x at a time
By plugging in the values, we can calculate the probability of getting 6 red plants and 2 white plants in each
cross. Then, we multiply this probability by the total number of crosses (640) to find the expected number of
crosses that would produce this outcome.
Therefore, the correct answer is 199 or 200 crosses.