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.

For simplicity, we will assign the letter "a" for the gene of interest. Thus, the heterozygous genotype is Aa. You may sketch a punnet square of the cross: Aa x Aa to help illustrate that the combinations result in 50% chance of heterozygous offspring.

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.

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. The probability of an offspring being dominant for both traits is thus nine out of sixteen.

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.

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.

Arbitrarily, we may assign the letter "b" for the gene of interest. The cross then is as follows: Bb x Bb. Each parent has a 50% chance of donating a recessive (b) allele to the offspring. We must multiply these probabilities to get the chance of a homozygous recessive offspring.

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.

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)