Genetics Probability: Albino Son Calculation
This question involves calculating the probability of a specific genetic outcome: an albino son from a couple heterozygous for the albino allele. We assume albinism is an autosomal recessive condition.
Allele Representation
Let 'A' denote the dominant allele for normal pigmentation and 'a' represent the recessive allele causing albinism.
Parental Genotypes
A couple who are heterozygous for the albino allele both have the genotype Aa.
Probability of Albino Genotype
To determine the probability of their offspring inheriting the albino genotype (aa), we analyze the cross between two Aa individuals:
- Parent 1's possible gametes: A, a
- Parent 2's possible gametes: A, a
| A | a | |
|---|---|---|
| A | AA | Aa |
| a | Aa | aa |
The resulting offspring genotypes are AA, Aa, and aa. The genotype for albinism is aa. Based on the Punnett square, the probability of an offspring having the aa genotype is $\frac{1}{4}$.
Probability of Male Offspring (Son)
The probability of any child being a boy (a son, possessing XY chromosomes) is $\frac{1}{2}$. This probability is independent of the autosomal gene inheritance for albinism.
Combined Probability Calculation
To find the probability of both events occurring (the child being albino AND being a son), we multiply their individual probabilities:
P(albino son) = P(albino genotype aa) $\times$ P(son)
P(albino son) = $\frac{1}{4} \times \frac{1}{2}$
P(albino son) = $\frac{1}{8}$
Final Answer
The probability that a couple heterozygous for the albino allele will give birth to an albino son is $\frac{1}{8}$.
Important Questions from Mendelian principles
In summer squash, white colour fruit (W) is dominant over yellow colour (w) and disc-shaped phenotype (D) is dominant over sphere-shaped phenotype (d). Determine the genotype of the parents if the cross between white, sphere crossed with white, sphere gives 3/4 white, sphere and 1/4 yellow, sphere.
Body weight of rabbits is determined by pairs of alleles at two loci, 'a' and 'b', that are additive and equal in their effects. Rabbits with genotype a-a-b-b- have average 1kg body weight, whereas individuals with genotype a+ a+ b+ b+ have animals that average 3.4 kg in weight. A male rabbit with a- a- b- b- is crossed with a female of genotype a+ a+ b+ b+. What will be predicted average weight of F1 progeny of this cross?
A plant that produces disc-shaped fruit is crossed with another plant that produces long fruit. All the F1 plants gave disc-shaped fruits. When the F1 were intercrossed, F2 progeny were produced in the following ratio : 9/16 plants with disc-shaped fruits; 6/16 plants with spherical fruits and 1/16 plants having long fruits. Which one of the following options gives correct genotype of spherical fruits obtained in F2?
In a mammal, coat colour is governed by gene B, The coat colour is either black or brown, depending on whether the genotype is BB or Bb. It is not known which of these genotypes lead to the black and brown colours. The genotype bb results in albino coat colour. Further, the genotype cc suppresses the expression of coat colour resulting in albino coat colour. An albino male was crossed with a brown female and the resulting progeny had individuals with either black or brown coats. From this observation it can be inferred that the genotype of the male and female that were crossed are:
Assuming that the A, B, C and D genes are not linked, the probability of a progeny being AaBBccDd from a cross between AABbccDd and aaBBccDD parents will be
