15. In fruit flies, brown bodies are dominant to black bodies. Cross two heterozygous fruit flies. Determine the phenotypic and genotypic ratios. Then determine how many fruit flies, if 200 are born, will have black bodies. 16. In humans, mid-finger hair is dominant to no mid-finger hair. A woman who has no finger hair wants to marry and have children with a man who does have mid-finger hair. They do not know his genotype though. Determine the likelihood that they have a child with no finger hair like their mom. Include all possible options. 17. In chimpanzees, straight fingers are dominant to bent fingers. If two chimps are crossed and have offspring that are \( 50 \% \) bent fingers and \( 50 \% \) straight fingers, what must have been the genotypes of the parents?

**15. Fruit Flies** Let the allele for brown body be \(B\) (dominant) and for black body be \(b\) (recessive). Two heterozygous fruit flies have the genotype \(Bb\). **Step 1. Draw the Punnett square for \(Bb \times Bb\):** \[ \begin{array}{c|cc} & B & b \\ \hline B & BB & Bb \\ b & Bb & bb \\ \end{array} \] **Step 2. Determine Genotypic Ratio:** - \(BB: 1\) - \(Bb: 2\) - \(bb: 1\) Thus, the genotypic ratio is \(1:2:1\). **Step 3. Determine Phenotypic Ratio:** Because brown is dominant over black: - The \(BB\) and \(Bb\) offspring have brown bodies. - Only the \(bb\) offspring have black bodies. Thus, the phenotypic ratio is: - Brown: \(1 + 2 = 3\) - Black: \(1\) So, the ratio is \(3:1\). **Step 4. Determine Number of Black-Bodied Fruit Flies if 200 are Born:** Since \( \frac{1}{4} \) of the offspring are \(bb\) (black), \[ \frac{1}{4} \times 200 = 50. \] So, 50 fruit flies will have black bodies. --- **16. Mid-Finger Hair in Humans** Let the allele for mid-finger hair be \(H\) (dominant) and for no mid-finger hair be \(h\) (recessive). The woman has no finger hair, so her genotype must be \(hh\). The man shows the dominant phenotype, so his genotype can be either \(HH\) or \(Hh\). **Option 1: Man is Homozygous Dominant (\(HH\))** - Cross: \(HH \times hh\) - All offspring will inherit one \(H\) from the father and one \(h\) from the mother, making them all \(Hh\). - Proportion with no mid-finger hair: \(0\%\). **Option 2: Man is Heterozygous (\(Hh\))** - Cross: \(Hh \times hh\) - Punnett square: \[ \begin{array}{c|cc} & H & h \\ \hline h & Hh & hh \\ h & Hh & hh \\ \end{array} \] - Offspring genotypes: - \(Hh: 50\%\) (with hair) - \(hh: 50\%\) (no hair) Thus, if the man is heterozygous, there is a \(50\%\) chance that a child will have no mid-finger hair. **Conclusion:** - If the man is \(HH\): \(0\%\) chance of a child with no finger hair. - If the man is \(Hh\): \(50\%\) chance of a child with no finger hair. --- **17. Chimpanzee Fingers** Let the allele for straight fingers be \(S\) (dominant) and for bent fingers be \(s\) (recessive). We are told crossing two chimps results in offspring with \(50\%\) bent fingers and \(50\%\) straight fingers. **Step 1. Analyze Possibilities:** - For an offspring to have bent fingers, it must have the genotype \(ss\). - A \(50\%\) occurrence of \(ss\) in the offspring means that half of the progeny are \(ss\). **Step 2. Find the Parental Genotypes:** Consider the cross \(Ss \times ss\): \[ \begin{array}{c|cc} & S & s \\ \hline s & Ss & ss \\ s & Ss & ss \\ \end{array} \] - Offspring: - \(Ss\): \(50\%\) (straight fingers) - \(ss\): \(50\%\) (bent fingers) This matches the observed ratios. The reverse situation \(ss \times Ss\) is equivalent. Thus, one parent must be heterozygous (\(Ss\)) and the other must be homozygous recessive (\(ss\)). --- **Final Answers:** - **15.** Genotypic ratio: \(1:2:1\). Phenotypic ratio (brown:black): \(3:1\). Out of 200 fruit flies, 50 will have black bodies. - **16.** If the man is \(HH\), none of the children will have no finger hair (0\%). If the man is \(Hh\), there is a 50\% chance that a child will have no finger hair. - **17.** The parental genotypes must be \(Ss\) and \(ss\).