The genotype for Sharon is rr, because blonde is a recessive trait therefore in order to have blonde hair she must be homozygous recessive. Her husband is Rr, because it states that he has brown hair, which is dominant, in addition to being heterozygous. When drawing out a punnet square, you will find the offspring will be Rr, Rr, rr and rr. Therefore, their daughter has 50% chance of having brown hair and 50% chance of having blonde hair.

The problem can be solved using the following Punnett square. Since short is recessive, the only genotype that will result in a short appearance is tt. tt occurs one time on the Punnett square, out of four possible combinations; therefore, there is a 1-in-4 chance of giving birth to a short child, or 25%.

Linda's mother has brown eyes: BB

Linda's father has blue eyes: bb

Using the pedigree (BB x bb), Linda has to be Bb (100%). Linda is marrying someone with blue eyes (bb). Doing the pedigree of Linda (Bb) and her partner (bb) gives you 0.50 Bb (brown eyes) and 0.50 bb (blue eyes).

Once you properly set up your punnet square of a dihybrid cross, you should obtain a phenotypic ratio of 9:3:3:1. There will be 9 plants with dark blue leaves/yellow flowers, 3 plants with light blue leaves/yellow flowers, 3 plants with dark blue leaves/orange flowers, and 1 plant with light blue leaves/orange flowers.

Since the parents are pure breeding, we can designate the purple flowers as PP and the white flowers as pp. Upon breeding, there will only be one possible plant created, which has a genotype of Pp.

PP x pp

Offspring: all offspring are Pp and will display the dominant phenotype (purple)

Upon self fertilization, there will be three genotypes in the second generation: PP, Pp, and pp. A punnet square will show that 50% of the second generation will be homozygous and 50% of the second generation will be heterozygous for the color trait.

Pp x Pp

Offspring: one PP, two Pp, one pp. PP and pp are homozygous, and the two Pp are heterozygous.

Both of the parent's genotypes are Kk. This means that the offspring can be either KK, Kk, or kk.

Parental genotype:

Offspring probability:

When dealing with two traits, it helps to approach each trait separately. The question asks for the fraction of plants in the second generation that have purple flowers and wrinkled peas.

First, we will look at the first generation. The parents are both pure breeding, meaning they are homozygous for each trait. One is purple (dominant) and wrinkled (recessive), while the other is white (recessive) and round (dominant). The result will be dihybrid offspring.

Parent cross: PPrr x ppRR

Offspring: All offspring will be PpRr and exhibit both dominant phenotypes (purple and round).

Now we will look at the first generation self-cross for each trait.

First generation: PpRr x PpRr

Offspring for color: 1 PP, 2 Pp, 1 pp; 3 purple and 1 white.

Offspring for seeds: 1 RR, 2 Rr, 1 rr; 3 round and 1 wrinkled.

We can see that three-fourths of the offspring will be purple, since they carry the dominant allele, and one-fourth of the offspring will be wrinkled, since only one of four will carry two recessive alleles.

Since we are looking for plants that have both of these traits, we multiply these two probabilities together.

In other words, of the second generation will have purple flowers and wrinkled peas.

Biologists use a diagram called a Punnett square to aid them in predicting traits that will be inherited by offspring. In this case both parents are heterozygous, meaning that they both carry one dominant allele (W) and one recessive allele (w). The Punnett square for this cross would look like this:

Because the W allele is dominant, squares with a W will produce the large wing trait. Three of the four possible genotypes from this cross carry the dominant allele, meaning that 75% of the offspring will have large wings. If the parents produce sixteen offspring, then twelve of them will show the large wing phenotype.

If either parent were homozygous dominant, all offspring would exhibit the dominant phenotype and the second phenotype would not be expressed.

Homozygous parent cross: RR x r_

Offspring: all offspring inherit a dominant R allele and will express the dominant phenotype.

If two heterozygotes are crossed, all three genotypes will be expressed, so all possible phenotypes will be expressed.

Both heterozygous: Rr x Rr

Offspring: 1 RR, 2 Rr, 1 rr with both dominant and recessive phenotypes expressed in a 3:1 ratio.

If one heterozygote and one homozygote are crossed the offspring will show only two genotypes, but both possible phenotypes.

One heterozygote: Rr x rr

Offspring: half Rr and half rr, with half showing dominant phenotype and half showing recessive phenotype.

The second and third examples give rise to both colors of offspring, regardless of which allele is dominant. The first example only shows the dominant phenotype, can cannot be possible for this question.

For better visualization of this explanation we can use eye color as an example. BB and BB will be brown eyes and bb will be blue eyes.

In a monohybrid cross you are dealing with only one gene. We can use "Bb" as an example for the gene as it is a hybrid of two alleles. BB & Bb are dominant and bb is recessive.

Setting up a punnet square of Bb x Bb we will get 4 results: BB, Bb, Bb, and bb.

The phenotype is what we see: eye color. Phenotype will be expressed the same for BB and Bb, therefore we have 3 that will express the dominant phenotype and 1 that will express the recessive phenotype; 3:1.

The genotype is the actual gene that creates the phenotype. So for this we have 3 different genes that arise from the monohybrid cross: BB, Bb, and bb. We get 1 homozygous dominant, 2 heterozygous, and 1 homozygous recessive; 1:2:1.