Mr. Arora is heterozygous for three autosomal gene M, N and T and homozygous recessive for 4th autosomal gene P. They have gene for colour blindness (c) on X chromosome. What will be the probability for sperm to carry mNTpc condition.

Options

A. `1/16`
B. `1/8`
C. 0
D. `1/32`

Answer

To solve the problem of determining the probability of Mr. Arora's sperm carrying the genotype mNTpc, we need to analyze the genetic contributions from his heterozygous and homozygous recessive genes. Here’s the step-by-step solution:

### Step 1: Identify the Genotypes
- Mr. Arora is heterozygous for genes M, N, and T. This means he has the following genotypes:
- M = Mm (heterozygous)
- N = Nn (heterozygous)
- T = Tt (heterozygous)
- He is homozygous recessive for gene P:
- P = pp (homozygous recessive)
- For color blindness (c), which is located on the X chromosome, we assume he has one X chromosome with the normal vision allele (C) and one with the color blindness allele (c):
- X = Xc (for color blindness)

### Step 2: Determine Gamete Formation
To find the probability of the sperm carrying the genotype mNTpc, we need to consider the possible alleles that can be contributed from each gene:
- For gene M (Mm), the possible alleles are:
- m (1/2 probability)
- M (1/2 probability)
- For gene N (Nn), the possible alleles are:
- N (1/2 probability)
- n (1/2 probability)
- For gene T (Tt), the possible alleles are:
- T (1/2 probability)
- t (1/2 probability)
- For gene P (pp), since he is homozygous recessive, the only allele is:
- p (1 probability)
- For the color blindness gene (Xc), the possible alleles are:
- c (1/2 probability)
- C (1/2 probability)

### Step 3: Calculate the Probability of Each Allele
Now we can calculate the probability for each allele in the desired genotype mNTpc:
- Probability of m from M = 1/2
- Probability of N from N = 1/2
- Probability of T from T = 1/2
- Probability of p from P = 1 (since it is homozygous recessive)
- Probability of c from color blindness = 1/2

### Step 4: Multiply the Probabilities
To find the overall probability of the sperm carrying the genotype mNTpc, we multiply the probabilities of each allele:
\[ P(mNTpc) = P(m) \times P(N) \times P(T) \times P(p) \times P(c) \]
\[ P(mNTpc) = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times 1 \times \left(\frac{1}{2}\right) \]
\[ P(mNTpc) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times 1 \times \frac{1}{2} = \frac{1}{16} \]

### Final Answer
The probability of Mr. Arora's sperm carrying the genotype mNTpc is **1/16**.

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