Lesson #5 - Beyond Simple Mendelian Genetics

So far we have been looking at very simple one gene systems that behave in a clear cut dominant-recessive manner with clear cut differences between the alternative states that the gene can assume. Unfortunately, genes don't exist in isolation and a clear cut dominant-recessive behavior isn't always evident. Part of this is that gene expression is often affected by the environment, a topic we will get to later, but also by genes interacting with one another within the organism we are studying and by the presence of more than two alleles for a given gene we are interested in.

In some cases there is no clear cut dominant-recessive behavior when crossing two individuals of different phenotype. The resulting hybrid is some blend of the two traits or intermediate between the two traits. This is called co-dominance. An example would be crossing a red flower with a white flower and getting a pink hybrid. In this case neither the red allele nor the white allele is dominant over the other. If we go back to the previous lesson that covered how genes act we can get a picture of what is happening.

Lets say the red allele is R and the white allele is r. A red flower would be genotype RR and a white flower would be genotype rr. The rr genotype doesn't produce any red pigment, so it is a white flower. So, how do we get a pink hybrid from the genotype Rr? What happens is that the R allele is not able to produce enough red pigment to produce a red flower. The lower concentration of red pigment then appears pink to us. We might be inclined to think that a pink flower was due to a different pigment than the red pigment, but dilution of pigment concentration can play tricks on our eyes. We can also run into problems when trying to study orange and yellow flowers as a diluted orange pigment may appear yellow and a blend of orange and yellow appears as gold.

Up to now we have only considered a single gene acting alone. Now, consider this case. We take a true breeding white flower; that is, when we self it all the resulting progenies are white, and cross it to another true breeding white flower. However, all the hybrid flowers are now red! So where did this "new" trait come from? The answer is dominant complementary genes - that is, two dominant genes are required for the presence of a specific trait.

Lets consider two genes, A and B on two different chromosomes. The first white flower would be genotype AAbb and the second white flower would be genotype aaBB. However, the red pigment is only produced when both the A allele in gene "A" and the B allele in gene "B" are both present. The first white flower is white because it lacks the B allele and the second white flower is white because it lacks the A allele. Now, when we cross these two white flowers we get a F1 hybrid of genotype AaBb. Since both the A and B alleles are present we get red flowers. Biochemically, we can think that gene "A" converts some precursor to a substrate and this substrate becomes the new precursor for gene "B" which produces for our example a red pigment. Precursor -> substance A -> substance B.

We can ascertain this dominant complementary gene system from examining the segregation ratios in the F2 and backcross generations as follows.

If we self, or in this case sib mate the F1 plants, we are making the 
cross AaBb x AaBb.  For the time being we are assuming the two genes 
are independent of each other.  The pollen parent would produce 
gametes AB, Ab, aB and ab in equal number - 1/4, 1/4, 1/4 and 1/4 when 
considered as a ratio.  The pod parent in this case would also produce 
the same gametes.  This gives the following Punnett Square. (Pollen 
gamets on top horizontal row, eggs cell gametes on vertical left 

        AB      Ab      aB      ab
AB|     AABB    AABb    AaBB    AaBb |
Ab|     AABb    AAbb    AaBb    Aabb |
aB|     AaBB    AaBb    aaBB    aaBb |
ab|     AaBb    Aabb    aaBb    aabb |

We can use some short hand when analyzing such cases. It makes no difference whether or not the "A" gene is genotype AA or Aa or if the "B" gene is genotype BB or Bb. We shorthand the AA or Aa combinations as A_ where the underscore can be either the dominant or recessive allele. Now we can simplify our Punnett Square as follows, leaving out the gametes and only looking at the progenies:

A_B_    A_B_    A_B_    A_B_
A_B_    A_bb    A_B_    A_bb
A_B_    A_B_    aaB_    aaB_
A_B_    A_bb    aaB_    aabb

Now, only plants with genotype A_B_ will produce red flowers. If you add up all the progenies with the A_B_ genotype you will see that 9 of the possible 16 different combinations have the correct genotype. Thus, when a dominant complementary plant of genotype AaBb is selfed or crossed to another plant of the same genotype we see a 9:7 segregation ratio for dominant:recessive in the F2 generation.

We can also get this result mathematically because each gene is segregating independently. Gene "A" in the F1 is producing gametes A and a in equal quantity. If we were looking at gene "A" as a single gene, 3/4 of the resulting progenies would be genotype A_, which would have the dominant trait, which was covered in a previous lesson. We want to know how many plants will have genotype A_B_. Since the two genes are independent all we have to do is multiply the frequency of plants that have the desired genotype at each location. In our case 3/4(A_) X 3/4(B_) = 9/16(A_B_).

The backcross of the F1 to the first white flower would be AAbb x AaBb. In this case the white parent is only going to produce gametes Ab while the F1 hybrid will produce gametes AB, Ab, aB and ab. This will give us a Punnett Square as follows:

        AB      Ab      aB      ab
Ab    | A_B_    A_bb    A_B_    A_bb  |

Thus, half the backcross progenies will be genotype A_B_ and thus be red. A 9:7 segregation ratio in the F2 generation and a 1:1 segregation ratio in a backcross to either recessive parent is a good indication for dominant complementary gene action.

To calculate the backcross ratio mathematically, notice that the pod parent (AAbb) is not segregating for the A allele; that is, all the resulting backcross progenies are going to be genotype A_ regardless of the genotype of the pollen. The frequency of A_ in the backcross progenies is going to equal 1. The "B" gene is homozygous recessive in the pod parent, so we are looking at a classic bb x Bb backcross of a single gene. One half of the progenies will be B_. Thus, we get 1(A_) X 1/2(B_) = 1/2(A_B_). Notice how in both the F2 generation and in the backcross we looked at each gene separately as if we were dealing with a single gene. In fact, we are dealing with single genes, but we are also looking at how these two single genes interact with each other.

There is one other gene interaction that we sometimes see that has some influence on how we look at tetraploid genetics, and that is duplicate genes. Duplicate genes occur when the same identical gene occurs at two different locations in the genome. For our discussion we will assume the duplicate genes are on separate chromosomes.

Lets go back to our true breeding red flower, but now it is genotype RRRR. We cross it to a true breeding white flower, rrrr to get the F1 RRrr. If we self this RRrr hybrid we get gametes RR, Rr, rR and rr. Since only one R allele is needed the only genotype that can produce a white flower is genotype rrrr. Only one of four pollen grains will carry the rr genotype and only one of four egg cells will carry the rr genotype. Thus, 1/4(rr) X 1/4(rr) = 1/16(rrrr). Thus, in the F2 generation duplicate genes will give a 15:1 ratio for dominant:recessive. In the backcross of the F1 to the recessive white parent we have rrrr X RRrr. The pod parent is going to produce 100% rr gametes while the pollen parent will produce gametes RR, Rr, rR and rr. 3/4 of these gametes will be R_ and 1/4 will be rr. 1(rr) x 1/4(rr) = 1/4(rrrr) in the backcross progenies. Thus, duplicate genes produce a 15:1 ratio in the F2 generation and a 3:1 ratio in the backcross progenies for dominant:recessive.

Note that a single gene behaving as a classic dominant - recessive gene will segregate 3:1 for the dominant trait in the F2 generation. However, if the gene is duplicated the segregation ratio is now 15:1 for the dominant trait. Thus, it becomes much more difficult to get the recessive trait. This can lead to phenotypic buffering for traits that are very desirable for a plant to have. Duplicate genes also allow the extra pair of genes to mutate and take on new functions while not effecting the original function of the gene.