Please help me with these genetics problems!!! asapp

Question

I got these questions wrong on a test and wanna know why...

1. In cats, there is an X-linked gene which controls coat color: genotypes c^(y)c^(y) and c^(y)Y are yellow, c^(b)c^(b) and c^(b)Y are black, and heterozygous c^(y)c^(b) females are calico. A calico female had a litter containing a yellow male, two balck males, a calico female, and a yellow female. What was the father's genotype?

*i know the answer is c^(y)Y but i dont understand why....

2. How many unique gametes cuold be produced by an individual with the genotype AAbbCCDDEe?

3.Assume a married couple has three children, all boys. The wife is pregnant for a fourth time. What is the probability that their fourth child will also be a boy???

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Do you know a strategy to solve these type of genetic problems??? i know they shouldn't be difficult but i dont know why i keep missing them O_O

Answer 1

1)  Since the last kitten is yellow female, it's c^(y)c^(y).  That means it got a c^(y) from both the mother and the father.  Therefore the father is c^(y)Y.

Did they say what the right answers to 2 and 3 are?

Answer 2

1.  Coat color is X-linked.  So, females have two copies of the gene and males only one.  A calico female is heterozygous and so is c^(y)c^(b).  Since the calico female had a yellow female kitten, the yellow female kitten had to be c^(y)c^(y), which means that the dad had to contribute a c^(y) allele and therefore had to be c^(y)Y.  (Because the dad gave a Y chromosome to all his male kittens, and the Y chromosome doesn't carry any c alleles, then dad just gave the Y to all sons.)

2.  Gametes have one of each gene in them.  Therefore, each gamete has to have an A (or a), B (or b), C (or c), D (or d), and an E (or e).  Since the individual can only provide A, b, C, and D, all gametes have to have these.  But, this individual can make gametes with both E and e.  Therefore, there are only two possible different gametes:  AbCDE and AbCDe.  

The easiest way to solve this is to take each gene by itself and multiply all the possibilities together.  For example, in this one, you would multiple 1 x 1 x 1 x 1 x 2 because there is only 1 possibility for A, 1 for b, 1 for C, 1 for D, and 2 for E/e.

3.  1/2  The previous results do not change the probability of the next event.  Think about it like a quarter.  If you flip the quarter three times an it's heads each time, does that affect the outcome of the next flip?

However, if the couple had not yet had any children and you asked, "What is the probability that they will have four boys in a row?" then the answer is 1/16.  In that scenario you multiple the probability of each event together.  So: 1/2 x 1/2 x 1/2 x 1/2