BIOLOGY QUESTION HELP BELOW?

A bit of careful thinking needed...

Call the normal-colouring gene N and the albino gene A

Because the gene is recessive, an albino must have A for each copy (AA).

Ross's first child is AA and must have got one gene from Ross. Since Ross is not an albino, Ross must be NA.

Since Rachel's parents are not albino but her sister is, each parent must be NA. I can’t draw Punnett squares here so I’ll do it 'long-hand'. The possibilities for Rachel’s genes are:

1. N from mum + N from dad = NN

2. N from mum + A from dad = NA

3. A from mum + N from dad = NA

4. A from mum + A from dad = AA

Since Rachel is not albino we exclude 4. She must be 1, 2 or 3. So there is a probability of 1/3 that Rachel is NN (which means a child ca not be albino) and 2/3 that she is NA.

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Call Ross’ and Rachel’s child ‘X’.

There is a probability of 2/3 that Rachel is NA. In this case the possibilities for X genes are:

1. N from mum + N from dad = NN

2. N from mum + A from dad = NA

3. A from mum + N from dad = NA

4. A from mum + A from dad = AA

The overall probability of each of these 4 possibilities is 2/3 x ¼ = 2/12 = 1/6

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The only case of AA is the last one which has a probability of 1/6.

Of course the answer can be much shorter if you don’t need to explain in detail.