Question

7. The frequency for ocular albinism is one in 200 in the Hopi population. What percentage of the individuals within this population will be in the homozygous dominant state? What is the approximate percentage of individuals who are in the heterozygous state? 8. In a certain population, one in 16 million (16,000,000) females is affected with factor VilI hemophilia How many males in this same population will be affected with this same condition? 9. Approximately one in 15,000 individuals in a specific population has porphyria. How many individuals in this population will be in the heterozygous state? (Hint: think about how this trait is inherited, ie. the genotype(s) that express the trait.) 10. In one population, the frequency of the M allele is 35% (.35). of individuals will express the N phenotype? What percentage of individuals will express the MN phenotype? 11. For an autosomal recessive trait, the frequency of those individuals expressing this recessive trait is 0.09 How many individuals in this population will express the dominant phenotype but be in the heterozygous genotype?

Answer 1

7.

• Ocular albinism is inherited as X-linked recessive state.

Representation:

Since the affected condition is caused by recessive state, representing ocular albinism affected condition as a, corresponding normal state as A.

Thus, genotypes may be represented as:

• Affected Male: X^aY (hemizygous, recessive)
• Normal Male: X^AY (hemizygous, dominant)
• Affected Female: X^aX^a (homozygous, recessive)
• Carrier female: X^AX^a (heterozygous, dominant)
• Normal female: X^AX^A (Homozygous dominant)

X^aY x X^AX^a

F1	Female
Male	XA	Xa	F2
Xa	XA Xa	Xa Xa	Female
Y	XAY	XaY	Male

• X^A X^a =Carrier Female = 25%
• X^a X^a= Affected Female = 25%
• X^AY = Normal Male = 25%
• X^aY = Affected Male = 25%

0% homozygous dominant (25% hemizygous dominant)

25% heterozygous dominant.

8.

• Factor VIII hemophilia is inherited as X-linked recessive state.

Representation:

Since the affected condition is caused by recessive state, representing Factor VIII affected condition as a, corresponding normal state as A.

Thus, genotypes may be represented as:

• Affected Male: X^aY (hemizygous, recessive)
• Normal Male: X^AY (hemizygous, dominant)
• Affected Female: X^aX^a (homozygous, recessive)
• Carrier female: X^AX^a (heterozygous, dominant)
• Normal female: X^AX^A (Homozygous dominant)

X^aY x X^AX^a

F1	Female
Male	XA	Xa	F2
Xa	XA Xa	Xa Xa	Female
Y	XAY	XaY	Male

• X^A X^a =Carrier Female = 25%
• X^a X^a= Affected Female = 25%
• X^AY = Normal Male = 25%
• X^aY = Affected Male = 25%

Since ratio: affected female: affected male is 1:1

Thus, if 1 out of 16,000,000 females are affected,

1 out of 16,000,000 males are affected.

9.

• Different types of porphyria have different modes of inheritance.
• In this case, as 1 out of 15,000 is affected with porphyria, considering the trait of autosomal recessive mode of inheritance (like congenital erythropoietic porphyra).

Representation:

• Normal = P= dominant: Genotype – PP (homozygous dominant), Pp (heterozygous dominant)
• Affected = recessive: Genotype – pp

Thus, for F1 generation (Pure breeding):

PP x pp

	P	P
p	Pp	Pp
p	Pp	Pp

In F2:

Pp x Pp

	P	p
P	PP	Pp
p	Pp	pp

PP 25%

Pp = 50%

Pp = 25%

50% of 15000 =7500 are heterozygous.

10.

Hardy-Weinberg equilibrium:

• Hardy-Weinberg equation describes: The most possible distribution of the population genotype of the population, if the frequencies of the alleles are known.
• If allele frequencies are p and q, then the genotype frequencies are p², q², 2pq.
• A population is considered to be in Hardy-Weinberg equilibrium when:

In a randomly mating population, allele frequency and genotype frequency remain constant.

Thus, under Hardy-Weinberg equilibrium:

• p+ q=1 and p²+2pq + q² = 1.

Given:

• Dominant allele (M) = 35% =0.35 = p
• Frequency of recessive allele (N) = q = 1-0.35 = 0.65
• Frequency of N phenotype = q² = (0.65)² = 0.4225= 0.4 = 40% approx.
• Frequency MN phenotype = 2pq = 2 x 0.35 x 0.65 = 0.455 = 45.5 %

11.

Given:

• Genotype frequency for recessive trait = q² =0.09
• Frequency for recessive = q = 0.3
• Frequency of dominant allele = p = 1-q = 1-0.3 = 0.7
• Heterozygous genotype = 2pq = 2 x 0.3 x 0.7 = 0.42