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I\'ll like you back and give a good feedback if you do good job. This has to do

ID: 3239964 • Letter: I

Question

I'll like you back and give a good feedback if you do good job. This has to do with "Markov Chain" please solve it only if you know what is that

Let's model the inheritance of a genetic trait, for example eye color. Every individual has a pair of chromosomes for eye color, and an offspring gets one chromosome from cach parent. Say a represents a brown-cyc gene, and Urepresents a blue-cyc gene, and that brown is dominant meaning the eye will be brown if the individual carries at least one a gene: and only bb pair will result in blue eyes). The order in the pair does not matter. We also know that throughout the population, the probability that a randomly closcm eye-color gene is a is given by p (so P 1 p), and that mate selcction is ependent of eye color. How can we model the gene passage from a parent to a child using Markov Chain? Please specify states and transition probabilities.

Explanation / Answer

The states are either blue eyes or brown eyes

Blue eyes are only possible in bb case.

P(blue eyes) = P(bb) = (1-p).(1-p) = (1-p)^2

P(brown eyes) = 1-(1-p)^2

P(ab) = p*(1-p) = P(ba)

Child can have blue eyes only in bb case, i.e. each parent has at least one b chromosome.

Possibilities are:

(bb,bb) - always blue, prob = 1*1 = 1

(bb,ab) - blue if b from 2nd parent chosen, prob = 1*1/2

(bb,ba) - blue if b from 2nd parent chosen = 1*1/2

(ab,bb) - blue if b from 1st parent chosen, prob = 1*1/2

(ab,ab) - blue if b from both parent chosen = 1/2*1/2

(ab,ba) - blue if b from both parent chosen = 1/2*1/2

Similarly we have for (ba,bb), (ba,ab), (ba,ba) as well

P(child has blue eyes|both parents have blue eyes) = 1

P(child has blue eyes|one parent has blue eyes and other brown) = 2*(2*1/2*(1-p)^2.p.(1-p)) = 2p(1-p)^3

P(child has blue eyes|both parents have brown eyes) = 4*1/2*1/2*p(1-p)p(1-p) = p^2.(1-p)^2

Similarly, we get

P(child has brown eyes|both parents have blue eyes) = 1-1=0

P(child has brown eyes|one parent has blue eyes and other brown) = 1-2p(1-p)^3

P(child has blue eyes|both parents have brown eyes) = 1- p^2.(1-p)^2

Hence transition probabilities are : P(blue|blue,blue) = 1, meaning that child has blue eyes given both parents have blue eyes

P(blue|blue,brown) = 2p(1-p)^3 and so on

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