Suppose that gene b is sex-linked, recessive, and lethal. A man marries a woman who is heterozygous for this gene. If this couple had many normal children, what would be he predicted sex ratio of these children?
Lethal when? If lethal after many years, the sex ratio is 1:1. If lethal in the womb, then one boy for every two girls, as the other boys died in the womb.
what would be the punnet square
i can it do it myself but what about genotype and phenotype
Woman = Bb
Man = By, since the lethal gene is sex-linked, it is not on the y chromosome. From this, you should be able to do the Punnet square to get the genotypes. From them, you can figure the phenotypes.
Ah, the gene b is quite the troublemaker! Well, my friend, since gene b is sex-linked, recessive, and lethal, it means that for a male child to have the lethal effect, he only needs to inherit one copy of the gene from his mother.
Now, to answer your question about the predicted sex ratio of their children, let's step into the clown car of probability. The couple has a 50% chance of having a male or female child with each pregnancy, just like any other ordinary genes.
However, in terms of the lethal gene b, if the couple were to have a male child, the chances of him being affected by the gene are quite high! Since the mother is heterozygous for the gene, there is a 50% chance she will pass on the lethal b gene to her son.
On the other hand, if the couple were to have a female child, she would have to inherit two copies of the gene, one from each parent, in order to be affected. This would only occur if both parents passed on the b gene. Statistically, this would be less likely to happen since the mother is only heterozygous for the gene.
So, to sum it up with a sprinkle of humor, if the couple has many normal children despite having the gene b, the predicted sex ratio would lean towards more female children. However, please remember that predictions are like juggling chainsaws - they can be entertaining, but not always precise!
To determine the predicted sex ratio of the children, we need to understand the inheritance pattern of the sex-linked, recessive, and lethal gene.
In humans, sex-linked genes are located on the sex chromosomes, specifically the X chromosome. Females have two copies of the X chromosome (XX), while males have one X and one Y chromosome (XY).
Given that the gene b is recessive, both copies of the gene must be present for the lethal effect to occur. This means that individuals who are heterozygous for the gene (XbX) are unaffected and considered normal, while those who are homozygous recessive (XbXb) will experience the lethal effect.
Now, let's evaluate the possible combinations of offspring for this couple:
1. Sons: The father will always contribute an X chromosome (since he is male), and the mother can contribute either an X or a Y chromosome. Therefore, the possible combinations are XY (no gene b) or XYb (heterozygous for gene b). In either case, the sons will be unaffected and normal.
2. Daughters: The father will always contribute an X chromosome, and the mother can contribute either an X chromosome without the gene b (X) or an X chromosome with the gene b (Xb). Therefore, the possible combinations are XX (normal) or XXb (heterozygous for gene b). The daughters who are XXb will be unaffected carriers of the gene b but will not experience the lethal effect.
Based on these possibilities, the predicted sex ratio of the children will be 50% sons (XY) and 50% daughters (XX and XXb).