As I said in part I of these tips and instructions, all genetics problems are solved according to the same pattern. When we complicate the problem by following more than one gene at a time, we aren't violating this promise. The logic of a problem involving two genes is exactly the same as for one involving only one gene. So let's see how the rules are applied.
Here's a reminder of those steps to follow:
|Step 1:||Figure out the genotypes of the parents.|
|Step 2:||Figure out what kinds of gametes the parents can produce.|
|Step 3:||Set up a Punnett Square for your mating.|
|Step 4:||Fill in the babies inside the table by matching the egg allele at the top of the column with the sperm allele at the head of the row.|
|Step 5:||Figure out the genotypic ratio for your predicted babies.|
|Step 6:||Figure out the phenotypic ratio for your predicted babies.|
|Step 7:||Answer the question you've been asked.|
Here's our problem: John and Elizabeth are have just had the first of what they hope will be several children. Both John and Elizabeth are brown eyed, but baby Elvis is blue eyed. John and Elizabeth also have free earlobes, though Elvis's tiny earlobes are completely attached.
We know the following about these genes. Brown eyes* (B) is completely dominant to blue eyes (b), and free earlobes (A) is completely dominant to attached earlobes (a).
[* Actually, the genetics of eye color is more complicated than this, but the old fashioned Brown/Blue system works almost all of the time.]
Here's our preliminary information:
So baby Elvis is homozygous recessive for both of these genes. Using our Five Rules, we can thus figure out that both John and Elizabeth are heterozygous for both of the genes. Thus, here is our mating:
Now comes the tricky part. This is where most students go astray on this kind of problem figuring out the gametes. Refer to Rule #1 from the Five Rules. Every gamete contains exactly one allele for every gene. So every one of Elizabeth's gametes must have one B and one A, but never two of either. And her eggs must include all possible combinations of the alleles that she carries. The same goes for John. So our gametes end up like this:
Now we plug these into a Punnett Square, and get on with the baby making part. Your Punnett Square must have a column for each kind of egg and a row for each kind of sperm, so you must build a four by four table. Here it is:
Note that, in the inside squares representing the kinds of babies made, the B's are grouped together and the A's are grouped together. Be consistent about this, and about the order in which you write the symbols for the two genes. It's easy to get confused if you get sloppy.
When you analyze the contents of this table, this is what you get for the genotypes:
1BBAA : 1BBaa : 1bbAA : 1bbaa : 2BbAA : 2Bbaa : 2BBAa : 2bbAa : 4BbAa
In other words, you got one of each kind of double homozygote, two of each kind of single heterozygote, and four of the double heterozygotes. This makes sense if you think about how many ways you can make each of these babies (i.e., how many different gamete combinations can produce each one).
When you ask yourself the phenotype question, "Do any of these genotypes look alike," you answer, "Yes; a lot of them do." After you figure out which ones look alike, you will get this phenotypic ratio:
9 Brown Eyes with Free Earlobes : 3 Blue Eyes with Free Earlobes : 3 Brown Eyes with Attached Earlobes : 1 Blue Eyes with Attached Earlobes.
And there you have it. All done exactly the same way our monohybrid cross was done. This kind of cross in which we are following two genes, and both parents are heterozygous for both genes is called a dihybrid cross (di = 2).
Epistasis: As soon as you introduce problems with more than one gene, you also introduce the possibility of interactions between genes. Epistasis is the interference of one gene with the expression of a totally different gene.
The classic example of an epistatic condition is albinism. The albino condition is caused by a recessive allele of a gene which we rarely pay any attention to until it isn't doing its job properly. This gene is one of those which controls the production of a molecule called melanin, which is the primary pigment molecule in human skin, hair and eye color. Melanin is brown in color. Yes, blue eyes are produced by a brown pigment. The color of the skin of fair skinned people which is actually orange, rather than pink or white is due to a combination of the light brown of low amounts of melanin, combined with the pink from blood vessels just under the skin.
An albino is unable to manufacture melanin, so all pigmentation is the result of other substances (like the blood in blood vessels). Albinos have pink-white skin and red eyes. The color in the eyes comes from the highly vascularized retina in the back of the eyeball. There is no pigmentation in the iris of the eye, where the melanin which makes blue or brown eyes is deposited.
If we use the same example we followed above, but substitute the albino gene for the earlobe gene, we can see the effect that epistasis has on the phenotypic ratio. The entire problem is worked exactly the same way, until the final determination of phenotypic ratio.
When it comes to that phenotype question, we have to recognize that any child with the aa genotype will be albino, and will have red eyes, no matter what the eye color alleles are. So our final ratio will be 9 Brown Eyed : 3 Blue Eyed : 4 Red Eyed (albino).
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Updated 25 September 2004