We can depict the parental genotypes by using to signify the dominant white-eye allele and to signify the recessive red-eye allele.

The mother would have genotype since she has red eyes. The father would have genotype since he has white eyes.

We can see that the possible genotype for offspring will be either for a daughter or for a son. Any daughters must inherit the X-chromosome from both parents, and must therefore inherit a white-eye allele from the father. We can conclude that all daughters will have white eyes.

While this question looks complicated it can be broken down into three individual Punnett squares, and the probability of each of the three individual allele genotypes can be multiplied to give the desired answer.

Starting with P: Parent 1 is homozygous dominant (PP) and parent 2 is homozygous recessive (pp), therefore 100% of their offspring will be heterozygous (Pp), or 1/1.

For Q: Both parents are heterozygous (Qq); therefore 1/4 of their offspring will be QQ, 1/2 will be Qq, and 1/4 will be qq.

For R: Parent 1 is heterozygous (Rr) and parent 2 is homozygous recessive (rr); therefore 1/2 their offspring will be Rr and 1/2 will be rr.

Because the question asks the probability that the offspring is PpQQRr, we can multiply 1/1 * 1/4 * 1/2 = 1/8.

The parent cross (P generation)  will be PP x pp, where P is the dominant allele (purple) and p is the recessive allele (white). All offspring from this cross will be heterozygous, Pp, and will represent the F1 generation.

The F2 generation will result from a cross of the F1 generation, meaning Pp x Pp. This cross will result in 1 PP, 2 Pp, and 1 pp genotypes. Only one out of every four F2 offspring will have the recessive phenotype of white flowers.

To be diagnosed with both cystic fibrosis and Prader-Willi, this child would have inherited either of the maternal alleles and the deleted paternal allele (50% chance of this happening). If the mother's chromosome 15 genotype is said to be AA and the father is Aa, where a is the chromosomal deletion, we can see that the possible combinations are: AA, AA, Aa, Aa. There is a 50% chance of inheriting the chromosome with the deletion.

He would have also had to inherit both dysfunctional copies of chromosome 7. Since both parents were healthy, they must be heterozygous carriers. The child has a 25% chance of inheriting one damaged chromosome from each parent. If each parent has genotype Cc, we can see that the possible offspring are: CC, Cc, Cc, cc. There is a 25% chance of inheriting both recessive alleles for cystic fibrosis.

Combining these probabilities tells us the chance of inheriting both disorders.

If Mendel's law of independent assortment applied to this case, we would expect to see a normal 9:3:3:1 phenotypic ratio of offspring. Also, we know that the recessive phenotypes are not lethal because we are told that approximately 25% of the progeny were homozygous recessive for both traits. The most likely explanation is that the genes are linked (located very closely on the same chromosome) and, thus, segregate together. Even if the insects had high rates of recombination it would not affect these genes that are, according to the offspring ratios, completely linked. 

The first step to this question is to determine the parental genotypes. We will use B to signify brown eyes and b to signify blue eyes.

The man in the question has blue eyes; he must be homozygous recessive in order to possess this genotype. The woman has brown eyes, but we are told that her mother had blue eyes. Her mother, then, must have been homozygous recessive and must have passed a recessive allele to the woman. Since she displays the dominant phenotype, but has a recessive allele, she must be a heterozygous carrier. The woman's father's phenotype is not necessary for this conclusion.

The parental cross will be Bb (woman) x bb (man). This will generate two possible offspring genotypes: half Bb (brown eyes) and half bb (blues eyes). To find the probability of three blue-eyed offspring, we need to multiply the probability for blue eyes three times.

A dihybrid cross has two genes that have two alleles: Aa and Bb. In order for the progeny to be homozygous for both alleles, both genes need to have two copies of the same allele. As a result, there are four possible genotypes that allow the offspring to be homozygous for both genes at the same time: AABB, AAbb, aaBB, and aabb. This means that there are four out of sixteen variations that will result in a homozygous organism.

The genetic principle best demonstrated here is incomplete dominance. Incomplete dominance means that if a given trait has two alleles, neither allele is individually dominant to the other allele, and therefore, if an organism is heterozygous for that trait, it will display a phenotype that is intermediate between that coded for by each individual alleles. 

In this case, the parent mice are most likely BB (black fur) and bb (white fur), therefore homozygotes for fur color. As a result, any offspring that they have, will have genotype Bb. If this trait was inherited in such a manner that black fur was completely dominant to white fur, or vice versa, the mice would have either one of those fur colors in the progeny. Because the resulting fur color is grey, which is intermediate between black and white, this is evidence that the genetic principle at play here is incomplete dominance. 

Co-dominance is not the correct answer because a heterozygote for a co-dominant trait would demonstrate both phenotypes simultaneously (e.g. black and white fur color), rather than a blending of the individual phenotypes (e.g. grey fur color) as in these mice. 

X-linkage of fur color would not describe the presence of an intermediate phenotype, nor would nondisjunction. 

Spontaneous mutation could, in theory, result in a phenotype different from that of the parent mice. However, given that the resultant phenotype for the progeny mice was intermediate to that of the parents' fur colors, and the fact that all progeny demonstrated this phenotype, rather than just one or two mice, this is a more compelling demonstration of incomplete dominance.