Question

multiple choice 1 point achondroplastic dwarfism is a dominant genetic trait that causes severe malformation of the skeleton. homozygotes for this condition are spontaneously aborted (hence, the homozygous condition is lethal) but heterozygotes will develop to be dwarfed. matthew has a family history of the condition, although he does not express the trait. jane is an achondroplastic dwarf. matthew and jane are planning a family of several children and want to know the chances of producing a child with achondroplastic dwarfism. the probability that matthew and jane’s first child will be an achondroplastic dwarf is 25% 50% 75% 100%

Explanation:

Step1: Determine Genotypes

Achondroplasia is dominant (\(D\)), homozygous dominant (\(DD\)) is lethal. Jane is a dwarf, so her genotype must be \(Dd\) (since \(DD\) is lethal). Matthew does not express the trait, so his genotype is \(dd\) (recessive homozygous, as dominant trait would be expressed if present).

Step2: Set Up Punnett Square

Cross \(Dd\) (Jane) and \(dd\) (Matthew). The possible gametes from Jane: \(D\) and \(d\); from Matthew: \(d\) and \(d\).

	\(d\)	\(d\)
\(d\)	\(dd\)	\(dd\)

Step3: Calculate Probabilities

From the Punnett square, the genotypes are \(Dd\) (dwarf, 2 out of 4) and \(dd\) (normal, 2 out of 4). But wait, wait—wait, no: wait, Jane is \(Dd\), Matthew is \(dd\). Wait, no, wait: the problem says Matthew does not express the trait, so he is \(dd\). Jane is a dwarf, so she must be \(Dd\) (since \(DD\) is lethal). So the cross is \(Dd \times dd\). The offspring genotypes: \(Dd\) (dwarf) and \(dd\) (normal), each with 50% chance? Wait, no—wait, the Punnett square: \(Dd\) (Jane) produces \(D\) and \(d\); \(dd\) (Matthew) produces \(d\) and \(d\). So the four possible combinations: \(Dd\), \(Dd\), \(dd\), \(dd\). So \(Dd\) (dwarf) is 2/4 = 50%, \(dd\) (normal) is 2/4 = 50%? Wait, but wait—the problem says homozygous dominant (\(DD\)) is lethal, but in this cross, \(DD\) isn't possible because Matthew is \(dd\) (can't contribute \(D\)). So the viable offspring are \(Dd\) (dwarf) and \(dd\) (normal), each with 50% probability. Wait, but wait—wait, the question is the probability of the first child being an achondroplastic dwarf (i.e., \(Dd\)). So from the Punnett square, 2 out of 4 (50%) are \(Dd\) (dwarf), 2 out of 4 (50%) are \(dd\) (normal). So the probability is 50%.

Answer:

50% (the option with 50%)