1) homozygous deep X homozygous shallow
                          DD X dd
 

2) In this question I have decided to let "S" be the dominant allele for short hair, while "s" will represent long hair. "D" will represent the dominent color dark while "d" will be the light color.

heterozygous length, recessive color X homozygous dominant length, heterozygous color
                                              Ssdd X SSDd
 

3) To determine if the straight winged fruit fly you had was herterozygous or homozygous you would have to breed it with another fruit fly that had curly wings. If the offspring from this cross all were born with straight wings then there is a good chance that the original straight winged parent was homozygous dominant. The straight winged parent would have donated the dominant straight winged allele to all his offspring. If any of the offspring were born with curly wings then the straight winged parent was heterozygous. He must have had the recessive allele in order for any of his offspring to have curly wings.

4) Yes it is possible for two black coat parents to produce a white coat offspring. This is true because if both the parents are heterozygous (Bb) the there is a 25% chance that one of their offspring will have a white coat. The reverse on the other hand is false, it is not possible for two white coat parents to produce a black coat offspring. Both parents would only have the recessive allele white for coat color therefore there is no way for the offspring to receive the dominant black.

homozygous rough black X smooth white
P                         RRBB X rrbb
F1                               RrBb (all black smooth)
P2                          RrBb X RrBb
 
 

5) Cross 1                    Cross 2
P  Hh X hh                    Hh X Hh
F1  hh                             hh

The genotype of the hornless bull was "Hh", the first cow was "hh", the first calf was "hh",the second cow was "Hh", and the second calf was "hh".

6a) Both Mr. and Mrs. Meadowmuffin must have the genotype "Tt" if one of their three children is a non-tatser. If one or both of them were "TT" then none of their children could have been a non-taster.

b i) heterozygous X hetereozygous
                       Tt X Tt
 

7) The inheritance pattern shown by the allele of the fruit fly is that in all probability normal wings shows complete dominance over curly wings. For this question I will let "N" represent normal wings while "n" will be curly wings. Since some of the offspring had curly wings and the parents had straight wings, then the genotype of both parents must have been "Nn".

8a) brown short X white short
                BbSs X bbSs
 

10) purple straight X purple curved
                    Ppcc X PPCc
 

b) The genotype of their straight haired child is "cc".

c) Two genotypes for curly haired people are "CC" and "Cc".

d) Cc X cc
  curly X straight
 

16) The simplest explanantion to explain the inheritance of color in chicks with those results is that the allele's experience some sort of incomplete dominance. Some of the chicks got two black allele's, others got two white ones, and the highest number probably got a black and a white allele. The black and the white allele both try to express themselves at the same time giving the chicks a grey color. If you mated a grey rooster with a black hen I would expect that you would probably find about 50% of the chicks were grey, while the other 50% would be black.

17) The best explanation for these results is that in cross 1 the black parent was homozyous dominant (BB), while in cross 2 the clack parent was heterozygous (Bb). 

For this problem I have decided to let "B" stand for the dominant black color, while "b" will give the albino appearance.

Cross 1                               Cross 2
Black X Albino                    Black X Albino
P  BB X bb                          Bb X bb
F1 Bb (15 of them)                    Bb ( 7 of them) and bb ( 5 of them)

18) pp    Pp    Pp (woman's father/husbands parents)
      Pp    pp (woman/husband)
      Pp    Pp    pp (their children two normal first, albino last)

The genotype of all the individuals is as follows;  the woman is "Pp", and her husband is "pp". The woman's father is "pp", and her husbands parents are both "Pp". Their two normal children are "Pp" while their albino child is "pp".

19)

20) There is a 50% that the son of a color-blind man and a normal woman (whose father was color blind) will also be color blind. Since males need a "X" and a "Y" chromosome to be males, and the female always gives an "X" chromosome, than there is a 50/50 chance that the "X" chromosome their son gets will be the one which causes color blindness. The female must have one "X" which contains the color blindness defect since her father was color blind. Their first daughter also has a 50/50 chance of being color blind since her father's "X" has the color blindness defect, while one of her mother's "X's" also does. Her father's "X" could join with her mother's color blindness causing "X" or with the one that is normal.

21)  The probability that their sons will have hemophilia is 50%. None of their daughters will have hemophilia, but 50% of them will be carriers of the disease. If the couple had four sons the probability that all four would be born with hemophilia is .

22)