People with albinism have little pigment in their skin, hair, and eyes. The gene that governs albinism has two forms (called alleles), which we denote by a and A. Each person has a pair of these genes, one inherited from each parent. A child inherits one of each parent's two alleles independently with probability 0.5. Albinism is a recessive trait, so a person is albino only if the inherited pair is aa. a. Beth’s parents are not albino, but she has an albino brother. This implies that both of Beth’s parts have type Aa. Why?b. Which of the type aa, Aa, AA could a child of Beth’s parents have? What is the probability of each type?c. Beth is not albino. What are the conditional probabilities for Beth’s possible genetic types, given this fact?

a)

If Beth’s parents are not albino, but she has an albino brother (aa), then each parent must have the allele a. In order for someone who is not albino to be able to have an albino child of albinism, they must have type Aa.

b)

P(AA) = 0.5(0.5) = 1/4  CHANCE

P(Aa) = 0.5(0.5) = 1/4  CHANCE

P(aa) = 0.5(0.5) = 1/4   CHANCE

c)

the new probabilities are:

P(AA | not aa) = 1/3  

P(Aa | not aa) = 2/3

Step-by-step explanation:

   Given the data in the question;

a) Beth’s parents are not albino, but she has an albino brother. This implies that both of Beth’s parts have type Aa. Why?

If Beth’s parents are not albino, but she has an albino brother (aa), then each parent must have the allele a. In order for someone who is not albino to be able to have an albino child of albinism, they must have type Aa.

To be fully albino you must be aa and must also have at least one a as well as have a child with someone who also has has at least a.

b) Which of the type aa, Aa, AA could a child of Beth’s parents have? What is the probability of each type?

                             A                 a

A                         AA                Aa

a                         aA                 aa

so, Since each type is equally likely, and each is inherited independently of each other then,

P(AA) = 0.5(0.5) = 1/4  CHANCE

P(Aa) = 0.5(0.5) = 1/4  CHANCE

P(aa) = 0.5(0.5) = 1/4   CHANCE

c) Beth is not albino. What are the conditional probabilities for Beth’s possible genetic types, given this fact?

First we rule out aa, since Beth is not an Albino.

the new probabilities are:

P(AA | not aa) = 1/3  

P(Aa | not aa) = 2/3

(n^3 - n)

= n(n-1)(n+1)

since it is divisible by 3 and 2 ,because it is a multiple 3 consecutive number

if, n-1 is odd then n+1 is also odd, then it is a multiple of 9

but if, n-1 is even then n+1 is also even, then it is a multiple of 4

so, largest multiple = max(9*2, 4*3) = max(18, 12) = 18

the answer is 175.4379

step-by-step explanation: