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NEET • BIOLOGY • PRINCIPLE OF INHERITANCE AND VARIATION

Principle of inheritance and variation practice · Hard

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31 Hard
31 of 66
Understanding +4 | -1
Sickle cell anaemia is an example of?
View answer and explanation
Correct option: C
Pleiotropy
Option C (Pleiotropy) is correct because sickle cell anemia is a single gene mutation that affects multiple organ systems and traits.
32 Hard
32 of 66
Analyzing +4 | -1
Assertion (A): An X-linked recessive trait shows transmission from carrier male to female progeny.

Reason (R): X-linked recessive trait shows non-crisscross inheritance.
View answer and explanation
Correct option: D
Both (A) and (R) are false
Option A (Sex-linked dominant) is correct because all daughters inherit the affected X chromosome from their father and express the disease, while sons inherit the Y chromosome and are unaffected.
33 Hard
33 of 66
Application +4 | -1
In dihybrid experiment on garden peas worked by mendel. What is the %\% of yellow round seed:
View answer and explanation
Correct option: A
56.25%56.25 \%
Option A (56.25%56.25 \% ) is correct because it represents the proportion of yellow round seeds in the F2 generation of a dihybrid cross.
34 Hard
34 of 66
Understanding +4 | -1
I. 100% parental combinations are found in F2{\mathrm{F}_2}-generation
II. F2{\mathrm{F}_2} phenotypic ratio is 3 : 1 in dihybrid cross
III. Dihybrid test cross ratio is 1 : 1 in F2{\mathrm{F}_2}-generation
IV. Linked genes tends to separate frequently
Choose the correct options from the above given statements
View answer and explanation
Correct option: D
I, II and III
Option D (I, II and III) is correct because the first three statements are generally true for linked genes and dihybrid crosses, while linked genes do not separate frequently unless crossing over occurs.
35 Hard
35 of 66
Understanding +4 | -1
14:12:14\frac{1}{4}:\frac{1}{2}:\frac{1}{4} ratio of TT : Tt : tt can be depicted mathematically binomial expression as (ideally)
View answer and explanation
Correct option: A
(ax+by)2{\left( {ax + by} \right)^2}
Option A ((ax+by)2{(ax + by)^2}) is correct because the 1:2:1 ratio represents the coefficients of the binomial expansion of (ax+by)2{(ax + by)^2}, which corresponds to the genotypic ratio of a monohybrid cross.
36 Hard
36 of 66
Understanding +4 | -1
Consider a gene that has two alleles and shows complete dominance. When two heterozygotes for this gene breed, they have a 25%25 \% chance of producing a homozygous recessive offspring. The next time they breed, what are the chances that they will once again have a homozygous recessive progeny?
View answer and explanation
Correct option: B
25%25 \%
Option B (25 %) is correct because each breeding event is independent, so the probability of having a homozygous recessive offspring remains constant at 25% for each event.
37 Hard
37 of 66
Remembering +4 | -1
XXY genotype shows:
View answer and explanation
Correct option: B
Hermaphrodite
Option A (Male) is correct because XXY genotype results in Klinefelter's syndrome, which is a male phenotype.
38 Hard
38 of 66
Analyzing +4 | -1
Genes A, B and C are linked. Genes A and B are more close than A and C. Find out the correct option for the given statements
I. A might be before B and C
II. B might be between A and C
III. C might be between A and B
IV. More crosses has occurred between A and C than A and B
View answer and explanation
Correct option: D
I, II and IV
Option D (I, II and IV) is correct because if A and B are closer than A and C, then A might be before B and C, B might be between A and C, and more crosses have occurred between A and C than A and B.
39 Hard
39 of 66
Application +4 | -1
How many gametes can be produced by a diploid organism, if it is heterozygous for 4 loci?
View answer and explanation
Correct option: C
1616
Option C (1616) is correct because the number of gametes produced by a diploid organism heterozygous for 4 loci is 24=162^4 = 16.
40 Hard
40 of 66
Understanding +4 | -1
Assertion (A): Multiple alleles can be found only when population studies are made.

