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Botany MCQs for NEET — Practice Questions with Answers

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The question consist of two statements each Assertion (A) and Reason (R) So answer these question choose anyone of the following four respones A:- Abraham Linconhas been analyzed for evidence of a genetic disorder called Marfanâ€™s syndrome R :- An additional benefit of DNAfingerprint technology is the diognos is of inherited disorders

A. If both (A) and (R) one 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) as false

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Explanation

Both Assertion (A) and Reason (R) are true. Abraham Lincoln has been analyzed for Marfan's syndrome, and DNA fingerprint technology can indeed be used to diagnose inherited disorders. However, the reason provided does not explain the assertion directly; it is an additional benefit of the technology.

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The question consist of two statements each Assertion (A) and Reason (R) So answer these question choose anyone of the following four respones A :- Helicase is called unwindase R :- DNA helix uncoils and splits into single strands by breaking of hydrogen bonds between complementary bases

A. If both (A) and (R) one 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) as false

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Explanation

Both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A). Helicase is called unwindase because it uncoils the DNA helix by breaking the hydrogen bonds between complementary bases, thereby splitting it into single strands.

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The question consist of two statements each Assertion (A) and Reason (R) So answer these question choose anyone of the following four respones A :- Formation of Lagging strand is slow R : - Formation of lagging strand begins bit later than that of leading strand

A. If both (A) and (R) one 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) as false

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Explanation

The formation of the lagging strand is indeed slower than the leading strand because it is synthesized in short, discontinuous segments known as Okazaki fragments. This is due to the fact that DNA polymerase can only synthesize DNA in the 5' to 3' direction, and the lagging strand template is oriented in the 3' to 5' direction. Therefore, the synthesis of the lagging strand begins a bit later than the leading strand as it must wait for the replication fork to open further. Hence, both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).

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The question consist of two statements each Assertion (A) and Reason (R) So answer these question choose anyone of the following four respones A : Doublet codons are inadequate for 20 types of aminoacids R : One aminoacid can be coded by many codons in triplet gentic code

A. If both (A) and (R) one 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) as false

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The question consist of two statements each Assertion (A) and Reason (R) So answer these question choose anyone of the following four respones A:- Mutations effect protein structure and function R :- Only one changed codon may be mis sense when it changes insertion of one aminoacid

A. If both (A) and (R) one 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) as false

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What would happen if in a gene encoding polypeptide of 50 aminoacids 25th codon (UAU) is mutated to UAA ? (AIPMT - 2003)

A. A Polypeptide of 24 aminoacids will be formed B. Two polypeptides of 24 and 25 aminoacids will be formed C. A polypeptide of 49 aminoacids will be formed D. A polypeptide of 25 aminoacids will be formed

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Explanation

In the given scenario, the original codon (UAU) codes for the amino acid Tyrosine. When it mutates to UAA, it becomes a stop codon. This means that translation will terminate prematurely at this point. Therefore, instead of a polypeptide of 50 amino acids, only the first 24 amino acids will be formed before the translation process stops at the 25th codon. Hence, a polypeptide of 24 amino acids will be formed.

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What does â€œLacâ€ refer to in what we call the lac operon ? (AIPMT - 2003)

A. Lactose B. Lactase C. Lac insect D. The number 1,00,000

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Explanation

The 'lac' in 'lac operon' stands for 'lactose'. The lac operon is a set of genes with a single promoter that are involved in the metabolism of lactose in Escherichia coli and some other bacteria. When lactose is present, it acts as an inducer by binding to the repressor and allowing transcription of the operon's genes to proceed.

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During transcription if the nucleotide sequence of the DNAstrand that is being coded is ATACG then the nucleotide sequence in the m RNA would be (AIPMT - 2004)

A. TATGC B. TCTGG C. UAUGC D. UATGC

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Explanation

During transcription, the DNA sequence is transcribed into mRNA. The base pairing rules are as follows: Adenine (A) pairs with Uracil (U) in RNA, and Cytosine (C) pairs with Guanine (G). Therefore, if the DNA sequence being coded is ATACG, the complementary mRNA sequence would be UAUGC.

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Which one of the following makes use of RNA template to synthesize DNA (AIPMT - 2005)

A. DNA polymerase B. RNA polymerase C. Reverse transcriptase D. DNA dependant RNA polymerase

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Explanation

Reverse transcriptase is an enzyme that transcribes RNA into DNA. It is used by retroviruses to integrate their genetic material into the host genome. This process is crucial in molecular biology, especially in the context of retroviruses like HIV.

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E coli cells with a mutant z gene of the lac operon cannot grow in mediumcontaining only lactose as the source of energy because (AIPMT - 2005)

A. the lac operon is constitutively active in these cells B. they cannot synthesize functional beta galactosidase C. in the presence of glucose E Coli cell do not utilize lactose D. they cannot transport lactose from the medium into the cell

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Explanation

The z gene in the lac operon codes for the enzyme beta-galactosidase. This enzyme is responsible for breaking down lactose into glucose and galactose. A mutation in the z gene means that E. coli cannot produce functional beta-galactosidase, thereby making it unable to utilize lactose as an energy source.

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