Problems and Solutions

A)  From Kindergarten Genetics:

What’s the genotype of a tri dog if all his puppies are always either black-and-tan or tricolor?

If he's had a reasonable number of puppies (over eight), you can assume he's "pure" for E (or "clear of e").  His genotype is atatEE.  Incidentally, he can only produce wholecolor puppies if bred to a wholecolor bitch:  his whole genotype is atatEEspsp so he gives all his puppies an allele for broken color.


What kind of cross will always give nothing but blenheim puppies?

Blenheim x Blenheim.  If you cross a recessive x a recessive, you should always get only recessive-type puppies.  Blenheim is recessive for both red and spotting.


If a black-and-tan bitch has one puppy of each color, what is her genotype and what can you say about the genotype of the stud you took her to?

She must have been atatEeSsp.  She could not have produced red puppies unless she had a red allele to pass on, though she's not expressing it herself; and ditto for broken colors.  The stud must have also had red and spotting alleles to pass on, but you can't tell what color he was just by looking at the puppies.


If a blenheim puppy was produced by a blenheim bitch mated to a tri male, what was the genotype of the male?

The male must have been carrying the red allele.  His genotype must have been atatEespsp


If a ruby bitch mated to a blenheim produces a tri puppy, does this mean that some other male must have gotten to her when you weren’t watching?

Yes.  The ruby bitch was ee and the blenheim stud was also ee.  They cannot have produced a E_ puppy.


If you want to find out whether your tri dog is carrying red, what should you breed him to?

Either a ruby or a blenheim will do.  If he produces any red puppies, he's carrying red, because they'd have to get a red allele from him as well as from the mother.  Incidentally, you'd want at least eight non-red puppies before you assume that your boy is not carrying red.


B)  From Punnett Squares and Probabilities:

1.  Dragons either prefer to eat sheep or princesses, never both.  The preference for princesses is completely dominant.  They also prefer to collect either silver or gold, never both.  The preference for gold is completely dominant.  These two traits are not linked and each is controlled by a single gene with two alleles.

One the one hand you have a dragon who prefers both gold and princesses.  His mother, however, preferred both silver and sheep.  On the other hand, you have a dragon who prefers gold and sheep.  Her father preferred silver and princesses.  When these two dragons get together, what is the probability that their dragonet will prefer gold and princesses?

You could do this with a Punnett square, if you prefer.  Choose any letters you like for your genes:  say, P for princesses and p for sheep, and G for gold and g for silver.  You will make your life infinitely easier if you remember that all alleles of the same gene get the same "base" letter (some variation of the letter "p" for both alleles that control dietary preferences, for example).

Then write down what you know about the parents:

P_G_ = prefers princesses and gold.  You leave blanks unless you have other information which can fill them in, which in this case you do.  If his mother preferred sheep and silver, then, she was ppgg and this dragon must have gotten a p and a g from her.  So here's Daddy Dragon:


With similar logic, you know Mom's genotype:  she is ppGg.

So your cross is PpGg x ppGg

And your initial Punnett square would look like this:

     PG    Pg    pG    pg

Mom only gets two lines because she can only make two types of gametes:  she can't give a baby a big P because she hasn't got one to give.  Then the filled-out Punnett square would look like this:

     PG    Pg    pG    pg
   pG  PpGG  PpGg  ppGG  ppGg
   pg  PpGg  Ppgg  ppGg  ppgg

The dragonet will prefer princesses and gold if it has any genotype that fits this basic pattern:  P_G_

Three out of eight boxes have some genotype that fits this pattern.  There is a 3/8 chance that the dragonet produced will have the preferences specified by the question.


You could also do this problem without the the Punnett squares.  If you did, you would choose letters and figure out the parental genotypes as above.  Here's your cross again:

PpGg x ppGg

The chance the baby dragonet will inherit a preference for princesses from Dad is 1/2.  Mom cannot pass on such a preference and is not relevant for this trait.                                                                                          1/2 P from Dad x 1 p from Mom = 1/2 P

The chance that the baby will also inherit a preference for gold from Dad is 1/2; similarly it is 1/2 for Mom.   1/2 G from Dad x 1/2 G from Mom = 1/4 GG.                                                                                           1/2 G from Dad x 1/2 g from Mom = 1/4 Gg.                                                                                               1/2 g from Dad x 1/2 G from Mom = 1/4 Gg

Put the probabilities together to build your dragonet:

1/2 P_ x 3/4 G_ = 3/8 P_G_


2.  In Labrador Retrievers, one gene controls black and brown pigment; black (B) is dominant to brown (b).  If a dog is chocolate, even its nose and eye rims will be brown, not black – it cannot form black pigment at all.  A second gene controls yellow pigment, and not-yellow (E) is dominant to yellow (e).  If a dog is yellow, it doesn’t matter what type of B or b alleles it has:  it is yellow.  Yellow dogs that can form black pigment will, however, have black noses and eye rims.

If you cross two black Labs, both heterozygous for both genes, what phenotypic ratio would you expect in the puppies?

Here's your cross:  BbEe x BbEe.

You would expect this ratio:

9/16  B_E_ black                                                                                                                                        3/16  B_ee yellow with a black nose                                                                                                                3/16  bbEe chocolate                    .                                                                                                              1/16  bbee yellow with a brown nose


3.  In Doberman Pinschers, one gene controls black and brown pigment; black (B) is dominant to brown (b), just as in Labs (except in Dobermans we call brown dogs red rather than chocolate.  A second independent gene imposes the blue dilute on top of whatever color was created by the b-locus.  Not-blue (D) is dominant to blue (d).  Blue in combination with black gives the blue color; blue in combination with red gives fawn (Isabella).

If you cross a BbDd dog with a bbDd bitch, what is the chance that you’ll get a fawn puppy?

1/2 b from Dad x 1 b from Mom = 1/2 bb

1/2 d from Dad x 1/2 d from Mom = 1/4 dd

1/2 bb x 1/4 dd = 1/8 bbdd = one-eight chance fawn


4.  In the same system, there is a problem called “color dilution alopecia”, in which an animal with the blue dilute (dd) loses a lot of its hair and basically becomes bald.  Let’s say that this condition is controlled by a third gene, where if the dominant allele A is present in combination with the dd genotype, the dog will develop alopecia.  [This is probably not what is actually going on with color dilution alopecia!]

Now, if you cross two dogs that are both red and both carrying d and a, then what is the chance that any given puppy born will develop alopecia?

This is now a three-gene system.  Here're our parents:

bbDdAa  x  bbDdAa

This is not difficult to work with because both parents are red (pure for b) and so this gene isn't going to contribute any kind of variability to the offspring (they'll all have a base color of red, futher diluted to fawn if they pick up a d from each parent).  A three-gene Punnett square can be up to 64 squares, but you wouldn't need more than 16 to do this cross.  Or you could use probability:

To have alopecia, the puppy must be homozygous recessive for both dd and aa.  Here's our only alopecia-affected genotype from this cross:  bbddaa.

1 b from Dad x 1 b from Mom = 1 bb                                                                                                          1/2 d from Dad x 1/2 d from Mom = 1/4 dd                                                                                                  1/2 a from Dad x 1/2 a from Mom = 1/4 aa

1 x 1/4 x 1/4 = 1/16 bbddaa (affected)