The correct answer is 1/16. A dihybrid cross results in a phenotypic ratio of 9:3:3:1. The 1 in the ratio is indicative of the double recessive phenotype, with a homozygous recessive genotype for both traits. Thus, only 1/16 of the offspring will display both recessive phenotypes.

This problem asks you to use concepts of inheritance and Mendelian genetics. The best approach to this problem is to rule out possiblities rather than to find the actual mode of inheritance, as the latter can be a much more difficult and time-consuming process. First off, we know that Y-linked inheritance could not explain this pattern because we see that in generation 1 (G1), the male is affected. If he is affected, all of his sons (who inherit his Y chromosome) would also be affected. There is one son in G2 who is not. Similarly, dominant X-linked inheritance could not explain this pattern; recall that the daughters inherit two copies of the X chromosome, and one is always inactivated. Were the trait X-linked dominant, then the girls of generation 3 (G3) would be affected, having received a copy of the affected gene from their father. Revisiting all other options, we see that any of the remaining inheritance patterns could possibly explain what we see.

Blood type is inherited as a codominant trait and relies on alleles for blood antibodies as well as Rh (Rhesus) factor. The father's blood type is A–, so he has no Rh factor and must be either AA or AO. The mother must be AB with an Rh factor.

Father possibilities: A^-A^- or A^-O^-

Mother possibilities: A^-B⁺ or A⁺B^- or A⁺B⁺

Based on these possibilities, we cannot conclude if the child will be positive or negative for Rh factor; however, since the mother has no allele for O blood type, we can conclude that the child cannot have O type blood. The child could receive AA, AO, BO, or AB.

The results of blood typing crosses can be calculated in the same way as those of traditional crosses: with Punnett squares. If we put the AB (maternal) genotype along one side of the square and the BB (paternal) genotype along the other, we see that only two genotypes are present in the offspring. 50% will be AB and 50% will be BB (a blood type of B); therefore, A is not one of the blood types present in the offspring, so this is our answer.

All of the other answer choices have a possibility of producing A type offspring.

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.