There will be twice the number of Aa individuals produced though because there are two ways to generate them.

Therefore we could say that the expected ratio is 1AA : 2Aa : 1aa. This is the genotypic ratio.

If A is dominant to a, AA and Aa will be indistinguishable in terms of appearance. Another term for appearance is phenotype.

The phenotypic ratio would therefor be 3:1. If this trait was for speed of swallowing for example, with fast swallowing being dominant to slow swallowing, the ratio would be 3 fast swallowing fish for every 1 slow swallowing fish.

We can also express the expected outcome in terms of proportions. In fact some of the answers for certain questions you get on your exam and in the homework must be converted to proportions to be meaningful. In the example shown above, the genotypic proportions would be 1/4AA, 2/4(or 1/2)Aa and 1/4aa. (What would be the expected phenotypic proportions?)

If a total of 4 fish hatched, it would be possible, but unlikely that 1 would be AA, 2 would be Aa and 1 would be aa; just as it would be unlikely but possible that if you tossed a die 6 times, it would land on each number, 1 through 6. However, if 400 fish hatched, approximately 100 would be AA, approximately 200 would be Aa and approximately 100 would be aa. How can we determine if the values we actually get agree with the numbers we predicted? We shall learn some appropriate statistical tests later in this chapter.

Example 3:

See the fish below?

The sperm and eggs will each be haploid and have just one allele each.

What type of sperm will the male produce? (Approximately half will be A ; the other half will be a.)

What type of eggs will the female produce? (All will be a.)

The eggs and sperm will unite with each other at random.

We then unite the sperm with the eggs to generate the next generation's progeny.

There will be some Aa and aa individuals generated from this cross, two different genotypic types or classes.

The expected ratio is 1Aa : 1aa.

The Phenotypic ratio would therefore be 1:1, and we could see this. If this trait was for speed of swallowing for example, with fast swallowing being dominant to slow swallowing, the ratio would be 1 fast swallowing fish for every 1 slow swallowing fish.

We can also express the expected outcome in terms of proportions. In the example shown above, the genotypic proportions would be 1/2Aa and 1/2aa. (What would be the expected phenotypic proportions?)

Example 4:

See the fish below?

The sperm and eggs will each be haploid and have just one allele each.

What type of sperm will the male produce? (Approximately half will have an A ; the other half will have an a.)

What type of eggs will the female produce? (All will have an A.)

The eggs and sperm will unite with each other at random.

Next, we unite the sperm with the eggs to generate the next generation's progeny.

There will be some AA and Aa individuals generated from this cross. These are the different genotypic types or classes.

The expected ratio is 1AA : 1Aa.

We can also express the expected outcome in terms of proportions. In the example shown above, the genotypic proportions would be 1/2AA and 1/2Aa. (What would be the expected phenotypic proportions?)

Crossing to a recessive is a useful way to determine the genotype of an individual expressing the dominant allele. Is the individual homozygous dominant or a heterozygous? This is called a test cross.

If the individual with the dominant phenotype is homozygous , all progeny from a cross with a recessive would be of the dominant phenotype.

Individuals with the dominant phenotype might be either homozygous dominant or heterozygous. You know the genotype of the individuals with the recessive phenotype; what you see is what they have.

If the individual is heterozygous, half of the progeny from a cross with a recessive would be of the dominant phenotype and the other half would be of the recessive.

In the problem above, there are only two possible genotypes for the unknown individual, either AA or Aa, since the unknown individual has the phenotype of the dominant allele.

As you are just becoming familiar, guess what the geneotype of the unknown might be, then test it. If it fails your simple Punnett square test, then the answer must be the only other possiblity.

This is a case where the heterozygotes have a phenotype that differs from both homozygous alternative genotypes. Often the phenotype is intermediate. Working with problems involving incomplete dominance is actually easier than with dominant/recessive. This is because the phenotype tells you the genotype. Problem solving simply follows the methods used for genotypic outcomes.

Codominance problems are somewhat similar to incomplete dominance problems. The phenotype reveals the genotype. Problem solving simply follows the methods used for genotypic outcomes.

These can get quite complex. Especially if there is a dominance heirachy, or some of the genes are of the dominant/recessive mode and others are of the codominance mode. Always remember that each individual can have just 2 copies of the gene. Use the Punnett square when possible.

Example:

A woman, whose mother was blood type o, and whose blood type is A, marries a man whose blood type is B. They have 5 children. One has blood type o, two have B, and the rest have A. What were the blood types of the woman and her husband?

Draw a Punnett square.

What do you know?

You know that the husband has a B gene. You know that the wife has an A gene. You even know that the wife has an o gene because her mother had 2, and she had to get one from her mother. You can also determine this from the progeny as we shall see next.

How did they get a child with the o blood type? The o blood type is recessive and so must be oo. Each parent must have had an o.

Why were there no children with AB? If the couple had more childeren, you would expect one or more.

Sex-linked is really X-linked. The alleles for sex-linked genes are located on the X-chromosomes. Males have just one X and therefore what you see is what they got. Females have two X chromosomes and a recessive allele can be masked by a dominant. Also, males always get their one and only X from Mom. Females get one X from Mom and the other from Dad. It is quite easy to solve sex-linked problems using a modified Punnett square.

Just place the alleles next to the X-chromosome symbol. The Y doesn't get an allele.

Example:

Blue-green color-blindness is a sex linked trait. A color-blind man marries a woman with normal vision, whose father was color-blind. What proportion of children would you expect to be color-blind?

What do you know?

You know that the husband has just one copy of the gene and it is the allele for color-blindness. You also know that the wife has a normal allele because she is not color-blind. However, she must be a carrier because her father was color-blind, he had only one X-chromosome to give, this is where the color-blind gene was located and he gave this X to his daughter (or she would have been a he). The genotype for the wife is Cc and for the husband, cy.

Now do the cross:

The Cc are female (2 X-chromosomes) that have normal vision. The cc are females that are color-blind. The Cy are males that have normal vision. The cy are males that are color-blind.

When doing sex-linked problems, always list the outcomes separately for the females and the males.

In this case 1/2 the females will be color-blind and the other 1/2 will be normal vision. Males will be 1/2 normal and 1/2 color-blind.

Forward to part 2: Solving problems involving more than one gene

Return to Genetics'Homepage.