In humans, albinism is controlled by a recess | vedclass

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(D) $1$. The parents are heterozygous $(Aa 	imes Aa)$.$2$. The probability of a normal child $(AA$ or $Aa)$ is $3/4$, and the probability of an albin

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(D) $1$. The parents are heterozygous $(Aa 	imes Aa)$.$2$. The probability of a normal child $(AA$ or $Aa)$ is $3/4$, and the probability of an albino child $(aa)$ is $1/4$.$3$. For a family of $4$ children, we use the binomial distribution formula: $P(x) = {^nC_r} p^x q^{n-x}$.$4$. Here, $n = 4$, $x = 1$ (normal child), $p = 3/4$ (probability of normal), and $q = 1/4$ (probability of albino).$5$. $P(1 	ext{ normal}, 3 	ext{ albino}) = {^4C_1} 	imes (3/4)^1 	imes (1/4)^3$.$6$. $P = 4 	imes (3/4) 	imes (1/64) = 12/256 = 3/64 = 12/256$.$7$. Since the options are given out of $256$, the value is $12/256$. However, checking the calculation: $4 	imes (3/4) 	imes (1/64) = 12/256$. The correct option is $12/256$, but since it is not listed, we re-evaluate the binomial coefficient for $3$ albinos: ${^4C_3} 	imes (1/4)^3 	imes (3/4)^1 = 4 	imes (1/64) 	imes (3/4) = 12/256$.

Column-$I$	Column-$II$
$(A)$ Dominance	$(1)$ Many genes govern a single character.
$(B)$ Co-dominance	$(2)$ In a heterozygous organism, only one allele expresses itself.
$(C)$ Pleiotropy	$(3)$ In a heterozygous organism, both alleles express themselves fully.
$(D)$ Polygenic inheritance	$(4)$ $A$ single gene influences many characters.

List $I$	List $II$
$A$. Two or more alternative forms of a gene	$I$. Backcross
$B$. Cross of $F_1$ progeny with homozygous recessive parents	$II$. Ploidy
$C$. Cross of $F_1$ progeny with any of the parents	$III$. Allele
$D$. Number of chromosome sets in plant	$IV$. Test cross

In humans, albinism is controlled by a recessive gene '$a$'. If both parents are known to be carriers $(Aa)$ for albinism, what is the probability of having $1$ normal and $3$ albino children in a family of $4$ (in $/256$)?

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