Hereditary and Population Genetics Problems With Solution

2021-05-07 12:24:31
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1. You have a female snail that coils to the right, but you do not know its genotype. You may assume that right coiling (D) is dominant to left coiling (d). You also have male snails at your disposal of known genotype. How would you determine the genotype of this female snail? In your answer, describe your expected results depending on whether the female is DD, Dd, or dd. (4 points)

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D- right coiling

d-left coiling

Cross female (unknown) with male (known)

male female

Dd xx

/\ /\

D d x xGenotype of f1 is Dx Dx dx dxCross f1 gen female with homozygous recessive male

Male female

dd Dx /\ /\

d d D x

genotype of female is Dd2. A variegated trait in plants is analyzed using reciprocal crosses. In the first cross, a variegated female plant is crossed with a normal male parent. From the progeny of this cross, 1024 plants are variegated and 52 plants are normal. In a second cross, a normal female plant is crossed with a variegated male plant. From the progeny of this cross, 1113 plants are normal and 61 plants are variegated. Explain the mode of inheritance for this trait. (4 points)

The mode of inheritance in this instance is mitochondrial inheritance. This happens because in this type of inheritance, only mitochondrial DNA from the mother are inherited, a case in which the number of progenies that inherit the variegation phenotype from the male organism are always limited to between 50 and 70. This type of inheritance does not follow mendelian laws of inheritance.

3. Both red-green color blindness and hemophilia A are X-linked recessive conditions in humans. If a man who exhibits both red-green color blindness and hemophilia A marries a woman with no family history of either disorder, what is the probability that their first daughter will be heterozygous for hemophilia A? (3 points)

Male heamophilic X*Y

Female Normal XX

Male Female

X*Y XX

/\ /\

X* Y X XF1 genotype is X*X X*X XY XY

Thus probability of the daughter being haemophilic is 1

a. What is the probability that their first daughter will be heterozygous for red-green color blindness? (3 points)

Male colorblind X+Y

Female normal XX

MALE FEMALE

X+Y XX

/\ /\

X+ Y X XF1 genotype is X+X, X+X , XY XYThus probability of daughter being heterozygous is 1

b. What is the probability that their first son will be affected with hemophilia A? (3 points)

The probability is 0 as the disease is sex linked and the mother is not a carrier of haemophilia s references in the above cross

c. What is the likelihood that their daughters will exhibit either disorder? (3 points)

The daughters wont exhibit the disorders . They are all heterozygous for the traits .The traits are only exhibited in homozygous state in females

s referenced in the above cross

4. How is the process of random X inactivation in mammals, similar to genomic imprinting? How is it different? (4 points)

The completeness and stability of genomic imprinting is lower as compared to that of random X inactivation. Also, DNA methylation contributes less in maintenance of the genomic imprinting than in random X inactivation. The similarity is that if either of them is inappropriately regulated, disease is the result. Another similarity is that in both, a single gene is expressed in a cell.

5. Herberdens nodes (bony excrescences on the fingers) is a variety of osteoarthritis found in humans. It is a sex-influenced trait that is dominant in females and recessive in males. The disease is not fully penetrant until after age 70. What is the probability that Mary will develop this disease if both of her parents are alive and over the age of 70 and her mother does not exhibit the disease but her father does? (2 points)

The probability is 1 as their Marys genotype is XHXhMary has been single all of her life. She meets Steve and thinks that he is the one. She does not know too much about him and his family at the moment. If she were to marry Steve, what is the probability that any children that they have together will develop this disease in their old age? (4 points)

Mary XHXh

Steve XhYMary Steve

XHXh XhY /\ /\

XH XhXh Y

Fi genotype is

XHXh, XHY, XhXh, XhYThe probability of their children expressing the trait is 1/2

It turns out that Mary and Steve were meant to be. On their honeymoon, Steve mentions that he noticed the arthritis on the fingers of Marys father. He tells Mary that his mother also has arthritis on her fingers. If the arthritis on their fingers is caused by the same gene, what is the new probability that any children that they have together will develop this disease in their old age? (4 points)

Mary XHXh

Steve XhYMary Steve

XHXh XhY /\ /\

XH XhXh Y

Fi genotype is

XHXh, XHY, XhXh, XhY

Probability is 1/2

Mary XHXh

Steve XHY

Mary Steve

XHXh XHY

/\ /\

XH XhXH Y

Fi genotype is

XHXH, XHY, XHXh, XhYThe probability is 3/4

Thus total probability is

1/23/4= 3/8

Thus total posibility is 3/8

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