A man marries a woman who is heterozygous for a recessive X-linked lethal gene "X" expressed in very early development. If the couple had many normal children, what would be the predicted sex ratio among their children?
I'm so confused. If it's a lethal allele, how would do you predict this?
To predict the sex ratio among the children of this couple, we need to understand the inheritance pattern and consequences of the X-linked lethal gene.
In humans, sex chromosomes determine the sex of an individual, with females having two X chromosomes (XX) and males having one X and one Y chromosome (XY). X-linked genes are located on the X chromosome. In this case, the woman is heterozygous for the X-linked recessive lethal gene, which means she carries one normal allele (XN) and one lethal allele (XL), where 'N' represents a normal gene and 'L' represents a lethal gene.
When considering the sex ratio of the couple's children, we need to consider the inheritance patterns of X-linked genes. In this case, if the father is unaffected and does not carry the lethal allele, he will pass on either his X chromosome with the normal allele (XN) to daughters or his Y chromosome to sons.
Since both parents contribute equally to the sex chromosomes of their offspring, the predicted sex ratio among their children would be half boys and half girls. This is because the presence or absence of the lethal allele on the X chromosome does not affect the probability of having a boy (XY) or a girl (XX).
The sex ratio would still be the same, 50% of each. However, if you are looking for those who would survive, use a Punnett square.
https://www.google.com/search?client=safari&rls=en&q=punnett+square&ie=UTF-8&oe=UTF-8&gws_rd=ssl
With x standing for the recessive lethal gene, mom would be Xx and dad would be XY. Assume chance variation.