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Heredity and Variation: MCQs Quiz - 4
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Question 1 |
If two pea plants having red (dominant) coloured flowers with unknown genotypes are crossed, 75% of the flowers are red and 25% are white. The genotypic constitution of the parents having red coloured flowers will be
Both heterozygous | |
One homozygous and other heterozygous | |
Both homozygous | |
Both hemizygous |
Question 2 |
In a given plant, red colour (R) of fruits is dominant over white fruit (r); and tallness (T) is dominant over dwarfness (t). If a plant with genotype RrTt is crossed with a plant of genotype rrtt, what will be the percentage of tall plants with red fruits in the next generation?
20% | |
40% | |
50% | |
60% |
Question 3 |
In Drosophila, the sex is determined by:
The ratio of pairs of X-chromosomes to the pairs of autosomes | |
X and Y chromosomes | |
The ratio of number of X-chromosomes to the sets of autosomes | |
Whether the egg is fertilized or develops parthenogenetically |
Question 4 |
In Mendel's experiments with garden pea, round seed shape (RR) was dominant over wrinkled seeds (rr), yellow cotyledon (YY) was dominant over green cotyledon (yy). What are the expected phenotypes in the F2 generation of the cross RRYY × rryy?
Only wrinkled seeds with green cotyledons | |
Only wrinkled seeds with yellow cotyledons | |
Only round seeds with green cotyledons | |
Round seeds with yellow cotyledons, and wrinkled seeds with yellow cotyledons |
Question 5 |
In pea plants, yellow seeds are dominant to green. If a heterozygous yellow seeded plant is crossed with a green seeded plant, what ratio of yellow and green seeded plants would you expect in F1 generation?
3 : 1 | |
50 : 50 | |
9 : 1 | |
1 : 3 |
Question 6 |
Mating of an organism to a double recessive in order to determine whether it is homozygous or heterozygous for a character under consideration is called
reciprocal cross | |
test cross | |
dihybrid cross | |
back cross |
Question 7 |
When a tall plant with round seeds (TTRR) crossed with a dwarf plant with wrinkle seeds (tfrr). The F1 generation consists of tall plants with round seeds. What would be the proportion of dwarf plant with wrinkle seeds in F1 generation?
0 | |
1/2 | |
1/4 | |
1/16 |
Question 8 |
Phenotype of an organism is the result of
Environmental changes and sexual dimorphism | |
Cytoplasmic effects and nutrition | |
Mutations and linkages | |
Genotype and environment interactions |
Question 9 |
Sickle cell anemia is
Characterized by elongated sickle like RBCs with a nucleus | |
An autosomal linked dominant trait | |
Caused by substitution of valine by glutamic acid in the β globin chain of haemoglobin | |
Caused by a change in a single base pair of DNA |
Question 10 |
Which one of the following conditions correctly describes the manner of determining the sex in the given example?
Homozygous sex chromosomes (XX) produce male in Drosophila | |
Homozygous sex chromosomes (ZZ) determine female sex in birds | |
XO type of sex chromosomes determine male sex in grasshopper | |
XO condition in humans as found in Turner Syndrome, determines female sex |
Question 11 |
Test cross of dihybrid ratio 1 : 1 : 1 : 1 then it proves that
F1 hybrid produces four different progenies | |
F1 hybrid produces two different progenies | |
F1 hybrid is homozygous | |
Four different progenies are produced by parents |
Question 12 |
The F2 generation offspring in a plant showing incomplete dominance, exhibit
variable genotypic and phenotypic ratios | |
a genotypic ratio of 1 : 1 | |
a phenotypic ratio of 3 : 1 | |
similar phenotypic and genotypic ratios of 1 : 2 : 1 |
Question 13 |
Two genes R and Y are located very close on the chromosomal linkage map of maize plant. When RRYY and rryy genotypes are hybridized, the F2 segregation will show:
Higher number of the parental types | |
Higher number of the recombinant types | |
Segregation in the expected 9 : 3 : 3 : 1 ratio | |
Segregation in 3 : 1 ratio |
Question 14 |
Two pea plants were subjected cross pollination. Of the 183 plants produced in the next generation, 94 plants were found to be tall and 89 plants were found to be dwarf. The genotypes of the two parental plants are likely to be
TT and tt | |
Tt and Tt | |
Tt and tt | |
TT and TT |
Question 15 |
Which of the following pair of features is a good example of polygenic inheritance?
Hair pigment of mouse and tongue rolling in humans | |
ABO blood group in humans and flower colour of Mirabilis jalapa | |
Human height and skin colour | |
Human eye colour and sickle cell anaemia |
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Thank you sir.It’s realy help us
Please make a correction in Q2, It has To be RrTt, instead of RRTt. That is if both is heterozygous then only the ratio is 50%.
Thank you sir.
sorry sir….but in Q 2 the genotype of plant should be RRTt….then only the ratio is 50% ….otherwise the no. of taal plants with red flowers is only 4 out of 16….so it cannot be 50%….it should be RRTt
in 9th que (d)is the correct ans and not (c),sickle cell anaemia is caused by the substitution of glutamic acid by valine but it is given opposite.consult NCERT
absolutely !!!
similarly there are few snag, but the problem is that after the user highlights it, they don’t do the needful!!!
Dear Dev/Priya, while I wholeheartedly appreciate your enthusiasm and participation on our blog and forum, simultaneously we need to verify the authenticity of student’s response as well. Consequently it might have delayed the response from publisher’s end. More importantly it is a free service ………., isn’t it? Dev thanking you honestly for your focus and replies.
ths test is helpful
In que. 10 in first option there will be male instead of make
Thanks Nisha for highlighting the errata. It has been corrected.