Reason (R): Occasionally, a single gene product may produce more than one effect.
View answer and explanation
Correct option: B
Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
Option B (Both (A) & (R) are true but the (R) is not the correct explanation of the (A)) is correct because while both statements are true, the reason does not explain why multiple alleles are found through population studies.
41 Hard
41 of 66
Remembering +4 | -1
How many matching are corrects:

a. Haemophilia - Queen victoria pedigree

b. Phenylketonuria - Inborn error of metabolism

c. Colour blindness - More appears in female than male

d. Sickel cell anaemia - Shape of RBC is biconvex disc
View answer and explanation
Correct option: C
2
Option C (2) is correct because Haemophilia is indeed associated with Queen Victoria's pedigree, and Phenylketonuria is an inborn error of metabolism. Colour blindness is more common in males, and sickle cell anemia results in sickle-shaped RBCs, not biconvex discs.
42 Hard
42 of 66
Understanding +4 | -1
In a mutational event, when adenine is replaced by guanine, it is the case of
View answer and explanation
Correct option: C
transition
Option C (transition) is correct because a transition is a point mutation where a purine base is substituted with another purine base (A to G or G to A) or a pyrimidine base is substituted with another pyrimidine base (C to T or T to C).
43 Hard
43 of 66
Understanding +4 | -1
Assertion (A): Male and female ratio is almost equal in this world.

Reason (R): Male and female genotype is XX and XY respectively.
View answer and explanation
Correct option: A
Both (A) & (R) are true and the (R) is the correct explanation of the (A)
Option A (Both (A) & (R) are true and the (R) is the correct explanation of the (A)) is correct because the equal male and female ratio is a direct consequence of the sex chromosome system (XX/XY).
44 Hard
44 of 66
Understanding +4 | -1
A common test to find the genotype of hybrid is by;
View answer and explanation
Correct option: D
Crossing of F1F_1 progeny with recessive parent.
Option D (Crossing of F1F_1 progeny with recessive parent) is correct because a test cross involves crossing the F1 generation with a homozygous recessive parent to determine the genotype of the F1.
45 Hard
45 of 66
Understanding +4 | -1
Failure of cytokinesis after telophase stage of cell division results in an increase in whole set of chromosomes in an organism__________ phenomenon is
View answer and explanation
Correct option: B
Polyploidy
Option B (Polyploidy) is correct because polyploidy is the condition where an organism has more than two complete sets of chromosomes, often resulting from the failure of cytokinesis after telophase.
46 Hard
46 of 66
Application +4 | -1
If haploid chromosomal number in a mammal is 24, the monosomic number will be
View answer and explanation
Correct option: B
47
Option B (47) is correct because a monosomic condition means one chromosome is missing from the diploid set. If the haploid number is 24, the diploid number is 48, and monosomic would be 48-1=47.
47 Hard
47 of 66
Analyzing +4 | -1
Consider the following statements about the ABOA B O blood grouping in human beings and select the incorrect pair from the given options

a. ABO blood groups are controlled by gene I, having three alleles IA,IBI^A, I^B and ii.

b. IAI^A and IBI^B are completely dominant over i but IAI^A and IBI^{\mathrm{B}} show incomplete dominance when they are present together

c. The alleles IA\mathbf{I}^{\mathrm{A}} and IB\mathbf{I}^{\mathrm{B}} produce almost similar form of sugar but the i\mathrm{i} allele produce a completely different sugar

d. A total of six different genotypes and four phenotypes are possible
View answer and explanation
Correct option: B
b and c
Option B (b and c) is correct because IAI^A and IBI^B show codominance, not incomplete dominance, and the i allele produces no sugar, not a completely different sugar.
48 Hard
48 of 66
Application +4 | -1
In a pea plants, yellow seeds are dominant to green. If a heterozygous yellow seeded plant is crossed with a green seeded plant, what ratio of green and yellow seeded plants would you except progeny.
View answer and explanation
Correct option: C
0.042361111
Option A (0.125694444) is incorrect. The correct ratio is 1:1, so the proportion of green seeded plants is 0.5.
49 Hard
49 of 66
Understanding +4 | -1
Assertion (A): There are six different combinations or genotypes of the human ABO blood types.

