Question

father
%kids expressing huntingtons=
%kids without huntingtons=
mother
= h = h = hh = hh = 0% = 25% = 50%

Explanation:

Step1: Identify parental genotypes

Huntington's disease is an autosomal dominant disorder, so the affected parent (father) has genotype $Hh$, and the unaffected mother has genotype $hh$.
Father's gametes: $H$, $h$
Mother's gametes: $h$, $h$

Step2: Fill Punnett square

Father\Mother	$H$	$h$
$h$	$Hh$	$hh$

Step3: Calculate trait percentages

• % with Huntington's ($Hh$): $\frac{2}{4} \times 100\% = 50\%$
• % without Huntington's ($hh$): $\frac{2}{4} \times 100\% = 50\%$

Answer:

• Father's genotype: $Hh$
• Mother's genotype: $hh$
• Punnett square cells (top-left, top-right, bottom-left, bottom-right): $Hh$, $hh$, $Hh$, $hh$
• %Kids expressing Huntington's: 50%
• %Kids without Huntington's: 50%