WEEK 2: INHERITED DISEASES

Question 1

State the effects of maternal ageing on oocyte division.

Answer 1

Maternal ageing can lead to the following

● Shortening of the Telomeres:

This often results in missegregation of chromosomes leading to aneuploids.

● Spindle Instability: leads to improper separation of chromosomes and sister chromatids

● Cohesion Dysfunctions:

chromosomal missegregation can also result if there is unstable cohesion between sister chromatids close to the centromere.

● Mitochondrial Dysfunction:

Mitochondria are important for energy production and if
this organelle dysfunctions the following processes can be impaired: spindle assembly.
chromosome segregation and cell cycle regulation.

Question 2

What is Mendelian Inheritance?

Answer 2

Genetic disorders due to a defect in a single gene follow the patterns of inheritance outlined by Mendel.

Question 3

Outline the 4 principles of Mendelian inheritance.

Answer 3

1) Genes in the offspring are inherited from each parent and thus an offspring has
two copies of a gene or two alleles of a gene
2) These alleles can act in a dominant or recessive manner
3) Each gamete receives only one allele from segregation of alleles during meiosis
4) Alleles at different chromosomal loci segregate independently

Question 4

Define the following terms:
1.Allele
2.Genotype
3.Phenotype

Answer 4

*Allele: is a different form a gene.

An allele can be composed of two dominant genes (AA), a combination of a dominant gene and a
recessive gene (Aa), or two recessive genes (aa).

*Genotype: could be used to represent a set of genes and hence alleles for an organism

*Phenotype: observable physical traits of an organism

Question 5

How is a dominant and recessive allele written?

Answer 5

Dominant gene (A) and recessive gene(a)

Question 6

If X-linked disorder is recessive, why are only males affected?

Answer 6

If X-linked disorder is recessive then males get affected since they will only have one X chromosome from the mother and Y from the father

Question 7

What are Pedigree Tree graphs?

Answer 7

These follow the inheritance of a genetic trait or disease through generations in a family tree.

Question 8

Differentiate between autosomal dominant trait inheritance.

Answer 8

Autosomal Dominant Trait
1) It doesn’t skip generations
2) Affected parents can have unaffected children

Autosomal Recessive Trait:
1) It skips generations
2) Unaffected parents can have affected children

Question 9

Describe X-linked Dominant Trait inheritance.

Answer 9

X-linked Dominant Trait:
1) Affected Father never gives son disease
2) Affected Father will give it to all his daughters

Question 10

Describe X-linked Recessive Trait inheritance.

Answer 10

X-linked Recessive Trait:
1) Affected Father never gives son disease
2) Males more affected due to having a single copy of X
chromosome
3) A female carrier will give it to 50% of her daughters and
50% of her sons

Question 11

Describe Y-linked inherited trait inheritance.

Answer 11

Y-linked Inherited Trait:
1) Affected Father gives it to all his son

Question 12

What is Hardy-Weinberg Equilibrium?

Answer 12

Genetic variation in a population will not change from one generation to the next in the absence of some factors.

Question 13

What are the 5 assumptions of the Hardy-Weinberg Equilibrium?

Answer 13

Assumptions are that:

1)Mating is random
2)Large population size
3)No natural selection
4)No migration
5)No mutations

Question 14

What do the following represent in the Hardy-Weinberg Equilibrium?
1.p
2.q
3.p^2
4.q^2
5.2pq

Answer 14

p = dominant trait proportion in population
q = recessive trait proportion in population

Homozygous Dominant frequency=p2
Homozygous recessive frequency=q2
Heterozygous frequency= 2pq

Question 15

State the 2 formulas of the Hardy-Weinberg Equilibrium.

Answer 15

p+q=1
p2+2pq+q2=1

Question 16

Question 1. Cystic fibrosis is an autosomal recessive disease which occurs in 1/2500 live births in the UK.

a. Calculate the frequency of the CF gene in the UK population using the Hardy-Weinberg Law.

b. What proportion of people in the UK are CF carriers?

Answer 16

Hardy-Weinbrg’s law; q2+2pq+p2=1 and q+p=1
Ans: q= frequency of the recessive allele
q2= 1/2500
q = √(1/2500)
q = 1/50

Ans: p + q=1
p + 1/50 =1, then p = 49/50
p2 + 2pq + q2 = 1, p2 =2401/2500 and q2 = 1/2500 therefore
2pq= 49/1250

Question 17

Huntingdon’s disease is an autosomal dominant disease which produces CNS degeneration. It does not usually manifest in an affected individual until after the age of 45 and sometimes much later.

a. A 28-year-old man discovers that his father has just been diagnosed with Huntingdon’s disease. What is the probability that he will develop the disease himself?

b. The man already has a baby girl. What is the probability that his daughter will develop the disease in later life?

