Mr rahul is suffering from hypertrichosis and phenylketonuria. The probability of a sperm having one recessive autosomal allele and holandric gene is?
Answers
➪We know that sum of three angles of a triangle is 180
➪Hence ∠A+∠B+∠C=180°
➪or A+B+C=180°
➪B+C=180° −A
➪Multiply both sides by 1/2
➪1/2 (B+C) = 1/2 (180° − A)
➪1/2 (B + C) = 90° - A/2...(1)
➪Now 1/2 (B+C)
➪Taking sine of this angle
➪sin (B + C /2)
➪[B + C /2 = 90° - A/2]
➪sin(90°-A/2)
➪cos A/2 [sin(90° - θ ) = cos θ]
➪Hence sin( B + C/2 )=cos A/2 proved
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Answer:
The probability of a sperm having one recessive autosomal allele and holandric gene is 1/2
Explanation:
Given that,
Mr rahul is suffering from hypertrichosis and phenylketonuria.
what is Hypertrichosis?
Hypertrichosis is a condition in which there is an excessive hair growth all over the body of the affected individual. It is also known as werewolf syndrome. It is an X linked genetic disorder showing dominant pattern of inheritance.
What is Phenylketonuria?
Phenylketonuria is a genetic disorder caused due to presence of recessive allele that results in decreased metabolism of the amino acid phenylalanine.
Holandric genes:-
Holandric genes are the genes present on the Y chromosome.
Now in this question, the patient is suffering from phenylketonuria and hypertrichosis.
- This means, all the sperms will carry the recessive autosomal allele responsible for phenylketonuria.
- Whereas, since hypertrichosis is X linked dominant disorder, half the sperms will be X (and would carry the disease causing gene), while rest half would be Y.
- In other words, half the sperms will get the holandric genes (present on Y chromosome).
Thus the probability of a sperm having one recessive autosomal allele and holandric gene is -
the probability of sperm having recessive allele (=1) multiplied with probability of a sperm having holandric gene (=1/2).
= 1*1/2
= 1/2
Hence,
The probability of a sperm having one recessive autosomal allele and holandric gene is 1/2