Question

1. georges aunt and uncle, zit and zilla have the most beautiful set of purple lips on the planet. zilla is believed to be heterozygous for her lip color. zits family brags that their blood line is pure. complete the punnett square to show the possibilities that would results if zit and zilla have offspring.

a. identify the genotype of each flobbit. zit _____ zilla _____
b. complete the punnett square to show the possibilities that would result if they had children.
punnett square grid
c. list the possible genotypes and phenotypes for the flobbit babies.
d. what is the probability that the kids will have purple lips? ______%
e. what is the probability that the kids will have green lips? ______%

1. commander shlob is famous for his three eyes. he recently found a good looking female flobbit who also has three eyes. only flobbits with three eyes can be commanders. would it be possible for them to have a flobbit baby with only one eye? why or why not? create a punnett square to help you answer this question.

punnett square grid

1. flobbits with two fangs make more money than flobbits with one fang. commander shlob (from question 6) is heterozygous for one fang, but wants to have a flobbit baby with two fangs. is it possible for him to have a flobbit baby with two fangs? what type of female flobbit would he need to marry in order to give him the best chance at having an offspring with two fangs? create a punnett square to help you answer this question.

punnett square grid

Explanation:

Step1: Analyze Genotypes (Question 5A)

Assume purple lips are dominant (\(P\)) and green are recessive (\(p\)). Zit has pure bloodline (homozygous dominant), so Zit: \(PP\). Zilla is heterozygous, so Zilla: \(Pp\).

Step2: Complete Punnett Square (Question 5B)

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

Step3: List Genotypes/Phenotypes (Question 5C)

Genotypes: \(PP\), \(Pp\). Phenotypes: All have purple lips (since \(P\) is dominant).

Step4: Calculate Purple Lips Probability (Question 5D)

All 4 squares have purple - related genotypes (\(PP\) or \(Pp\)). Probability = \(\frac{4}{4} \times 100 = 100\%\).

Step5: Calculate Green Lips Probability (Question 5E)

No square has \(pp\) (green genotype). Probability = \(\frac{0}{4} \times 100 = 0\%\).

Step6: Analyze Three - Eye Trait (Question 6)

Assume three eyes are dominant (\(E\)), one eye is recessive (\(e\)). Both parents have three eyes. If both are homozygous dominant (\(EE\)), Punnett square:

	\(E\)	\(E\)
\(E\)	\(EE\)	\(EE\)

All offspring have \(EE\) (three eyes). So no one - eye baby. If one is \(Ee\) and one \(EE\), still no \(ee\). Only if both are \(Ee\), but problem says "only three - eyed can be commanders", so parents are likely \(EE\). Thus, impossible to have one - eye baby.

Step7: Analyze Fang Trait (Question 7)

Let two fangs be dominant (\(F\)), one fang recessive (\(f\)). Commander is heterozygous (\(Ff\)). To maximize two - fang offspring, marry a homozygous dominant (\(FF\)) female. Punnett square (Ff x FF):

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

Probability of two fangs (\(FF\) or \(Ff\)) is 100% (if female is \(FF\)) or 75% (if female is \(Ff\)), so best is \(FF\) female.

Answer:

Question 5A:

Zit: \(PP\), Zilla: \(Pp\)

Question 5B:

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

Question 5C:

Genotypes: \(PP\), \(Pp\); Phenotypes: Purple lips (all offspring)

Question 5D:

\(100\)

Question 5E:

\(0\)

Question 6:

No, because if both parents have three eyes (dominant trait, likely homozygous dominant), all offspring will have three eyes (no \(ee\) genotype for one eye). Punnett square (assuming \(EE\times EE\)) shows all \(EE\).

Question 7:

Yes, possible. He should marry a homozygous dominant (\(FF\)) female for two fangs. Punnett square (Ff x FF) gives all offspring with two - fang - related genotypes (\(FF\) or \(Ff\)), maximizing the chance.