In man, brown eyes (B) are dominant over blue eyes (b) and normal skin pigmentat

Question

In man, brown eyes (B) are dominant over blue eyes (b) and normal skin pigmentation (N) is dominant over albinism (n). A brown eyed albino man whose mother was a blue eyed albino marries a brown eyed woman with normal skin pigmentation whose mother was a blue eyed albino. A. What is the probability that the first child born to the couple will have blue eyes and normal skin pigmentation? B. What is the probability that the first four children of the couple will have brown eyes and normal skin pigmentation? C. What is the probability that two of the first three children will be brown eyed albinos and the other a blue eyed child with normal skin pigmentation? D. What is the probability that the first child will be either a blue eyed albino or a brown eyed albino? E. What is the probability that the first child will have brown eyes and normal skin, the second will have blue eyes and normal skin, and the third will have blue eyes and be an albino?

Explanation / Answer

Please follow the images..

Since Brown eyes dominate over Blue eyes, Brown eyes are BB or Bb.

Blue eyes = bb

Similarly Normal pigmentation is dominant and therefore = NN or Nn.

Albino = nn

Since both the mothers are blue eyed albino they are = bbnn.

Therefore albino man with Brown eyes = Bbnn ( since he had to have inherited a dominant Brown or 'B' allele from his father to become a heterozygous normal)

Also similarly for his wife, who is a Normal woman with Brown eyes = BbNn ( she had to have inherited both her normal alleles 'B' and 'N' from her father)

So a cross between a Bbnn male and a BbNn female creates the Bn and bn gametes from male and the BN,Bn,bN,bn gametes from female.

A punnette square of the F1 progenies gives :

3 Brown eyes + Normal

3 Brown eyes + albino

1 Blue eyes + normal

1 Blue eyes + albino

Total possibilities = 8

Therefore for (A) first child Blue eyes+ Normal = 1/8 = 0.125

(B) first child Brown eyes + Normal = 3/8 and probabilities of each child is independent of the other child..

Therefor First four children Brown eyes + Normal = 3/8 * 3/8 * 3/8 * 3/8= 81/4096 = 0.0198

(C) One child Brown eyes + albino = 3/8

First and Second child both Brown eyes + albino = 3/8 * 3/8

First three children with two of them Brown eyes + albino and the other Blu eyes + Normal = 3/8 * 3/8 * 1/8= 9/512 = 0.0178

(D) Blue eyes + albino = 1/8

Brown eyes + albino =b 3/8

Therefore Blu eyes + albino OR Brown eyes + albino = 1/8 + 3/8 = 4/8 = 1/2 = 0.5

( here its plus'+' because the too possibilities are not independent of each other)

(E) First child = Brown eyes + Normal = 3/8

Second child = Blu eyes + NOrmal = 1/8

Third child = Blu eyes + albino = 1/8

Therfore all these 3 conditions occuring at the same time = 3/8 * 1/8 * 1/8 = 3/512 = 0.00586

Sorry I wasn't able to upload the images.. Hope this helps. Good Luck!

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