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NEET Practice Questions (MCQs) with Answers & Solutions

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All Physics Chemistry Botany Zoology

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Out of the following , Find out the reason for occurrence of free martin in cattles ?

A. Both the twins are connected by a common umbilical cord B. Gonad of the male develop earlier then those of the female C. Male hormones reach the female embryo and influence the development of male sex in the female embryo D. All of the above

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Explanation

The occurrence of Free Martin in cattle is due to multiple factors: both twins being connected by a common umbilical cord, the earlier development of male gonads, and the influence of male hormones on the female embryo. These combined factors result in the female embryo developing intersex characteristics and becoming sterile.

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Which factor in nature causes discontinuous Variation in a population ?

A. Recombination B. Shuffling of parental genes C. Mutation D. Geneflow

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Explanation

Discontinuous variation in a population is primarily caused by mutations. Mutations introduce new genetic variations that can lead to distinct phenotypic changes, which do not follow a continuous range but appear as distinct categories.

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This is reversal of mutation

A. Spontneous mutation B. Zygotic mutation C. Back mutation D. Recessive mutation

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Explanation

A back mutation, or reversion mutation, is a genetic change that reverses the effect of a previous mutation. This type of mutation restores the original sequence of DNA, thus reversing the mutation.

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State the cause of Philadelphia syndrome

A. Deletion from the long arm of chromosome number 20 B. Duplication in long arm of chromosome number 20 C. Deletion from the long arm of chromosome number 22 D. Deletion from the short arm of chromosome number 22

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What happens to excessive phenylalanine accumulation in the blood of phenylketonuria patient ?

A. Get deposited on synovial membrane B. excreted in urine C. Get deposited in storage organs D. All of the above

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Explanation

In patients with phenylketonuria (PKU), excessive phenylalanine accumulates in the blood and is converted to phenylketones, which are then excreted in the urine. This is a diagnostic feature of PKU.

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Name the enzyme needed for the following biochemical reaction Tyrosine Melanin

A. Melanase B. tyrosinase C. Tyrosine aminase D. transaminase

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Explanation

The enzyme needed for the conversion of Tyrosine to Melanin is Tyrosinase. Tyrosinase is a crucial enzyme in the pathway of melanin synthesis. It catalyzes the oxidation of tyrosine to dihydroxyphenylalanine (DOPA) and subsequently to DOPAquinone, which eventually leads to the formation of melanin.

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Match the following

Column I

Single gene inheritance

2. Double gene inheritance

3. Test cross

4. Incomplete dominance

Column II

a. 1 : 1: 1 : 1

b. 1 : 2 : 1

c. 3 : 1

d. 9 : 3 : 3 : 1

(1 - c) (2 - d) (3 - a) (4 - b)

(1 - d) (2 - c) (3 - b) (4 - a)

(1 - a) (2 - b) (3 - d) (4 - c)

(1- d) (2 - b) (3 - a) (4 - c)

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Explanation

The correct matching is: (1 - c), (2 - d), (3 - a), (4 - b). Here's the reasoning:

Single gene inheritance follows a 3:1 ratio.
Double gene inheritance follows a 9:3:3:1 ratio.
Test cross results in a 1:1:1:1 ratio.
Incomplete dominance results in a 1:2:1 ratio.

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Match the following Column I Blood groups Possible Genotypes

A
B
AB
O Column II (a) Id^Bd Id^Bd, Id^Bd i (b) ii (c) Id^Ad Id^Ad, Id^Ad i (d) Id^Ad Id^Bd

A. (1 - a) (2 - b) (3 - c) (4 - d) B. (1 - c) (2 - a) (3 - b) (4 - d) C. (1 - c) (2 - b) (3 - d) (4 - a) D. (1 - c) (2 - a) (3 - d) (4 - b)

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Explanation

The correct matching is: (1 - c), (2 - a), (3 - d), (4 - b). Here's the reasoning:

Blood group A can have the genotypes IAIA or IAi.
Blood group B can have the genotypes IBIB or IBi.
Blood group AB has the genotype IAIB.
Blood group O has the genotype ii.

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The cross between heterozygous A blood group mother with B blood group father heterozygous).What will be the expected blood group of First filial generation.

A. AB group B. A group C. B group D. O group E. All of the above

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Explanation

The cross between a heterozygous A blood group mother (IAi) and a B blood group father (IBIB or IBi) results in various possible genotypes for the offspring: IAIB (AB blood group), IAi (A blood group), IBi (B blood group), and ii (O blood group). Hence, the expected blood groups of the First filial generation can be A, B, AB, or O. Therefore, the correct answer is not listed in the provided options.

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The value of X/A is given in column A and sex of the fly is written in column B. Match the following :- Column I

1
1.5
0.67
0.33
0.5 Column II (a) Normal Male (b) super male Sterile (c) inter sex (d) super Female sterile (e) Normal Female

A. (1 - a) (2 - d) (3 - c) (4 - e) (5 - b) B. (1 - e) (2 - d) (3 - c) (4 - b) (5 - a) C. (1 - b) (2 - a) (3 - c) (4 - d) (5 - e) D. (1 - e) (2 - c) (3 - d) (4 - a) (5 - b).

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Explanation

In Drosophila, the sex of the fly is determined by the ratio of X chromosomes to autosomes (X/A ratio). The correct matches are:

X/A = 1 corresponds to a normal female (e).
X/A = 1.5 corresponds to a super female sterile (d).
X/A = 0.67 corresponds to an intersex (c).
X/A = 0.33 corresponds to a super male sterile (b).
X/A = 0.5 corresponds to a normal male (a).

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