Biology Question Bank – 131 MCQs on “Genes & Chromosomes” – Answered!

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131 Questions with Answers and Explanations on Genes & Chromosomes for Biology Students.

1. Genetic code consists of

(a) adenine and guanine

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(b) cytosine and uracil

(d) all the above.

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Answer and Explanation:

1. (d): The genetic information is transferred from DNA to m-RNA to protein. The proteins are made up of some 20 amino acids whose sequence is hidden in the sequence of nucleotides of mRNA. Hence, genetic code consists of all 20 amino acids. Thus genetic code is the relationship of amino acids sequence in a polypetide and nucleotide/base sequence in mRNA antisense strand and DNA.

2. Haploids are able to express both recessive and dominant alleles/mutations because there are

(a) many alleles for each gene

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(b) two alleles for each gene

(d) only one allele in a gene.

Answer and Explanation:

2. (c): Haploids are able to express both recessive and dominant alleles/ mutations because there are only one allele for each gene in the individual. Hybrid is a organism containing two different alleles or individual containing both dominant and recessive genes of an allelic pairs.

3. A family of five daughters only is expecting sixth issue. The chance of its beings a son is

(a) zero

(b) 25%

(d) 100%.

Answer and Explanation:

3. (c): A family of five daughter only is expecting sixth issue. The chance of its beings a son is 50%. Human have 22 pairs chromosomes which are XX in females and XY in males. So if we cross the parents there is 1 : 1 chance for boy and girl.

4. Mutations used in agriculture are commonly

(a) induced

(b) spontaneous

(d) recessive and lethal.

Answer and Explanation:

4. (a): Mutations used in agriculture are commonly induced. Induced mutations can be produced artificially by certain mutagenic agents called as mutagens. Ethyl methyl sulphonate (EMS) is the most extensively used chemical mutagen in different crop plants nowadays.

5. DNA replication is

(a) conservative and discontinuous

(b) semiconservative and semidiscontinuous

(d) conservative.

Answer and Explanation:

5. (b): DNA replication is semi-conservative and semi-discontinuous. Meselson and Stahl, 1958 by using ¹⁴N and ¹⁵N confirmed that the replication of DNA in E- coli is semi conservative. Carin, 1963 by using ³H in thymidine also confirmed semi-conservative mode of DNA replication in this bacterium.

6. Diploid chromosome number in human is

(a) 46

(b) 44

(d) 42.

Answer and Explanation:

6. (a): Diploid chromosome number in human is 46. All animals including man have a characteristics number of chromosomes in their body cells called diploid (or 2n) number. These occur as homologous pairs, one member of each pair have been acquired from the gamete of one of the two parents of the individual whose cells are being examined. It is actually 44 + XX or 44 + XY.

7. In the genetic dictionary, there are 64 codons as

(a) 64 amino acids are to be coded

(b) 64 types of tRNAs are present

(d) genetic code is triplet.

Answer and Explanation:

7. (d): Codon is a sequence of three nucleotides coding for one aminoacid in a polypeptide chain. There are four different types of nucleotides – A, T, G and C and since a codon is a triplet thus 64 (4 x 4 x 4) distinct triplets of purines or pyrmidine bases determine the 20 amino acids. But out of these 64 codons, only 61 codons code for individual amino acids during protein synthesis. The three codons that do not code for amino acids are the termination or nonsense codons. These are UAA, UAG and UGA. So the remaining 61 codons code for the 20 amino acids.

8. An octamer of 4 histones complexed with DNA forms

(a) endosome

(b) nucleosome

(d) centromere.

Answer and Explanation:

8. (b): An octamer of 4 histones complexed with DNA forms nucleosome. The association of histones with DNA is very characteristic. It involves the formation of linear array of spherial structures called nucleosomes. This structure contains four pairs of histones (H₂A, H₂B, H₃and H₄) in a ball; around which is wrapped a stretch of about 150 base pairs of DNA.

9. Bateson used the terms coupling and repulsion for linkage and crossing over. Name the correct parental of coupling type along with its cross over or repulsion

(a) coupling AABB, aabb; Repulsion AABB, aabb

(b) coupling AAbb, aaBB; Repulsion AaBb, aabb

(d) coupling AABB, aabb : Repulsion AAbb, aaBB.

Answer and Explanation:

9. (d): Bateson and Punnet explained that when two dominants enter from the same parent-they try to remain together, called coupling. When two dominants enter from different parents they try to remain seperate called repulsion. Bateson and Punnett (1906) used the term coupling and repulsion in sweet pea (Lathyrus odoratus) for linkage and crossing over. The correct parental of coupling type along with its cross over or repulsion is coupling AABB, aabb : Repulsion AAbb; aaBB.

10. The process of transfer of genetic information from DNA to RNA/formation of RNA from DNA

(a) transversion

(b) transcription

(d) translocation.

Answer and Explanation:

10. (b): The process in living cells in which the genetic information of DNA is transferred to a molecule of messenger RNA (mRNA) is the first step in protein synthesis (see also genetic code). Transcription takes place in the cell nucleus or nuclear region and is regulated by transcription factors.

