Genetics MCQ 2 Free Online Test 2026

Since the genes for coat color do not show a dominant-recessive relationship and one gene is incompletely dominant over the other gene, they show a blended progeny. Thus, it is an example of incomplete dominance.

When the genotypes of all the progeny is listed, 6 out of 16 possible genotypes have either of the dominant alleles.

Only those genes which are located on different chromosomes undergo independent assortment. Those genes which are located on the same chromosome are said to represent a linkage group and are inherited together. They experience independent assortment only if they are located far apart and crossing over can occur between them.

he probability of having a male child is ½ . The probability of a male child being colorblind is also ½. Since the child has to be a male and color blind at the same time, we apply the product rule. Since we need to find out the probability of both the children being male and colorblind, So, 1/2*1/2*1/2*1/2 = 1/16 or 6.25%.

The trait skips a generation and both males and females are affected by it. So it is most likely an autosomal recessive trait. Since the mother in the 2nd generation is affected by the trait, she would have the genotype aa. One of their children is also affected, so the father must be heterozygous for the trait i.e. Aa.

According to the question
B → Brown hair
b → bloude hair
The two mating partners are heterozygous

∴ Probability for a brown haired progeny (p) = 3/4 
Probability for a blonde haired progeny (q) = 1/4
According to the question, at least one of the next three progenies have blonde hair. Therefore, following combinations of binomials will satisfy the condition-
(1) Only one progeny of the three has blonde hair
(2) Two of the three progenies have blonde hair
(3) All of them have blonde hair
Using binomial expression. the Probabilities of each possible combination can be calculated as
(1) n = 3 = Total progenies
x = 1 = No. of progenies with blonde hair
y = 2 = No. of progenies with blonde hair
p= 1/4 = Probability of having blonde hair
q = 3/4 = Probability of having brown hair 

A large population size is necessary to accommodate rare phenotypes in the population which would otherwise go amiss.

Given that the genes A and B are linked by 50 map unit apart. So the frequency of recombinant is 50%. That is crossing the AaBb with aabb will results 50% recombinants and 50% parental gametes (AB : aB : Ab : ab : : 1 : 1 : 1 : 1). Thus the genes will seggregate by independent assortment as proposed by Mendel.
According to Mendelian dihybrid cross selfing of AaBb will result in aabbprogeneis in the ratio of 1/16.

Phenylketonuria is an example of plieotropy in which mutation in a gene can cause multiple unrelated phenotypic symptoms and reduced hair pigmentation is one of them. 

The genes complement each other when they are present on two different chromosomes and produce a wild type product. This can be seen using a complementation test. 

Recessive epistasis (or supplementary genes interaction) is the condition when homozygous recessive of one gene mask the effect of other gene in the dominant or recessive condition. Presence of B allele is dominant against E or e, results in black and E produces brown in the absence of B allele, whereas recessive homozygous ‘ee’ is epistatic to B or b and thus, gives yellow. The cross can be explained as follows   

Based on the given statements the inheritance of the trait can be explained as X-linked recessive. This is because.
1. Males are heterozygous X-chromosome (XY). So X linked trait usually shows in males due to single X- chromosome.
2. Affected mother (X^aX^a) gives all sons affected (X^aY) and daughter as carrier(X^aX).
Female progenies develop disease when the father is affected (X^aY) and mother is earner (X^aX) for the allele.

3. Males are heterozygous X-chromosome (XY). So X linked trait usually shows in males due to single X- chromosome.
4. Affected mother (X^aX^a) gives all sons affected (X^aY) and daughter as earner (X^aX).
Female progenies develop disease when the father is affected (X^aY) and mother is earner (X^aX) for the allele.

Mendel’s laws do not apply to these traits as several genes are responsible for a single trait and it is impossible to track each on them on the basis of Mendel’s principles. These genes may or may not necessarily be present on the same chromosome. The genotype as well as the environment influence these traits. 

The expression of a phenotype is limited to one sex in a sex limited trait but is not true for  a sex-influenced trait.

The species is a diploid and the gametes would be haploid and contain n number of chromosomes i.e. 10. The somatic cells would contain 2n number of chromosomes i.e 20. Only 10 different types of chromosomes are present in the species. 

The counting and identification of chromosomes in the karyotype of an individual can diagnose aneuploidies, diseases caused by an alteration in the number of chromosomes in relation to the normal number in the species. It does not detect diseases caused due to mutations.

Males are affected in Klinefelter’s syndrome. They have 2 X and 1 Y chromosome and are sterile. 

When AaBb and AaBb are crossed with each other, the probability of getting a genotype Aa will be ½. Same way the probability of getting genotype Bb will be ½.
Using the product rule, to get an individual with both Aa and Bb, we will multiply the individual probabilities. This gives us ¼ as the probability of getting a baby with genotype AaBb.

The mother will have genotype XX’ and father will be XY. The genotypes of their babies could be XX, XY, XX’ and X’Y. Only the male child will be affected. Therefore, the probability is ¼ or 25%.

q2 = 0.09, then q = 0.3
p = 1-q = 0.7
heterozygous population = 2pq = 2*0.7*0.3 = 42%

If q2 = 0.85
Then q = “0.85 = 0.921
p = 1 – q = 1 – 0.921 = 0.079