When a homozygous tall pea plant and a homozygous short pea plant are crossed, the next generation are all heterozygous and the dominant allele is expressed in all the plants, and the recessive trait seems to disappear. Since we are not told whether the tall or short allele is dominant, we cannot assume either.

Dominant alleles are neither better, nor worse than recessive alleles. They are simply expressed in the phenotype of a heterozygous pair. Several deleterious, and even deadly disorders like Huntington's disease, are inherited via an autosomal dominant pattern.

Note that the frequency of an allele or genotype in a population must be determined experimentally, and can be linked to environmental influences, autosomal versus sex-linked patterns, and numerous other factors.

The genotype describes the genetic makeup or alleles of the individual, while the phenotype describes the physical characteristic of the individual.

The parent generation refers to the generation in which the two organisms are crossed; the F1 generation is the first filial generation, or the offspring produced from the cross; the F2 generation is the second filial generation, or the offspring produced from a cross between two F1 organisms.

When attempting to determine if an organism is heterozygous or homozygous for a dominant trait, it is best to use a test cross. A test cross involves crossing the flower in question with a homozygous recessive flower. Since a white flower can only contribute a white allele, we can determine if the purple flower in question is heterozygous or homozygous. Any white flowers in the next generation will confirm that the purple flower is heterozygous. If they are all purple, we can confirm that the flower is homozygous for the trait.

The F₁ generation will be entirely heterozygous, thus the F₂ generation is the result of a heterozygous self-cross. A Punnet square reveals that 75% of the generation will be purple (PP or Pp) and 25% will be white (pp). Of the three purple flowers in the punnett square, two of them are heterozygous for color (Pp). The other flower is homozygous for the purple allele (PP). In addition, the white flower is homozygous for the recessive white allele (pp).

There are two homozygous flowers, and two heterozygous flowers in the punnett square, so 50% is the correct answer.

The possible genotypes of the unknown plant are GG, Gg, or gg. To create the F1 generation, it is crossed with a plant with genotype GG.

Scenario 1: GG x GG, result is all GG in F1; F2 cannot possibly contain a yellow (gg) plant

Scenario 2: Gg x GG, result is half Gg and half GG; F2 will contain both green and yellow plants of all genotypes

Scenario 3: gg x GG, result is all Gg in F1; F2 will be 75% green and 25% yellow

In order to have the ratios described in the question, the unknown plant must be homozygous recessive.

This question requires us to do a dihybrid cross. We can represent the gene for beak color with the symbol "A" for dominant yellow and "a" for the recessive orange. Likewise, for feather color, we can use "B" for blue feathers and "b" for black.

The problem states that both birds are heterozygous for each trait, implying that our cross is between two birds with the genotype AaBb.

Now we look at the gametes that can be produced by these parents: AB, Ab, aB, and ab. These gametes can then be used to make a punnet square.

Offspring: 1 AABB, 3 Aabb, 8 AaBa, 3 aaBb, 1 aabb

There are 16 total offspring. 12 of them carry the dominant A allele, giving them the yellow beak phenotype.

Yellow beaks: 1 AABB, 3Aabb, 8 AaBb

Finally, of these 12, 9 carry the dominant B allele for blue feathers.

Yellow beaks and blue feathers: 1 AABB, 8 AaBb

This gives a total of 9 out of the 16 offspring that will express both the yellow beak and blue feather phenotypes.

You should be familiar with the 9:3:3:1 phenotypic ratio resulting from dihybrid crosses.

Because we are only told the phenotype of the naturally colored parent, we do not know if it is homozygous dominant or heterozygous. To account for this, we must anticipate the phenotypic ratios of both dominant genotypes. Two crosses must be performed: one between a homozygous dominant parent and a homozygous recessive parent, and one between a heterozygous parent and a homozygous recessive parent. The resulting ratios would be 100% natural and 50% albino/50% natural respectively. 

AA x aa: all offspring Aa (natural)

Aa x aa: half offspring Aa (natural), half offspring aa (albino)

This question requires that we do a dihybrid cross. The cross in question is AaBb x AaBb, using A to represent dominant green color and B to represent dominant long shape. The parents are heterozygous for both traits, meaning they will carry one dominant color allele and one dominant shape allele.

The result of a punnet square for a dihybrid cross is: 1 AABB, 3 Aabb, 8 AaBb, 3 aaBb, 1 aabb.

This gives a total of sixteen different offspring. Two different genotypes carry dominant alleles for both traits: AABB and AaBb. There are a total of nine offspring between these two genotypes that will be green and long. The other genotypes all represent different phenotypes. Aabb will be green and round. aaBb will be white and long. aabb will be white and round. Together, this gives a final phenotypic ratio of 9:3:3:1.