Puedes hacerlo mas facil y con el punett square echo
(Question 3): \( \frac{1}{2} \) (or 50%) ### Question 4
Pregunta
Turn 1 Question
section 4: non mendelian genetics
- 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?
- 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
Pasos de solución
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Understand the question
Turn 1 Question
section 4: non mendelian genetics
- 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?
- 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
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Response
Turn 1 Answer
Question 3
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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:
- \( X^C X…
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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%).
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Final answer
(Question 3):
\( \frac{1}{2} \) (or 50%)Question 4
Respuesta
Response
Turn 1 Answer
Question 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:
- \( X^C X^C \) (daughter, normal)
- \( X^C X^c \) (daughter, carrier, normal)
- \( X^C Y \) (son, normal)
- \( X^c Y \) (son, colorblind)
So for a daughter (cases 1 and 2), both have normal vision. The probability of having a daughter is \( \frac{2}{4} = \frac{1}{2} \), and given it's a daughter, the probability of normal vision is 1. So the probability of a daughter with normal vision is \( \frac{2}{4} = \frac{1}{2} \)? Wait, no—wait, the question is "what is the probability of having a daughter with normal vision". So the chance that a child is a daughter AND has normal vision. The number of such children is 2 (the two daughters), out of 4 total children. So \( \frac{2}{4} = \frac{1}{2} \)? Wait, no—wait, the two daughters are both normal, so the probability is \( \frac{2}{4} = 0.5 \) or 50%? Wait, no—wait, the daughter genotypes: both are normal (carrier is not colorblind). So all daughters have normal vision. So the probability of a daughter (any daughter) having normal vision is 1 (since if it's a daughter, it's normal). But the probability of having a daughter with normal vision (i.e., the chance that a child is a daughter and has normal vision) is \( \frac{2}{4} = 0.5 \) (since 2 out of 4 children are daughters with normal vision).
Answer
(Question 3):
\( \frac{1}{2} \) (or 50%)
Question 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%).
Answer
(Question 4):
\( \frac{1}{2} \) (or 50%)
Turn 2 Answer
¡Claro! Aquí lo dejo súper sencillo con los cuadros de Punnett listos:
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Pregunta 3: Daltonismo (ligado al X)
- Genotipos: Mujer portadora = \( X^C X^c \) (donde \( X^C \) = visión normal, \( X^c \) = daltonismo); Hombre con visión normal = \( X^C Y \)
- Cuadro de Punnett:
| | \( X^C \) (mujer) | \( X^c \) (mujer) |
|--------|------------------|------------------|
| \( X^C \) (hombre) | \( X^C X^C \) (hija, visión normal) | \( X^C X^c \) (hija, portadora, visión normal) |
| \( Y \) (hombre) | \( X^C Y \) (hijo, visión normal) | \( X^c Y \) (hijo, daltónico) |
- Resultado: De los 4 hijos posibles, 2 son hijas y AMBAS tienen visión normal. La probabilidad de tener una hija con visión normal es 50% (o 1/2).
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Pregunta 4: Vacas roanas (codominancia)
- Genotipos: Ambos padres roanos = \( RW \) (R = pelo rojo, W = pelo blanco; RW = pelaje roano, ambos colores)
- Cuadro de Punnett:
| | \( R \) (padre) | \( W \) (padre) |
|--------|----------------|----------------|
| \( R \) (madre) | \( RR \) (ternera roja) | \( RW \) (ternera roana) |
| \( W \) (madre) | \( RW \) (ternera roana) | \( WW \) (ternera blanca) |
- Resultado: De los 4 crías posibles, 2 son roanas. La probabilidad de tener una cría roana es 50% (o 1/2).
Question Analysis
Subject
natural science
Sub Subject
biology
Education Level
high school
Difficulty
unspecified
Question Type
with image, calculation
Multi Question
Yes
Question Count
2
Analysis Status
completed
Analyzed At
2026-02-06T14:51:43
OCR Text
Show OCR extraction
Puedes hacerlo mas facil y con el punett square echo
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