Answer 17

Ans: let H be the dominant allele and h be the recessive allele, Hh in this case is the father and Hh is the mother. Therefore, the probability that the man will not develop the disease is 1/2.

½ x ½= ¼

Question 18

A family has four children:

a. What is the probability that at least one will be a boy?

b. What is the probability that the children will be born in the order GBGB?

c. What is the probability that the children will all be boys?

Answer 18

1.The probability that at least one child will be a boy is 15/16 or 0.9375. This is because the probability of having four girls is 1/16, so the probability of having at least one boy is 1 - 1/16 = 15/16

2.The probability that the children will be born in the order GBGB is 1/16 or 0.0625. This is because the probability of having a girl is 1/2 and the probability of having a boy is also 1/2. So, the probability of having a girl followed by a boy followed by a girl followed by a boy is (1/2) * (1/2) * (1/2) * (1/2) = 1/16

3.The probability that all children will be boys is 1/16 or 0.0625. This is because the probability of having a boy is 1/2, so the probability of having four boys is (1/2) * (1/2) * (1/2) * (1/2) = 1/16

Question 19

Problem 1. Kitso and Lorato recently got married to each other and plan on having 6 children. Lorato is a carrier for a rare X-linked recessive disorder called Androgen Insensitivity Syndrome (AIS).

a) What are the three types of AIS?
b) What is Lorato’s karyotype?
c) If they have 3 sons and 3 daughters, how many will be carriers?
d) What is the probability that a son will have the disease?
e) What is the probability that a daughter will have the disease?
f) If Kitso also had the abnormal gene on the X-chromosome and was fertile (very rare), what
proportion of their children will have AIS and how many will be carriers?

Answer 19

a.
-Complete androgen insensitivity syndrome (CAIS), with typical female external genitalia
-Partial androgen insensitivity syndrome (PAIS) with predominantly female, predominantly male, or ambiguous external genitalia
-Mild androgen insensitivity syndrome (MAIS) with typical male external genitalia

b. 46, XY

c. 1 son affected, 2 carriers and 1 daughter affected, 2 carriers.

d.0%

e.25%
f. All daughters will have AIS.
All the sons will be unaffected.

Question 20

Problem 4. Give an example of a molecular genetic test for Sickle cell disease and explain how it is used to identify the disease.

Question 21

Problem 5.
The frequency of alleles for Cystic Fibrosis in Molepolole are 0.15 (F) and 0.85 (f). Assume that the population in Molepolole is in Hardy-Weinberg equilibrium.

a) What is the percentage of individuals that are heterozygous in the population?
b) What is the percentage of individuals that are homozygous recessive in the population?

Answer 21

The Hardy-Weinberg equation is p^2 + 2pq + q^2 = 1, where p is the frequency of one allele and q is the frequency of the other allele. In this case, p = 0.15 and q = 0.85.

To calculate the percentage of individuals that are heterozygous in the population (2pq), we can substitute p and q into the equation:

2(0.15) (0.85) = 0.255
Therefore, 25.5% of individuals in the population are heterozygous.

b.To calculate the percentage of individuals that are homozygous recessive in the population (q^2), we can substitute q into the equation:

(0.85)^2 = 0.7225

Therefore, 72.25% of individuals in the population are homozygous recessive.

Question 22

Problem 6.
In Francistown, 0.3% of individuals are homozygous recessive for Gaucher’s disease. Given that the population of Francistown is 147122, calculate the number of individuals that are homozygous dominant and the number of individuals that are heterozygous.

Answer 22

p^2 + 2pq + q^2 = 1
where p is the frequency of the dominant allele and q is the frequency of the recessive allele.

Since we know that 0.3% of individuals are homozygous recessive for Gaucher’s disease, q^2 = 0.003.

Therefore, q = sqrt(0.003) = 0.0548 (rounded to four decimal places).

Since p + q = 1, p = 1 - q = 0.9452.

Number of individuals that are homozygous dominant = p^2 * population
= (0.9452) ^2 * 147122
= 129,684.6

Number of individuals that are heterozygous = 2pq * population
= 2 * 0.9452 * 0.0548 * 147122
= 17,992.4

Question 23

Problem 7.
You have just inherited 10 heads of cattle; all cattle are black. However, one of the cows carries a recessive allele for red coat (b) (genotype is Bb) and the rest do not.

a) Calculate the frequencies of the alleles B and b.

b) After 50 years, your cattle population has increased to 4000. Assuming Hardy-Weinberg equilibrium for the alleles, how many heads of cattle would have a red coat?

Answer 23

a.
B= 19/20
b=1/20

b.