In humans, short fingers and widow's peak are dominant over long fingers and straight hairline. A heterozygote in both regards reproduces with a similar heterozygote in both regards reproduces with a similar heterozygote. What is the chance of any one child having the phenotype as the parents?
To determine the chance of any one child having a particular phenotype, we should first understand the principle of Mendelian inheritance. In this scenario, we are dealing with two traits: finger length and hairline type. According to the given information, short fingers and a widow's peak are dominant over long fingers and a straight hairline.
The parents are described as heterozygotes, which means they have one dominant allele and one recessive allele for both traits. Let's represent the dominant alleles as "S" for short fingers and "W" for widow's peak, while the recessive alleles will be represented as "s" for long fingers and "w" for straight hairline.
The possible genotypes of the parents can be represented as follows:
Parent 1: SsWw (short fingers, widow's peak)
Parent 2: SsWw (short fingers, widow's peak)
To determine the chance of any one child having the phenotype as the parents, we need to examine the possible combinations of alleles that can be inherited from each parent. This can be done using a Punnett square.
Crossing the alleles of Parent 1 (SsWw) with the alleles of Parent 2 (SsWw) in a Punnett square would look like this:
| s | s |
_________|________|________|
W | SW | SW |
_________|________|________|
W | SW | SW |
_________|________|________|
| s | s |
_________|________|________|
W | SW | SW |
_________|________|________|
w | Sww | Sww |
_________|________|________|
The resulting genotypes can be read from the Punnett square as follows:
- 25% chance of offspring being SSWW (short fingers, widow's peak)
- 25% chance of offspring being SSWw (short fingers, widow's peak)
- 25% chance of offspring being SsWW (short fingers, widow's peak)
- 25% chance of offspring being SsWw (short fingers, widow's peak)
Therefore, there is a 100% chance that any one child will have the phenotype of short fingers and a widow's peak as both parents have the same phenotype and are heterozygotes for the respective traits.