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
1/4
Probability of Progeny Genotype
The question asks for the probability of obtaining a specific progeny genotype (AaBBccDd) from a genetic cross between two parents (AABbccDd and aaBBccDD). We are told that the genes A, B, C, and D are not linked. This is a key piece of information because it means that the inheritance of alleles at one gene locus is independent of the inheritance of alleles at the other gene loci. Therefore, we can calculate the probability of the desired genotype at each locus separately and then multiply these probabilities together to find the overall probability of the combined genotype.
Let's analyze each gene locus independently:
- Gene A: The cross is AA × aa.
- Parent 1 (AABbccDd) is AA for gene A. It produces gametes with only allele A.
- Parent 2 (aaBBccDD) is aa for gene A. It produces gametes with only allele a.
- When an A gamete combines with an a gamete, the resulting progeny genotype is Aa.
- All progeny from this cross (AA × aa) will have the genotype Aa.
- The probability of a progeny being Aa is \(P(Aa) = 1\).
- Gene B: The cross is Bb × BB.
- Parent 1 (AABbccDd) is Bb for gene B. It produces gametes with allele B (probability 1/2) and allele b (probability 1/2).
- Parent 2 (aaBBccDD) is BB for gene B. It produces gametes with only allele B (probability 1).
- To get the genotype BB in the progeny, a gamete with B from Parent 1 must combine with a gamete with B from Parent 2.
- Probability of B from Parent 1 = 1/2.
- Probability of B from Parent 2 = 1.
- The probability of progeny being BB is \(P(BB) = P(\text{B from Parent 1}) \times P(\text{B from Parent 2}) = \frac{1}{2} \times 1 = \frac{1}{2}\).
- Gene C: The cross is cc × cc.
- Parent 1 (AABbccDd) is cc for gene C. It produces gametes with only allele c.
- Parent 2 (aaBBccDD) is cc for gene C. It produces gametes with only allele c.
- When a c gamete combines with a c gamete, the resulting progeny genotype is cc.
- All progeny from this cross (cc × cc) will have the genotype cc.
- The probability of a progeny being cc is \(P(cc) = 1\).
- Gene D: The cross is Dd × DD.
- Parent 1 (AABbccDd) is Dd for gene D. It produces gametes with allele D (probability 1/2) and allele d (probability 1/2).
- Parent 2 (aaBBccDD) is DD for gene D. It produces gametes with only allele D (probability 1).
- To get the genotype Dd in the progeny, a gamete with d from Parent 1 must combine with a gamete with D from Parent 2.
- Probability of d from Parent 1 = 1/2.
- Probability of D from Parent 2 = 1.
- The probability of progeny being Dd is \(P(Dd) = P(\text{d from Parent 1}) \times P(\text{D from Parent 2}) = \frac{1}{2} \times 1 = \frac{1}{2}\).
Calculating Combined Probability
Since the genes are unlinked, the probability of the combined genotype AaBBccDd is the product of the probabilities of obtaining the desired genotype at each locus:
\(P(\text{AaBBccDd}) = P(Aa) \times P(BB) \times P(cc) \times P(Dd)\)
Substituting the probabilities we calculated:
\(P(\text{AaBBccDd}) = 1 \times \frac{1}{2} \times 1 \times \frac{1}{2}\)
\(P(\text{AaBBccDd}) = \frac{1}{4}\)
Thus, the probability of a progeny being AaBBccDd from the cross AABbccDd × aaBBccDD is 1/4.
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:
- What is the probability that a couple heterozygous for albino allele will give birth to an albino son?
