When working with more than one gene, I recommend that you work with each pair separately.
I have never found a problem that failed to be solved working with one pair of alleles at a time.
For help with working with one pair of alleles, see Solving Mendel's Problems Part I.
After you have solved each individually, combine them.
Example 1:
Several dihybrid pea plants, with alleles Aa and Bb, are selfed. Assuming independent assortment of the alleles and a dominant mode of inheritance, what would be the expected genotypic and phenotypic ratios?
1st step: The Aa monohybrid cross is done as described in Part I. Solve for the Bb cross, also.
2nd step: Combine them:
Multiply them using a matrix:
First multiply the top line by 1BB. You will get 1AABB : 2AaBB : 1aaBB.
Then, multiply the top line by 2Bb. You will get 2AABb : 4AaBb : 2aaBb.
Finally, multiply the top line by 1bb. You will get 1AAbb : 2Aabb : 1aabb.
I usually continue adding the results to a line. My results would look like this:
1AABB : 2AaBB : 1aaBB : 2AABb : 4AaBb : 2aaBb : 1AAbb : 2Aabb : 1aabb.
Now if it was a trihybrid, we can just add the next monohybrid cross to the first two.
An example is shown below where the effect of the next gene is that it produces either blue or white flowers.
3 Blue
1 White
First T or S is tall or short, second B or N is broad or narrow, and third B or W is blue or white. What if the next gene produced a 1 to 1 ratio? (Hint: when solving problems of this nature, it is always simplest to add any 1:1 ratios, last.) Here is a useful shortcut for solving certain problems. Example: Plants with the genotype AABbccdd are crossed to plants with the genotype aaBbCcDd. What proportion of plants will have the genotype AaBbDd?
The ratio for Aa is 1, for Bb is 2 and for Dd is 1. Multiply these to get 2. What about the proportions?
Multiply these: 1 X 4 X 2 X 2 = 16 to get the denominator. Answer: 1/8. Another method is to simply multiply the probabilities of the individual outcomes. 1/1 X 1/2 X 1/2 X 1/2 = 1/8 How many different genotypic classes would you expect?
If you multiply these out, 1 X 3 X 2 X 2 = 12, you get the number of different genotypic classes expected. You would use the same algorithm for the number of different phenotypic classes, but be sure that you include your knowledge of the mode of expression (complete dominance , incomplete dominance or codominance) for each of the traits. How do you solve genetics problems that have all the information combined? Example: Two pure-bred strains of a certain plant are crossed. One strain has purple flowers, broad leaves and smooth seeds. The other has white flowers, narrow leaves and wrinkled seeds. The F1 plants are crossed to the parent strain that has purple flowers. Among the progeny are counted 101 purple, broad, smooth; 107 purple, narrow wrinkled; 96 purple, broad, smooth; and, 99 purple, narrow, wrinkled. Which alleles of each of the genes were dominant? Solution Method: Work out the problem looking at each gene, separately. First, the color gene, all F2 progeny were purple, even though we know some had to have gotten the white allele because it was in the F1. Therefore, purple must have been dominant. Second, the broad leaf, 101 + 96 of the progeny had broad leaves. This equals 197. Narrow leaves where found in 107 + 99 of the progeny. This equals 206. If broad leaves were dominant, and we crossed to broad-leafed plants, all would have had broad leaves. Instead, we got about half. This is consistent with broad leaves being recessive, so narrow is dominant. Solve the last part yourself. Epistasis Please look at the lecture notes for chapters 3 and 4 for a discussion on solving problems having to do with epistasis.
1 Aa
1 BB
2 Bb
1 bb
1 Cc
1 cc
1 Dd
1 dd
1 Aa
1 BB
2 Bb
1 bb
1 Cc
1 cc
1 Dd
1 dd