Question

punnett square practice
complete the punnett squares below about mendels peas.
underlined letters indicate a recessive trait for p, s, and w alleles.
p - purple flower
p - white flower
t - tall plant
t - short plant
g - green pod
g - yellow pod
s - yellow seed
s - green seed
w - round seed
w - wrinkled seed
t - tall plant
t - short plant
yellow seed - ______ %
green seed - ______ %
round seed - ______ %
wrinkled seed - ______ %
tall plant - ______ %
short plant- ______ %
p - purple flower
p - white flower
s - yellow seed
s - green seed
g - green pod
g - yellow pod
homozygous dom - ______ %
heterozygous- ______ %
homozygous rec - ______ %
homozygous dom - ______ %
heterozygous- ______ %
homozygous rec - ______ %
homozygous dom - ______ %
heterozygous- ______ %
homozygous rec - ______ %

Explanation:

Step1: Complete the Punnett square for \(S - s\)

The Punnett square for \(S - s\) (where both parents are \(S\)):

	\(S\)	\(S\)
\(S\)	\(SS\)	\(SS\)

All offspring have \(SS\) (yellow seed as \(S\) is dominant). So yellow seed \(= 100\%\), green seed \(= 0\%\)

Step2: Complete the Punnett square for \(W - w\)

The Punnett square for \(W - w\) (parents \(Ww\) and \(ww\)):

	\(W\)	\(w\)
\(w\)	\(Ww\)	\(ww\)

Out of 4 offspring, \(2\) are \(Ww\) (round, since \(W\) is dominant) and \(2\) are \(ww\) (wrinkled). So round seed \(= 50\%\), wrinkled seed \(= 50\%\)

Step3: Complete the Punnett square for \(T - t\)

The Punnett square for \(T - t\) (parents \(TT\) and \(Tt\)):

	\(T\)	\(t\)
\(T\)	\(TT\)	\(Tt\)

All offspring have at least one \(T\) (tall, as \(T\) is dominant). So tall plant \(= 100\%\), short plant \(= 0\%\)

Step4: Complete the Punnett square for \(P - p\) (second set)

The Punnett square for \(P - p\) (parents \(Pp\) and \(Pp\)):

	\(P\)	\(p\)
\(p\)	\(Pp\)	\(pp\)

Out of 4 offspring: \(1\) \(PP\) (homozygous dom), \(2\) \(Pp\) (heterozygous), \(1\) \(pp\) (homozygous rec). So homozygous dom \(= 25\%\), heterozygous \(= 50\%\), homozygous rec \(= 25\%\)

Step5: Complete the Punnett square for \(S - s\) (second set)

The Punnett square for \(S - s\) (parents \(ss\) and \(ss\)):

	\(s\)	\(s\)
\(s\)	\(ss\)	\(ss\)

All offspring are \(ss\) (homozygous rec). So homozygous dom \(= 0\%\), heterozygous \(= 0\%\), homozygous rec \(= 100\%\)

Step6: Complete the Punnett square for \(G - g\) (second set)

The Punnett square for \(G - g\) (parents \(GG\) and \(Gg\)):

	\(G\)	\(G\)
\(g\)	\(Gg\)	\(Gg\)

Out of 4 offspring: \(2\) \(GG\) (homozygous dom), \(2\) \(Gg\) (heterozygous). So homozygous dom \(= 50\%\), heterozygous \(= 50\%\), homozygous rec \(= 0\%\)

Answer:

Yellow seed - \(100\)%, Green seed - \(0\)%, Round seed - \(50\)%, Wrinkled seed - \(50\)%, Tall plant - \(100\)%, Short plant - \(0\)%, Homozygous Dom (\(P - p\)) - \(25\)%, Heterozygous (\(P - p\)) - \(50\)%, Homozygous Rec (\(P - p\)) - \(25\)%, Homozygous Dom (\(S - s\) second set) - \(0\)%, Heterozygous (\(S - s\) second set) - \(0\)%, Homozygous Rec (\(S - s\) second set) - \(100\)%, Homozygous Dom (\(G - g\) second set) - \(50\)%, Heterozygous (\(G - g\) second set) - \(50\)%, Homozygous Rec (\(G - g\) second set) - \(0\)%