- Punnet Squares -

Question

1. A heterozygous woman carrying the recessive gene for colorblindness marries a man who is color-blind. Do a Punnet square to determine the answers to the following questions.

A) Genotype of parents X^CX^c (X)  X^cY

1. a. What percent of the children will be colorblind:
2. b. What is the percent colorblind children by sex: % Males % Females
3. c. What percentage will be carriers %
4. d. What is the sex of the carriers?

Answer 1

The X^cX⁰ individual is the mother in this scenario due to the XX chromosome code, while the X^cY individual is the father due to the XY code. (I'm using c to indicate the the colorblind allele and 0 to represent the typical color vision allele to make the notation a bit clearer.)

Here are the possible genotypes and their associated phenotypes for simple X-linked recessive colorblind inheritance:

X⁰X⁰ - female with typical color vision (not a carrier for the colorblind allele)

X^cX⁰ - female with typical color vision (is a carrier for the colorblind allele)

X^cX^c - female with colorblindness

X^cY - male with colorblindness

X⁰Y - male with typical color vision

Because this trait is X-linked recessive, individuals are only affected if all of their X chromosomes carry the allele (the unaffected X chromosome acts as a backup, preventing colorblindness). The males only have one X chromosome, all male individuals who inherit the colorblind allele will be affected (i.e., males are either affected or unaffected, but they cannot be carriers, unlike the females).

In the question's pairing of a carrier female and an affected male, the Punnett square for the parents would produce the following offspring possibilities:

X⁰ X^c

X^c | X^cX⁰ | X^cX^c |

_________

Y | X⁰Y | X^cY |

As expected, 50% of the of the offspring are female (XX) and 50% are male (XY). Because both parents have at least one affected X chromosome, all female offspring are either affected or carriers. In total, 50% of the affected female offspring are colorblind, 50% of the female offspring have color vision but are carriers, 50% of male offspring are colorblind, and 50% of male offspring have color vision (and cannot pass the colorblindness trait onto their own offspring). Another way of saying that is that we have a genotype ratio of 25% X^cX⁰, 25% X^cX^c, 25% X⁰Y, and 25% X^cY.

*Note that we are assuming a simplified setup; real-world scenarios can get much more complex. For example, there are multiple forms of colorblindness. While red-green colorblindness, the most common type, is usually X-linked recessive, some of the rarer forms can be linked to autosomes rather than sex chromosomes. Some colorblindness can be caused by diseases or non-genetic injuries that lead to degeneration or malformation of the cone cells. Individuals with chromosome abnormalities, such as those with rare X0 or XXY chromosome sets, will often experience slightly different inheritance than typical XX or XY individuals, but the genes themselves will still follow the expected dominance rules on relation to the X chromosomes.

Answer 2

Julia, 

Colorblindness is caused by a mutation in the gene C. The mutated allele is coded c. The alleles C and c are carried by the X chromosome. The allele c is recessive. Colorblindness is therefore X-linked. A female with both mutated alleles (c/c) would be colorblind whereas a male would be colorblind as long as his X chromosome carries the mutated allele c. 

1) Punnet square

                                                    Father’s gametes for sex chromosomes and gene C / allele c

                                                      X^c                                             Y

Mother’s

gametes  for sex chromosomes 

and gene C / allele c                                   

X^C                                                     X^C / X^c                                                               X^C / Y

                                                      [female not color blind]               [male not color blind]

                                                         

X^c                                                      X^c / X^c                                                              X^c / Y

                                                         [female color blind]                      [male color blind]

1a) Percentage of children that can be colorblind. 

From the Punnet square, we see that there are 4 possible genotypes (combining sex and alleles C or c) for a child that this couple would have.

Of the 4 possible genotypes, 2 (X^c/X^c or X^c/Y) would have a phenotype of colorblindness since c is recessive.

2/4 = 1/2 = 0.5. In percentage, it is 50%. 

There is a 50% of chance that this couple have a color blind child. 

1b) Percentage of colorblind children by sex. 

The parents have 50% of chance of having a boy (X/Y) and 50% of having a girl (X/X).

In term of colorblindness, if the child is a girl, there is 50% of chance that the baby girl would be colorblind.

50% is 50/100 = 0.5. 

0.5*0.5 = 0.25 or 25%. 

There is 25% of chance that those parents will have a girl who is color blind. 

The same reasoning applies for the boy. Those parents have 25% of chance of having a baby boy who is color blind. 

1c) Percentage of having a child who is a carrier. 

A carrier is an individual who is not affected (not colorblind) but who can transmit the gene. 

From the Punnet square, only 1 possibility to be a carrier, it is the individual with the genotype X^C/X^c. Hence, those parents have a 25% of having a child who is a carrier. This child would be a girl. 

1d) Sex of the carrier.

The carrier genotype in respect to sex chromosomes and alleles C or c is X^C/X^c. This is a girl.  

A little long-winded.... but I am not sure how much you already knew!