Answer:
There are two traits in this world problem, so it is a dihybrid problem. The probability of these two individuals having children with mid-digit hair and attached earlobes is 1/4.
Step-by-step explanation:
First, we must figure out how many traits are involved in this problem. We can see that the problem introduces mid-digit hair and attached earlobes, so we can conclude that there are two traits. Since there are two traits, this is considered to be a dihybrid problem. If there were only one trait, it would be a monohybrid problem.
To solve this problem, we should begin by determining the parent's genotypes from the given information. Let's let the letter H represent mid-digit hair (dominant) and h represent no mid-digit hair (recessive). Let's let the letter A represent attached earlobes (dominant) and a represent unattached earlobes (recessive).
The mother is homozygous recessive for mid-digit hair and heterozygous for attached earlobes, so she has the genotypes hh and Aa. The father is heterozygous for mid-digit hair and homozygous recessive for unattached earlobes, so he has the genotypes Hh and aa.
To find the probability of their child having mid-digit hair (dominant) AND attached earlobes (dominant), we can find the probability of each of these events occurring independently and then multiply them together (by the Law of Multiplication). Their child has a 50% chance of having mid-digit hair (Hh) and 50% chance of having no mid-digit hair (hh). Similarly, their child has a 50% chance of having attached earlobes (Aa) and a 50% chance of having unattached earlobes (aa). Therefore, the chance that their child would have mid-digit hair and attached earlobes is 1/2 * 1/2 = 1/4.
Note that it may be helpful for you to draw out Punnett squares to see these probabilities more clearly.
Hope this helps!