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.

The question tells us that green is dominant to blue. This means that the C allele must correspond to green and the c allele must correspond to blue. We are looking for the cross that will produce the most blue offspring. Since blue is recessive, all blue offspring will be homozygous recessive; thus, we are essentially looking for the cross that carries the most recessive alleles.

Of the given crosses, cc x cc contains the greatest number of recessive alleles. Every offspring from this cross will be homozygous recessive and display the blue phenotype. This cross thus gives the highest percentage of blue offspring, with that percentage being 100%.

Heterozygous organisms carry one dominant allele and one recessive allele. The dominant allele is expressed over the recessive allele, giving the organism the dominant phenotype. If the heterozygous rat in the question is brown, then we can conclude that brown is dominant to white.

The cross of these two rats would be:

Parents: BB (brown) x Bb (brown)

Offspring: half BB (brown), half Bb (brown)

Though half of the offspring will be homozygous and half will be heterozygous, all offspring will be brown. None of the offspring from this cross will show the white phenotype.

By creating a Punnett square in which the male carries the label Dd and the female has the label dd, one can see the possible combinations of alleles that the child may have. D will symbolize the allele for dimples, and d will symbolize the allele for no dimples. The cross will be Dd x dd.

There is a 50% chance that the child does not obtain the allele needed for dimples (dd), and a 50% chance that the child is heterozygous (Dd). Because dimples is an autosomal dominant trait, heterozygosity will express dimples, leading to a 50% chance that the child will have dimples. 

We are told that the yellow parent is heterozygous, so we can set up a simple cross. We know that the orange insect must be homozygous recessive due to the fact that it displays the orange phenotype.

In our cross, we will use A as the dominant allele and a as the recessive allele. This gives the yellow parent a genotype of Aa and the orange parent a genotype of aa.

Aa x aa

Possible offspring: Aa, Aa, aa, aa

We can see that there are only two possible genotypic results. Half of the offspring will be Aa and half will be aa. The Aa offspring will be yellow and the aa offspring will be orange, giving a 1:1 phenotypic ratio.

Let's look at each trait individually. We can see that all of the offspring from this cross will be tall; the tall allele must be dominant over the recessive allele. We also know that one parent must be homozygous dominant for the tall allele, since it is passed to every single offspring. The parent genotypes for height are TT (tall) and tt (short). The short plant must be homozygous recessive to show the recessive phenotype.

Now, let's look at color. Half of the offspring are green and half are yellow. This tells us that whichever parent displays the dominant phenotype is also heterozygous, allowing the recessive phenotype to appear in the offspring. Since we know that green is dominant, we know that the green plant must be heterozygous. In order to display the recessive yellow phenotype, the yellow plant must be homozygous recessive. The parent genotypes for color must be Gg (green) and gg (yellow).

Together, we can see that the tall-green plant will be TTGg and the short-yellow plant will be ttgg.

The ratio of kittens with black fur to kittens to white fur is 3:1. Since the recessive phenotype is shown in the offspring, we know that both of the parents are carriers of the recessive genotype and are thus both heterozygous, Bb.

Each parent will contribute one allele at random from each of the two genes. The odds of having brown fur, given the parental genotypes, is 100% because all children will receive a dominant allele for brown fur (B) from the first parent. This means that the odds of having brown straight fur will depend only on the odds of having straight fur. Having straight fur requires one recessive allele (c) from each parent. The second parent will always contribute a recessive allele while the first will contribute a recessive allele half the time. Thus, the odds of straight fur is 50%, as are the odds of straight, brown fur. Alternatively, a Punnett square may be used. The square below shows the 50% offspring combinations with straight, brown fur highlighted in red.