Sampling of 200 persons for their ABO blood group was done from an urban area. The types of blood group observed in the given population are as follows: A = 60, B = 32, AB = 10 and O = 98 Which of the following gives the correct frequency of blood group determining alleles IA , IB and IO in the given population?
IA = 0.19, IB = 0.11, IO = 0.7
The ABO blood group system in humans is a classic example of multiple alleles and codominance. It is determined by three alleles: IA, IB, and IO. The alleles IA and IB are codominant with respect to each other, and both are dominant over the allele IO.
The genotypes and corresponding phenotypes (blood groups) are as follows:
We are given data from a sample of 200 persons in an urban area, showing the distribution of ABO blood groups:
| Blood Group | Number of Persons | Observed Frequency |
|---|---|---|
| A | 60 | 60/200 = 0.30 |
| B | 32 | 32/200 = 0.16 |
| AB | 10 | 10/200 = 0.05 |
| O | 98 | 98/200 = 0.49 |
| Total | 200 | 1.00 |
We want to find the frequencies of the alleles IA, IB, and IO in this population. Let the frequency of allele IA be p, the frequency of allele IB be q, and the frequency of allele IO be r. According to the Hardy-Weinberg principle, the sum of allele frequencies is 1, i.e., $p + q + r = 1$.
The frequencies of the genotypes are related to the allele frequencies as follows:
The frequencies of the phenotypes are the sum of the frequencies of the corresponding genotypes:
We can estimate the allele frequencies from the observed phenotype frequencies. The easiest allele frequency to calculate is that of IO (r), because only the genotype IOIO results in blood group O.
Frequency(Blood Group O) = $r^2 = 0.49$
So, the frequency of allele IO is:
$r = \sqrt{0.49} = 0.7$
Next, we can use the cumulative frequencies of blood groups A and O, and B and O:
Frequency(Blood Group A) + Frequency(Blood Group O) = $(p^2 + 2pr) + r^2 = (p + r)^2$
Frequency(Blood Group B) + Frequency(Blood Group O) = $(q^2 + 2qr) + r^2 = (q + r)^2$
Using the observed frequencies:
$(p + r)^2 = 0.30 + 0.49 = 0.79$
$p + r = \sqrt{0.79} \approx 0.8888$
$(q + r)^2 = 0.16 + 0.49 = 0.65$
$q + r = \sqrt{0.65} \approx 0.8062$
Now we can find p and q using the value of r = 0.7:
$p = (p + r) - r \approx 0.8888 - 0.7 \approx 0.1888$
$q = (q + r) - r \approx 0.8062 - 0.7 \approx 0.1062$
Let's check if the sum of frequencies is close to 1:
$p + q + r \approx 0.1888 + 0.1062 + 0.7 = 0.995$
This sum is close to 1, with the slight difference due to rounding the square roots. The frequencies are approximately $p \approx 0.19$, $q \approx 0.11$, and $r = 0.7$.
These values correspond to the frequencies of alleles IA, IB, and IO, respectively.
Frequency of IA $\approx$ 0.19
Frequency of IB $\approx$ 0.11
Frequency of IO = 0.70
These calculated allele frequencies match one of the given options.
Mendel selected which of the following traits for his studies?
Two strains of mice which are genetically identical except for a single genetic locus or region are said to be:
Interacting genes which are involved in producing continuous variation in phenotypes in a population are known as/constitute
The table given below shows the Lod score values of three different pairs of genes studied for assessing if they are linked pairs:
| Gene Pair '1' | Gene Pair '2' | Gene Pair '3' | |
| Lod Score | 1 | 2 | 3 |
The following conclusions were made from the data given above:
A. For gene pair 1 the probability of the genes being linked is 10 times more likely than them assorting independently.
B. For gene pair 2, the Lod score 2 indicates that the probability of the genes being linked is twice more likely than assorting independently.
C. The genes of pair 1 and 2, can both be considered as linked, while the genes of pair 3 exhibits independent assortment.
D. The genes of pair 3 can be considered as linked.
Which one of the following options represents statement(s) that is/are correct?
Columns X and Y of the following table list some treatment methods, reagents, and events that are related to human lymphocyte culture, and banding/ karyotyping of human chromosomes.
Column X | Column Y | ||
A. | 5% barium hydroxide treatment at 50°C | I. | R-banding |
B. | Trypsin treatment | II. | C-banding |
C. | Phytohaemagglutinin | III. | Mitotic stimulation |
D. | Phosphate buffer treatment at 80°C | IV. | Nucleolar Organizer Regions (NOR) |
E. | Silver Staining | v. | G-banding |
Which one of the following options represents all correct matches between Column X and Column Y?