A man who is heterozygous for tongue rolling and has a normal vision and a woman who cannot roll her tongue and is heterozygous for colorblindness have a child. What is the probability that it will be able to roll his/her tongue and have normal vision? Tongue-rolling is due to a dominant allele. Color Blindness is caused by an X-linked recessive gene. ...?

To solve this problem, we need to use Punnett squares to predict the probabilities of the child inheriting the traits of tongue-rolling and normal vision.

For tongue-rolling, let R represent the dominant allele (able to roll the tongue), and r represent the recessive allele (unable to roll the tongue). The man is heterozygous (Rr), and the woman cannot roll her tongue, so she must have two recessive alleles (rr).

1. Setting up Punnett square for tongue rolling: Man (Rr) Woman (rr)

R | r ----------- r | Rr | rr ----------- r | Rr | rr

From the Punnett square, we see that there is a 50% chance (2 out of 4) for the child to be able to roll their tongue (Rr) and a 50% chance to be unable to roll their tongue (rr).

For colorblindness, since it is X-linked, let's denote the normal vision allele as X^N, and the colorblind allele as X^c. Since the woman is heterozygous for colorblindness (carrier), her genotype is X^NX^c. The man has normal vision therefore he can only contribute an allele for normal vision on his X chromosome, X^N, and his Y chromosome (which doesn't carry the vision trait) is denoted as Y.

2. Setting up Punnett square for color vision: Man (X^NY) Woman (X^NX^c)

X^N | Y ------------ X^N | X^NX^N | X^NY ------------ X^c | X^NX^c | X^cY

From the Punnett square, for a daughter (XX), the chance of having normal vision is 100% because even though she could inherit one colorblind allele (X^c), she would always have one normal allele (X^N) from her father which is enough to give her normal vision. For a son (XY), the probability is 50% because he would have normal vision if he inherited the X^N allele from the mother, but he would be colorblind if he inherited the mother's X^c allele.

Since color vision and tongue-rolling are inherited independently, we calculate the probability of both traits occurring together by multiplying their independent probabilities.

Assuming equal probability of having a son or a daughter (which in reality carries a slight male bias), we can do the calculations for both scenarios:

- For a daughter: The probability is 50% (tongue-rolling) * 100% (normal vision) = 50%. - For a son: The probability is 50% (tongue-rolling) * 50% (normal vision) = 25%.

Thus, if the child is a daughter, there is a 50% chance she will be able to roll her tongue and have normal vision. If the child is a son, there is a 25% chance he will be able to roll his tongue and have normal vision.

Extra: Punnett squares are a very useful tool in genetics used to predict the genotypes and phenotypes of offspring given the genotypes of the parents. Each square represents a possible combination of one parental allele with one from the other parent.

Dominant alleles are those that express a trait even when only one copy is present. In contrast, recessive alleles only express the trait when two copies are present. In an X-linked condition, the gene causing the condition is located on the X chromosome, which is one of the two sex chromosomes (X and Y). Males (XY) and females (XX) have different probabilities for X-linked traits since males have only one X chromosome and females have two.

When doing genetic problems: - Independent assortment law: traits are transmitted to offspring independently of one another. - The probability of two independent events both occurring is the product of their individual probabilities.