# LIFESCI 4 Lecture 20: LIFESCI 4 - Lecture 20 - Review Lecture

LIFESCI 4 - Lecture 20 - Review and Cancer

Clicker 1:

A family that lives in a population where the frequency of an autosomal recessive genetic allele

si 1 in 100 (0.01). A pedigree is shown below. What is the probability that the child will have the

disease?

What is the probability that child C will have the disease?

A) 1/100

B) 1/200

C) 1/400

D) 1/1000

E) 1/10000

Explanation:

- Population is in equilibrium

- p = frequency(a) = 1/100

- q = frequency(A) = 1 - 1/100 = 0.99

- Both parents must be heterozygous for C to have the disease

- P[C is aa] = P[Dad is Aa] X P[Mom is Aa] X ¼

- Can approximate 0.99 as 1 and 0.01 as 0

- is approximately

1−(0.01)2

2(0.99)(0.01) q

1−0

2(1)(0.01) = 2

- Same approximation for mother

- = q q /4 q

2pq

1−q2×2pq

p−2pq

2×4

1=1−(0.01)2

2(0.99)(0.01) ×2pq

p−2pq

2×4

1≈ 2 × 2 × 1 = 2

-0.01)(0.01) 1/10000q2= ( =

Clicker 2:

Below is a pedigree from a family that lives in a population where the frequency of an

autosomal recessive genetic allele is 1 in 3 (0.33). A pedigree is shown below. What is the

probability that the child will have this disease?

What is the probability that child B will have the disease?

A) 1/4

B) 1/9

C) 1/16

D) 1/20

E) 1/100

Explanation:

-⅓ = frequency for (q)

-⅔ = frequency for (p)

- For child B to be homozygous recessive, both parents must be heterozygous

- P(B is aa) = P[Dad is Aa] X P[Mom is Aa] X P[both parents pass down small (a) =

¼]

- P [Dad is Aa] = (total is not since he cannot be

2pq

p + 2pq

2pq qp2+ 2 + 2q2

- P [Mom is Aa] = 2pq

p + 2pq

2

- Plug in all values for (p) and (q) into the equation

- P[B is aa] = ) /16( 2pq

p + 2pq

2

2× 4

1= 1

Clicker 3:

Below is a pedigree from a family that lives in a population where the frequency of an

autosomal recessive genetic allele is 1 in 100 (0.01)

. A pedigree is shown below. What is the

probability that the child will have this disease?

What is the probability that child D will have the disease?

A) ¼

B) 1/9

C) 1/16

D) 1/100

E) 1/300

Explanation:

- Both parents have history of disease in family

- Need to calculate probability that parents are carriers

- P(D is aa) = P[Dad is Aa] X P[Mom is Aa] X P[both parents pass down small (a) =

¼]

- P[Dad is carrier (Aa) ] = ⅔

- P[Mom is carrier (Aa) ] = ⅔

-⅔ X ⅔ X ¼ = 1/9

## Document Summary

Lifesci 4 - lecture 20 - review and cancer. A family that lives in a population where the frequency of an autosomal recessive genetic allele si 1 in 100 (0. 01). What is the probability that child c will have the disease: 1/100, 1/200, 1/400, 1/1000, 1/10000. Population is in equilibrium p = frequency(a) = 1/100 q = frequency(a) = 1 - 1/100 = 0. 99. Both parents must be heterozygous for c to have the disease. P[c is aa] = p[dad is aa] x p[mom is aa] x is approximately. 1 2 2 1 q. Below is a pedigree from a family that lives in a population where the frequency of an autosomal recessive genetic allele is 1 in 3 (0. 33). What is the probability that child b will have the disease: 1/4, 1/9, 1/16, 1/20, 1/100. Explanation : (cid:987) = frequency for (q) (cid:988) = frequency for (p) For child b to be homozygous recessive, both parents must be heterozygous.