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問題

解法ステップ

1. Understand the question

   Turn 1 Question

   section 4: non mendelian genetics

   1. colorblindness is an x - linked recessive trait. a female carrier marries a male with normal vision. what is the probability of having a daughter with normal vision?
   2. in cows the genes for red and white hair are codominant. in cows that have both red and white alleles, both colors are seen (a roan coat). if two roan cows are bred, what is the probability of having roan offspring?

   Turn 2 Question

   Puedes hacerlo mas facil y con el punett square echo

2. Response

   Turn 1 Answer

   Question 3

3. Explanation

   Step 1: Determine Genotypes

   Let \( X^C \) be the normal allele and \( X^c \) be the colorblind allele. The female carrier has genotype \( X^C X^c \), and the male with normal vision has genotype \( X^C Y \).

   Step 2: Set Up Punnett Square

   The possible gametes for the female are \( X^C \) and \( X^c \). The possible gametes for the male are \( X^C \) and \( Y \). The Punnett square combinations for daughters (from \( X^C \) or \( Y \); we only consider \( X^C \) for daughters) are:

   • \( X^C \) (female) × \( X^C \) (male) → \( X^C X^C \) (normal)
   • \( X^c \) (female) × \( X^C \) (male) → \( X^C X^c \) (carrier, normal vision)

   Step 3: Calculate Probability for Daughters

   For daughters, the possible genotypes are \( X^C X^C \) and \( X^C X^c \), both with normal vision. The total number of daughter genotypes from the Punnett square (considering only female offspring) is 2 out of 2 (since male gamete \( Y \) gives sons). So the probability of a daughter with normal vision is \( \frac{2}{2} = 1 \)? Wait, no—wait, the Punnett square for all offspring: total 4 cells. Daughters are 2 cells (\( X^C X^C \), \( X^C X^c \)), sons are 2 (\( X^C Y \), \( X^c Y \)). Both daughter genotypes have normal vision (carrier is not colorblind, just a carrier). So the probability of a daughter (any daughter) having normal vision is \( \frac{2}{2} \) (since 2 daughter cells, both normal) or \( \frac{2}{4} \) (total offspring) but we want probability of daughter (so given it's a daughter, what's the chance of normal vision). Since all daughters (2 out of 4 total) have normal vision, the probability is \( \frac{2}{4} \) for a daughter (i.e., the chance that a child is a daughter with normal vision: probability of daughter is \( \frac{1}{2} \), and given daughter, probability of normal is 1. So combined: \( \frac{1}{2} \times 1 = \frac{1}{2} \)? Wait, no. Let's re-express:

   The female is \( X^C X^c \), male \( X^C Y \).

   Possible offspring:

   • \( X^C X^C \) (daughter, normal)
   • \( X^C X^c \) (daughter, carrier, normal)
   • \( X^C Y \) (son, normal)
   • \( X^c Y \) (son, colorblind)

   So out of 4 offspring, 2 are daughters, both with normal vision. So the probability of having a daughter with normal vision is \( \frac{2}{4} = \frac{1}{2} \)? Wait, no—wait, the question is "probability of having a daughter with normal vision". So first, probability of having a daughter: \( \frac{1}{2} \). Then, given it's a daughter, probability of normal vision: 1 (since both daughter genotypes are normal). So by the law of total probability: \( P(\text{daughter}) \times P(\text{normal} | \text{daughter}) = \frac{1}{2} \times 1 = \frac{1}{2} \)? Wait, no—wait, the two daughter genotypes are both normal (one is homozygous normal, one is carrier). So regardless, a daughter will have normal vision. So the probability is \( \frac{2}{4} = \frac{1}{2} \) (since 2 out of 4 offspring are daughters with normal vision) or \( \frac{1}{2} \) (since 50% chance of daughter, and 100% chance that daughter has normal vision). So the probability is \( \frac{1}{2} \)? Wait, no—wait, the Punnett square has 4 cells. The daughters are 2 cells, both with normal vision. So the probability is \( \frac{2}{4} = \frac{1}{2} \)? Wait, no, the question is "probability of having a daughter with normal vision". So the possible outcomes are 4 (all offspring). The favorable outcomes are the two daughter cells (both normal). So \( \frac{2}{4} = \frac{1}{2} \)? Wait, no—wait, the female has \( X^C X^c \), male \( X^C Y \). The cross:

   Gametes:

   • Female: \( X^C, X^c \)
   • Male: \( X^C, Y \)

   Offspring:

   1. \( X^C X…
4. Explanation

   Step 1: Determine Genotypes

   Let \( R \) be the red allele and \( W \) be the white allele. Roan cows have genotype \( RW \) (codominant, so both colors are expressed).

   Step 2: Set Up Punnett Square

   Both parents are \( RW \). The possible gametes for each are \( R \) and \( W \). The Punnett square is:

   | | \( R \) | \( W \) |
   |-------|--------|--------|
   | \( R \) | \( RR \) | \( RW \) |
   | \( W \) | \( RW \) | \( WW \) |

   Step 3: Identify Roan Offspring

   Roan offspring have genotype \( RW \). From the Punnett square, the \( RW \) genotypes are in two cells (top-right and bottom-left).

   Step 4: Calculate Probability

   Total number of offspring genotypes: 4. Number of roan (\( RW \)): 2. So the probability is \( \frac{2}{4} = \frac{1}{2} \) (or 50%).

5. Final answer

   (Question 3):
   \( \frac{1}{2} \) (or 50%)

   Question 4

答え

Question Analysis

OCR Text

Puedes hacerlo mas facil y con el punett square echo

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