You're on a roll — Here are the solutions to the genetics problem: a) What are the two possible genotypes for a brown-eyed rabbit? Step 1: Identify the dominant allele. The allele for brown eyes is dominant, represented by $E$. Step 2: Determine genotypes for the dominant phenotype. A dominant phenotype can be homozygous dominant or heterozygous. The two possible genotypes are: $EE$ (homozygous dominant) $Ee$ (heterozygous) The two possible genotypes for a brown-eyed rabbit are $\boxed{EE \text{ and } Ee}$. b) What is the only possible genotype for a blue-eyed rabbit? Step 1: Identify the recessive allele. The allele for blue eyes is recessive, represented by $e$. Step 2: Determine the genotype for the recessive phenotype. A recessive phenotype must be homozygous recessive. The only possible genotype for a blue-eyed rabbit is $\boxed{ee}$. c) Draw a genetic diagram to show the expected phenotypes in the offspring from a cross between a homozygous brown-eyed rabbit and a blue-eyed rabbit. Step 1: Identify the genotypes of the parents. Homozygous brown-eyed rabbit: $EE$ Blue-eyed rabbit: $ee$ Step 2: Determine the gametes produced by each parent. Parent 1 ($EE$): $E$ Parent 2 ($ee$): $e$ Step 3: Construct a Punnett square. $$ \begin{array}{|c|c|} \hline \text{Gametes} & E \\ \hline e & Ee \\ \hline \end{array} $$ Step 4: Determine the genotypes and phenotypes of the offspring. All offspring will have the genotype $Ee$. All offspring will have brown eyes. The genetic diagram shows that all offspring will be $\boxed{\text{brown-eyed}}$. d) Draw a genetic diagram to show the expected phenotypes in the offspring from a cross between a heterozygous brown-eyed rabbit and a blue-eyed rabbit. Step 1: Identify the genotypes of the parents. Heterozygous brown-eyed rabbit: $Ee$ Blue-eyed rabbit: $ee$ Step 2: Determine the gametes produced by each parent. Parent 1 ($Ee$): $E, e$ Parent 2 ($ee$): $e$ Step 3: Construct a Punnett square. $$ \begin{array}{|c|c|c|} \hline \text{Gametes} & E & e \\ \hline e & Ee & ee \\ \hline \end{array} $$ Step 4: Determine the genotypes and phenotypes of the offspring. Genotypes: $Ee$ (50%), $ee$ (50%) Phenotypes: Brown eyes (50%), Blue eyes (50%) The genetic diagram shows that the expected phenotypes are $\boxed{50\% \text{ brown-eyed and } 50\% \text{ blue-eyed}}$. e) A brown-eyed rabbit was crossed with a blue-eyed rabbit. They had nine young, of which four had brown eyes and five had blue eyes. What was the genotype of the brown-eyed parent? Step 1: Identify the known parent genotype and offspring phenotypes. Blue-eyed parent: $ee$ Offspring: 4 brown-eyed, 5 blue-eyed (approximately a 1:1 ratio). Step 2: Consider the two possible genotypes for the brown-eyed parent. Case 1: If the brown-eyed parent was $EE$. Cross: $EE \times ee$ Offspring: All $Ee$ (brown-eyed). This does not match the observed offspring (some blue-eyed). Case 2: If the brown-eyed parent was $Ee$. Cross: $Ee \times ee$ Offspring: $50\%$ $Ee$ (brown-eyed) and $50\%$ $ee$ (blue-eyed). This is a 1:1 ratio. Step 3: Compare expected ratios with observed offspring. The observed ratio of 4 brown-eyed to 5 blue-eyed offspring is approximately 1:1, which matches the expected ratio from an $Ee \times ee$ cross. The genotype of the brown-eyed parent was $\boxed{Ee}$. f) A rabbit breeder crossed a brown-eyed rabbit with a blue-eyed rabbit. The rabbits had three young, all of which had brown eyes. The breeder said this proved that the brown-eyed rabbit must have the genotype EE, but her daughter said she could not be certain and she should do the cross again to make sure. Who was right – the breeder or her daughter? Step 1: Identify the known parent genotype and offspring phenotypes. Blue-eyed rabbit: $ee$ Offspring: 3 young, all brown-eyed. Step 2: Consider the two possible genotypes for the brown-eyed parent. Case 1: If the brown-eyed parent was $EE$. Cross: $EE \times ee$ Offspring: All $Ee$ (brown-eyed). This is consistent with observing 3 brown-eyed offspring. Case 2: If the brown-eyed parent was $Ee$. Cross: $Ee \times ee$ Offspring: $50\%$ $Ee$ (brown-eyed) and $50\%$ $ee$ (blue-eyed). Step 3: Evaluate the certainty based on the observed offspring. If the brown-eyed parent was $Ee$, there is a $\frac{1}{2}$ probability for each offspring to be brown-eyed. The probability of having 3 brown-eyed offspring in a row from an $Ee \times ee$ cross is $(\frac{1}{2})^3 = \frac{1}{8}$. This is a low probability, but it is not impossible. Step 4: Conclude who is right. Since there is still a chance (1 in 8) that an $Ee$ parent could produce 3 brown-eyed offspring, observing only 3 brown-eyed offspring does not conclusively prove that the brown-eyed parent is $EE$. More offspring would be needed to increase certainty. Therefore, the $\boxed{\text{daughter}}$ was right. Got more? Send 'em! 📸