Final answer:
The chances that the couple has a child with the disease can be determined using a Punnett square, resulting in a 50% chance.
Step-by-step explanation:
The chances that the couple has a child with the disease can be determined using a Punnett square. If one parent is a carrier for a sex-linked recessive disease (X^cX) and the other parent has the disease (X^cY), the Punnett square would show that there is a 50% chance of having a son with the disease (X^cY) and a 50% chance of having a daughter who is a carrier for the disease (X^cX). Therefore, the answer is 3) 50.
The chance that a child will either be a carrier of or be affected by the sex-linked recessive disease is 100% when one parent has the disease and the other is a carrier.
The question relates to a sex-linked recessive disease and the calculation of the probability of a child inheriting it when one parent is a carrier and the other parent has the disease. To calculate the chances, we can use a Punnett square to determine possible genetic combinations.
Let's denote the normal X chromosome as 'X' and the affected X chromosome as 'Xd' (where 'd' stands for the disease). Females have two X chromosomes and males have one X and one Y chromosome. In this scenario, the affected parent will pass on the 'Xd' chromosome and the carrier parent will pass on either a normal 'X' chromosome or an 'Xd'.
Thus, for a female child (XX), she could inherit 'XXd' (carrier) or 'XdXd' (affected). For a male child (XY), he could inherit 'XdY' (affected) or 'XY' (unaffected, since males only have one X chromosome and will get the Y chromosome from the father). Consequently, since one parent has the disease 'XdXd' or 'XdY', all offspring will inherit at least one 'Xd' and thus there is a 100% chance that any child will either be a carrier or affected by the disease.