Objectives 12

How to set up genetics problems

1) Always write the known information before starting your cross

Write down a key for the symbols you are using for the allelic variants of each gene.

Use the same letter to designate both alleles for a gene (e.g. A and a).

Use uppercase to designate a dominant allele (A).
Use lowercase to designate a recessive allele (a).

Always use standard designations for the generations.

The beginning generation is called the P, or parental generation.
The offspring of this generation is called the F₁, first filial generation.
The offspring resulting when two F₁ individuals is bred is called the F₂, second filial generation.

2) Figure out the genotypes of the parents. You will either be given this information, or you'll be given clues.

Are they from true breeding lines? If so, they should be homozygous.
If they express the recessive phenotype, you can assume that they are homozygous recessive.
Is the phenotype of the offspring given? Sometimes that will give you a clue.

3) Figure out what kinds of gametes the parents can produce. Remember that every gamete carries one allele for every gene.

If it is a on gene cross, we must apply the principle of segregation:

A homozygote for a particular gene has only one form of the allele and makes only one kind of gamete (e.g. AA forms A and aa forms a).
A heterozygote for a particular gene has two allele types and forms two different gametes (Aa forms A and a).

If it is a two gene cross we must apply both the principle of segregation and the principle of independent assortment. Remember that every gamete carries one allele for every gene. The gamete has to have an “A” type allele and a “B” type, but never two of either.

A homozygous for two genes (e.g. AABB) has one allele type for each gene.

There is only one possible kind of gamete (AB).

A heterozygous for two genes AaBb has two types of alleles for each gene: A and a, B and b. The assortment of "A" and "a" into gametes is independent of the assortment of "B" and "b".

There will be four types of gametes possible: AB, Ab, aB, and ab.

4) Set up a Punnett Square for your mating.

Place the possible types of gametes (sperm) from one parent (father) down the left side and the gametes (eggs) from the other parent (mother) across the top.
Your Punnett Square must have a column for each kind of egg, and a row for each kind of sperm.
Fill in the Punnett Square with the offspring by matching the sperm allele on the side, with the egg allele across the top.
Avoid confusion by consistently placing the dominant allele first, and the recessive allele second in heterozygotes. (Aa, not aA).
If it is a dihybrid cross, it is very important to always write the two alleles of one gene first and the two alleles of the other gene second.

It doesn't matter which gene you write first, but once you've decided on the order, it is critical that you maintain it consistently.

5) Figure out the genotype ratio for your predicted offspring.
6) Figure out the phenotypic ratio for your predicted offspring.
7) Answer the question you've been asked.

Practice for gamete determination and Punnett Square set up.
One gene setup:
1) In peas, yellow color is dominant to green. Predict the phenotypes (and their proportions) of the offspring of the following crosses.

A) homozygous yellow and green.
B) heterozygous yellow and green
C) heterozygous yellow and homozygous yellow
D) heterozygous yellow and heterozygous yellow

First, write out the known information and designate the letter for your gene.
        In this case I will call the gene Y for yellow.
            Y - dominant
            y - recessive

A) For homozygous yellow and green: Notice, these are true breeders. This is the same P generation Mendel used.
P YY x yy

        Next, determine your gametes from each parent:
                YY would produce all gametes with the Y allele
                yy would produce all gametes with the y allele

Set up your Punnett Square

Y Y

y Yy Yy

y Yy Yy

In this case, there is only one type of offspring for the F₁,all with a heterozygous genotype (Yy), and a yellow phenotype.

B) heterozygous yellow and green
P Yy x yy (This would have been a test cross if we had not been told that the yellow pea was heterozygote.)

Y y

y Yy yy

y Yy yy

The genotype of the offspring are half Yy to half yy (1:1).
Their phenotype is half yellow to half green (1:1).

C) heterozygous yellow and homozygous yellow
P Yy x YY

Y y

Y YY Yy

Y YY Yy

The genotype of the offspring are half YY to half Yy (1:1).
Their phenotype is all yellow.

D) heterozygous yellow and heterozygous yellow
P Yy x Yy

Y y

Y YY Yy

y Yy yy

The genotype of the offspring are one YY to two Yy to one yy (1:2:1).
Their phenotype is three yellow to one green (3:1).

Two gene setup:
2) In guinea pigs, black fur is dominant to brown, short fur is dominant to long fur. Predict the phenotypes (and their proportions) of the offspring of the following crosses.

A) (female) homozygous black and short fur with (male) brown and long fur.
B) (female) heterozygous black and short fur with (male) brown and long fur.
C) (female) homozygous black, heterozygous short with (male) heterozygous black and long fur.
D) (female) heterozygous black and short fur with (male) heterozygous black and short fur.
B - black fur (BB or Bb) S - short fur (SS or Ss)
b - brown fur (bb) s - long fur (ss)

A) (female) homozygous black and short fur with (male) brown and long fur. Notice that both parents are true breeders.
P BBSS x bbss (we know the genotype of the male because we were told that these are the recessive traits)
F₁

	BS	BS
bs	BbSs	BbSs
bs	BbSs	BbSs

The genotype of the offspring are all heterozygous for both genes, BbSs.
Their phenotype is all black, short fur.

B) (female) heterozygous black and short fur with (male) brown and long fur.
P BbSs x bbss (This could be a test cross had we not been told that the female was heterozygote.)
F₁

BS Bs bS bs

bs BbSs Bbss bbSs bbss

bs BbSs Bbss bbSs bbss

The genotype of the offspring are one BbSs, to one Bbss, to one bbSs, to one bbss (1:1:1:1).
Their phenotype is one black, short fur to one black, long fur to one brown, short fur, to one brown, long fur (1:1:1:1)

C) (female) homozygous black, heterozygous short with (male) heterozygous black and long fur.
P BBSs x Bbss
F₁

BS Bs

Bs BBSs BBss

bs BbSs Bbss

The genotype of the offspring are 1:BbSs, 1:BBss, 1:BbSs, 1:Bbss (1:1:1:1)
Their phenotype is two black, short fur, to two black, long fur (1:1) (notice that they are all black but the fur length varies).

D) (female) heterozygous black and short fur with (male) heterozygous black and short fur.
P BbSs x BbSs (Notice that this is a dihybrid cross We started the P generation with two dihybrids. Mendel always started his P generation with two true breeders (as in part 2A) resulting in two dihybrids in his F₁generation. He crossed the two dihybrids for his F₂ generation.)

F₁

BS Bs bS bs

BS BBSS BBSs BbSS BbSs

Bs BBSs BBss BbSs Bbss

bS BbSS BbSs bbSS bbSs

bs BbSs Bbss bbSs bbss

The genotype of the offspring are one BBSS, two BBSs, one BBss, two BbSS, four BbSs, two Bbss, one bbSS, two bbSs, one bbss (9 different genotypes)
Their phenotype is nine black, short fur, to three black, long fur, to three brown short fur, to one brown, long fur (9:3:3:1).