Reason (R): The gene (I)(I) has three alleles IA,IBI^A, I^B and IO.I^{O}.
View answer and explanation
Correct option: A
Both (A) & (R) are true and the (R) is the correct explanation of the (A)
The correct answer is A (Both (A) & (R) are true and the (R) is the correct explanation of the (A)) because the ABO blood group system has three alleles, leading to six possible genotypes.
50 Hard
50 of 66
Analyzing +4 | -1
In rabbit, two recessive genes produce a solid body colour and long hair in contrast to dominant spotted body colour and short hair. The result of a cross between heterozygous spotted short haired rabbit to solid long haired rabbit gives spotted, short haired 48, spotted long haired 5, solid short haired 7, solid long haired 40, total 100. In terms of cross over units, how far are these two genes on the chromosome?
View answer and explanation
Correct option: C
12 map units
Option C (12 map units) is correct because the recombination frequency, calculated from the recombinant offspring, directly corresponds to the map distance between the genes.
51 Hard
51 of 66
Understanding +4 | -1
Number of Barr body which will found in case of Turner's
syndrome will be
View answer and explanation
Correct option: C
0
Option C (0) is correct because Turner's syndrome is characterized by the genotype XO, meaning there is only one X chromosome and therefore no Barr body.
52 Hard
52 of 66
Understanding +4 | -1
The reason why some mutations which are harmful do not get eliminated from gene pool is that:
View answer and explanation
Correct option: B
They are recessive and carried by heterozygous individuals
Option B (They are recessive and carried by heterozygous individuals) is correct because recessive mutations can persist in a population if they are masked by dominant alleles in heterozygous carriers.
53 Hard
53 of 66
Understanding +4 | -1
The phenotype of an individual may be affected if the modified allele produces

A. The normal / less efficient enzyme

B. No enzyme at all

C. A non-functional enzyme
View answer and explanation
Correct option: D
B and C are correct
Option D (B and C are correct) is correct because if the modified allele produces no enzyme or a non-functional enzyme, the phenotype of the individual will be affected.
54 Hard
54 of 66
Understanding +4 | -1
Assertion (A): In co-dominance and incomplete dominance, the genotypic & phenotypic ratios are same.

Reason (R): In case of co-dominance the F1F_1 generation resembles both parents.
View answer and explanation
Correct option: B
Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
Option B (Both (A) & (R) are true but the (R) is not the correct explanation of the (A)) is correct because while both statements are true, the reason doesn't explain the assertion.
55 Hard
55 of 66
Remembering +4 | -1
In Down syndrome:
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because broad palms, characteristic palm creases, and furrowed tongues are all common characteristics of Down syndrome.
56 Hard
56 of 66
Understanding +4 | -1
Assertion: To study mutation, haploids are used.

Reason: Most of the mutations are recessive.
View answer and explanation
Correct option: A
If both Assertion and Reason are true and
Reason is the correct explanation of Assertion
Option A is correct because haploids are used to study mutations because most mutations are recessive and will be expressed in haploids.
57 Hard
57 of 66
Understanding +4 | -1
Genetic variation cannot be effectively brought about by
View answer and explanation
Correct option: D
Amplification
Option D (Amplification) is correct because amplification refers to increasing the number of copies of a gene, which doesn't inherently introduce new genetic variation.
58 Hard
58 of 66
Analyzing +4 | -1
Assertion (A): Gynacomastia is observed in Turner's syndrome.

Reason (R): Such a disorder is caused due to the presence of extra copy of X-chromosome.
View answer and explanation
Correct option: D
Both (A) and (R) are false
Option D (Both (A) and (R) are false) is correct because Gynaecomastia is observed in Klinefelter's syndrome, not Turner's syndrome, and is caused by an extra X chromosome in males (XXY).
59 Hard
59 of 66
Understanding +4 | -1
Mendel work later formulated into laws of
I. Linkage
II. Segregation
III. Incomplete dominance
IV. Independent assortment
Choose the correct option
View answer and explanation
Correct option: B
II and IV
Option B (II and IV) is correct because Mendel's work was formulated into the laws of segregation and independent assortment.
60 Hard
60 of 66
Application +4 | -1
What is the probability of production of dwarf offsprings in a cross between two heterozygous tall pea plants.
View answer and explanation
Correct option: C
0.25
Option C (0.25) is correct because in a cross between two heterozygous tall pea plants (Tt x Tt), the probability of getting a dwarf offspring (tt) is 1/4 or 0.25.
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