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Hereditary and Variation MCQs for NEET — Botany Questions with Answers

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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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The Questions consist of two statements each - Assertion (A) and Reason (R).
To answer these questions choose any one of the following four responses. Assertion (A) - chromosomes undergo Segregation and independent assortment.
Reason (R) - During mitosis, their number is reduced into half.

A. If both (A) and (R) are true and (R) is the correct explanation of (A) B. If both (A) and (R) are true but (R) is not the correct explanation of (A) C. If (A) is true but (R) is false D. If both (A) and (R) are false.

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Explanation

Assertion (A) is true because chromosomes do undergo segregation and independent assortment according to Mendel's laws. Reason (R) is false because the reduction in chromosome number by half occurs during meiosis, not mitosis. Therefore, (A) is true but (R) is false.

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The Questions consist of two statements each - Assertion (A) and Reason (R).
To answer these questions choose any one of the following four responses. Assertion (A) :- pedigree is same in colourblindness and hacmophila Reason (R) :- Colourblindness and hacmophilia are X -linked recessive traits.

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Explanation

Colourblindness and hemophilia are indeed both X-linked recessive traits, meaning they are carried on the X chromosome and manifest primarily in males. The pedigrees for these conditions will look similar because they follow the same inheritance pattern. Therefore, both the Assertion (A) and Reason (R) are true, and (R) is the correct explanation of (A).

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The genes Controlling the seven characters of a pea plant studied by Mendel are now known to be located on how many different chromosomes? AIPMT - 2003

A. Seven B. Six C. Five D. Four

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