FREE NEET Biology Practice MCQ Test: Molecular Basis Of Inheritance Exercise 3 Questions Answers With Detailed Explanations [PDF]

Answer: (b)

β -galactosidase enzyme will be produced in a cell in which non-sense mutation takes place in the lac y gene. Non-sense mutation stops polypeptide synthesis due to the formation of non-sense codon. In lac operon, sequence of structural genes is z (codes for β -galactosidase), y (permease) and a gene (transacetylase). Due to non-sense mutation at y-gene, permease synthesis will be stopped resulting in non-expression of both y and successive gene a also. Thus, only β -galactosidase enzyme will be produced.

Answer: (d)

The genetic code consists of 64 triplets of nucleotides. These triplets are called codons. With three exceptions, each codon encodes for one of the 20 amino acids used in the synthesis of proteins. That produces some redundancy in the code. Most of the amino acids being encoded by more than one codon. The genetic code can be expressed as either RNA codons or DNA codons.

Answer: (c)

Answer: (c)

Answer: (d)

The process of formation of protein sequence from DNA strand is called transcription which requires RNA polymerase chain. RNA polymerase chains are of 3-typesin eukaryotes

1. RNA polymerase-I
2. RNA polymerase II
3. RNA polymerase-III

Answer: (d)

Ribosomal RNA (rRNA) is most abundant in animal cell. It constitutes 80% of total RNA of the cell.

Answer: (b)

Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Adenine cannot pair with cytosine. Adenine pairs with thymine and cytosine pairs with guanine. This occurs because adenine pairs with thymine with two hydrogen bonds, i.e. have only two hydrogen donor / hydrogen acceptor sites whereas cytosine has three hydrogen donor / hydrogen acceptor sites. Thus, due to lack of complementarity between the hydrogen donor and hydrogen acceptor sites between adenine and cytosine, these cannot pair.

Answer: (d)

Prokaryotic chromosomes are circular and contain termination sequences.

Answer: (d)

It was given by Geneticists George W. Beadle and E. L.Tatum which states that each gene in an organism controls the production of a specific enzyme. It is these enzymes that catalyze the reactions that lead to the phenotype of the organism.

Answer: (a)

The phosphorylated nucleotides are deATP (deoxy Adenosine Triphosphate), deCTP (deoxy Cytidine Triphosphate), deTTP (deoxy Thymidine Triphosphate). These triphosphates serve dual purpose. These act as substrate as well as provide energy for polymerisation of nucleotides by releasing energy after dissociating the phosphate group.

Answer: (a)

Answer: (d)

Answer: (a)

The incorrect match and can be corrected as

Sequence annotation is simply sequencing the whole set of genome that contained all the coding and non-coding sequence and later assigning different regions in the sequence with functions.

Rest of the matches are correct.

Answer: (c)

According to Chargaff purines and pyrimidines are in equal amounts. Purine (adenine) is equimolar with pyrimidine (thymine) and purine (guanine) is equimolar with pyrimidine (cytosine). Base ratio is specific for species.

Answer: (d)

Answer: (c)

Genetic code is the depiction of codon by which the information in RNA is decoded in a polypeptide chain. The information is transferred in the form of triplet of bases coding for one amino acid. It is triplet, universal, non-ambiguous and degenerate in nature.

Answer: (c)

Answer: (c)

Answer: (a)

Regulator is a gen e which forms a biochemical for suppressing the activity of operator gene. Promoter is the gene which provides the point of attachment to RNA polymerase required for transcription of structural genes.

Answer: (c)

At the 5′ end of eukaryotic mRNA, 7 mG (7-methyl guanosine) is present. In eukaryotes, primary transcript is often larger than the functional RNAs. Therefore, post-transcription processing is required to convert primary transcript of all types of RNAs into functional RNAs. It is of four types; cleavage, splicing, terminal addition and nucleotide modification.

The terminal additions include capping and tailing. In capping, 7 mG is added to 5′ end of mRNA.