Question

4.) tongue rolling in humans
f = can roll tongue
f = cannot roll tongue
ff = 25% = 25%
ff = ff = 50%
ff = ff = 25%
cannot roll tongue: 25 % chance
can roll tongue: 75 % chance

Explanation:

Step1: Fill in the Punnett square

The Punnett square for \(Ff\times Ff\) (assuming parents are heterozygous). The possible gametes from each parent are \(F\) and \(f\).

	\(F\)	\(f\)
\(f\)	\(Ff\)	\(ff\)

Step2: Calculate genotypes and phenotypes

• Genotype \(FF\): There is \(1\) out of \(4\) squares (\(\frac{1}{4}=25\%\))
• Genotype \(Ff\): There are \(2\) out of \(4\) squares (\(\frac{2}{4} = 50\%\))
• Genotype \(ff\): There is \(1\) out of \(4\) squares (\(\frac{1}{4}=25\%\))
• Phenotype:
• Cannot roll tongue (\(ff\)): \(25\%\)
• Can roll tongue (\(FF + Ff\)): \(\frac{1 + 2}{4}=75\%\)

Answer:

• \(FF=\frac{1}{4}\times100\% = 25\%\)
• \(Ff=\frac{2}{4}\times100\%=50\%\)
• \(ff=\frac{1}{4}\times 100\%=25\%\)
• Cannot roll tongue: \(25\%\) Chance
• Can roll tongue: \(75\%\) Chance