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.

Because we are told that the blue parent is homozygous dominant, we can set up a simple cross. We know that the black beetle must be homozygous recessive to present the black phenotype. Using B as the dominant blue allele and b as the recessive black allele, we can see that the parent beetles have genotypes of BB and bb.

BB x bb

Offspring: All offspring will be Bb and present the dominant (blue) phenotype.

All resulting offspring will have at least one dominant allele, giving us a 100% blue ratio.

We know that the yellow dog must be homozygous recessive and that the black dog must be either heterozygous or homozygous dominant. When crossed, their offspring can show either the dominant (black) phenotype or the recessive (yellow) phenotype. In order for offspring to show the recessive phenotype, they must inherit a recessive allele from each parent. To have this outcome, the black dog must carry a recessive allele even though it expresses the dominant trait; this makes the black dog heterozygous.

We can look at a punnett square to verify the result. We will use B as the dominant allele and b as the recessive allele.

Bb (black dog) x bb (yellow dog)

Offspring: Half Bb (black) and half bb (yellow)

To obtain the overall probability, multiply the individual probabilities for each locus.

Parent cross: AABbccDdEeFF x AaBBCCDdEeff

Offspring: AaBbCcddEEFf

A locus cross is AA x Aa. Half of the offspring will be AA and half will be Aa. The probability of Aa is one half.

B locus cross is Bb x BB. Half of the offspring will be BB and half will be Bb. The probability of Bb is one half.

C locus cross is cc x CC. All offspring will be Cc. The probability of Cc is one.

D locus cross is Dd x Dd. One-fourth of the offspring will be DD, half will be Dd, and one-fourth will be dd. The probability of dd is one-fourth.

E locus cross is Ee x Ee. One-fourth of the offspring will be EE, half will be Ee, and one-fourth will be ee. The probability of EE is one-fourth.

F locus cross is FF x ff. All offspring will be Ff. The probability of Ff is one.

Multiply all the probability values.

This question requires that we do a dihybrid cross. If A represents dominant color and B represents dominant shape, then both parent plants have the genotype AaBb.

Parent cross: AaBb x AaBb

Possible offspring: AABB, AABb, AAbb, Aabb, aaBB, aaBb, AaBB, aabb, AaBb

The phenotypic ratios resulting from a dihybrid cross are always 9:3:3:1, where 9 represents both dominant phenotypes, each 3 represents one dominant and one recessive phenotype, and 1 represents both recessive phenotypes. In this cross, there would be 9 long yellow seeds, 3 round yellow seeds, 3 long gray seeds, and 1 round gray seed.

Only one out of the potential sixteen offspring from the cross will carry both recessive phenotypes.

We know that the yellow dog must be homozygous recessive and that the brown dog must be either heterozygous or homozygous dominant. If B is used to represent the dominant brown allele and b is used to represent the recessive yellow allele, this means that the yellow parent must be bb and the brown parent could be either BB or Bb.

We know that all of the puppies must carry a dominant allele, since they all express the dominant phenotype. The yellow parent only carries the recessive allele. This indicates that every puppy must have inherited a dominant allele from the brown parent. The most likely genotype of the brown parent would be homozygous dominant, which would lead to all puppies being heterozygous and brown.

BB x bb

All offspring will be Bb and express the dominant brown phenotype.

Note, however, that we cannot determine the genotype of the brown parent for certain. Due to independent assortment, it is possible that a heterozygous brown parent gave the recessive allele to all offspring by chance.

Bb x bb

Half offspring will be Bb and half will be bb.

The probability of getting six brown puppies from this cross would be equal to one-half to the sixth power, or 1.56%. While this is a very small chance, it is not impossible and we cannot rule out a heterozygous genotype for the brown dog.