11. Escherichia coli fully labelled with ¹⁵N is allowed to grow in ¹⁴N medium. The two strands of DNA molecule of the first generation bacteria have

(a) different density and do not resemble parent DNA

(b) different density but resemble parent DNA

(d) same density but do not resemble parent DNA.

Answer and Explanation:

11. (a): E-coli fully labelled with ¹⁵N is allowed to grow in ¹⁴N medium. The two strands of DNA molecule of the first generation bacteria have different density and do not resemble parent DNA. Meselson and Stahl, 1958 by using ¹⁴N and ¹⁵N confirmed that the replication of DNA in E-coli is semi-conservative in nature.

12. Khorana first deciphered the triplet codons of

(a) serine and isoleucine

(b) cysteine and valine

(d) phenylalanine and methinonine.

Answer and Explanation:

12. (b): Khorana synthesised a chain of alternate nucleotide GUGUGUGUGU. He found that it stimulated synthesis of a peptide having alternate valine-cysteine- valine-cysteine.

13. Experimental material in the study of DNA replication has been

(a) Escherichia coli

(b) Neurospora crassa

(d) Drosophila melanogaster.

Answer and Explanation:

13. (a): Experimental material in the study of DNA replication has been Escherichia coli. E-coli fully labelled with ¹⁵N is allowed to grow in ¹⁴N medium. The two strands of DNA molecule of the first generation bacteria have different density and do not resemble parent DNA. Meselson and Stahl, 1958 by using ¹⁴N and ¹⁵N confirmed that the replication of DNA in E-coli is semi-conservative in nature.

14. Out of 8 ascospore formed in Neurospora the arrangment is 2a : 4a : 2a showing

(a) no crossing over

(b) some meiosis

(d) first generation division.

Answer and Explanation:

14. (c): Out of 8 ascospore formed in Neurospora the arrangement is 2a : 4a : 2a showing second generation division. In Neurospora, products of meiosis remain linearly arranged and undergo one mitosis. Crossing over occurs in four strand stage.

15. When a certain character is inherited only through female parent, it probably represents

(a) multiple plastid inheritance

(b) cytoplasmic inheritance

(d) mendelian nuclear inheritance.

Answer and Explanation:

15. (b): When a certain character is inherited only through female parent, it probably represents cytoplasmic inheritance. Plasma genes present in the maternal cytoplasm are transmitted to offspring hence this type of inheritance is termed as cytoplasmic or maternal inheritance.

16. Nucleotide arrangement in DNA can be seen by

(a) X-ray crystallography

(b) electron microscope

(d) light microscope.

Answer and Explanation:

16. (a): Nucleotide arrangement in DNA can be seen by X-ray crystallography. Watson and Crick, 1953 proposed the double helical model for DNA. They were awarded Nobel prize in 1962. This model was developed by them on the basis of several previous observations including the of-helix of Pauling, 1951 and X-ray reflection studies of Franklin and Gosling, 1953.

17. A DNA with unequal nitrogen bases would most probably be

(a) single stranded

(b) double stranded

(d) four stranded.

Answer and Explanation:

17. (a): A DNA with unequal nitrogen bases would most probably be single stranded. Nitrogenous bases are unequal in number in single stranded DNA, because they do not posses complementary base pairs.

18. The process of translation is

(a) ribosome synthesis

(b) protein synthesis

(d) RNA synthesis.

Answer and Explanation:

18. (b): The process of translation is protein synthesis. Emil Fischer, a German chemist established that the proteins are polymers of amino acids. There are some twenty amino acids involved in protein synthesis. In translation, the message coded by DNA on m-RNA is translated into a specific protein.

19. During DNA replication, the strands separate by

(a) DNA polymerase

(b) topoisomerase

(d) gyrase.

Answer and Explanation:

19. (c): During DNA replication, the strands separate by unwindase/helicase. The molecule is unwound by DNA unwinding proteins called helicases. The helicases II and III get attached to logging strand and protein to the leading strand. The formation of bands is avoided by single stranded DNA binding proteins (SSB).

20. Because most of the aminncids are represented by more than one codon, the genetic code is

(a) overlapping

(b) wobbling

(d) generate.

Answer and Explanation:

20. (c): Because most of the amino acids are represented by more than one codon, the genetic code is degenerate. Certain amino acids are identified by more than one codons. This phenomenon is called as degeneracy e.g. only AUG codes for methionine and UGG tryptophan.

21. Who proved that DNA is basic genetic material?

(a) Griffith

(b) Watson

(d) Hershey and Chase.

Answer and Explanation:

21. (d): Hershey and Chase proved that DNA is a basic genetic material. Hershey and Chase, 1952, by using s³²and s³⁵ with a T-2 type phage concluded that DNA is the genetic material.

22. The transforming principle of Pneumococcus as found out by Avery, MacLeod and McCarty was

(a) mRNA

(b) DNA

(d) polysaccharide.

Answer and Explanation:

22. (b): The transforming principle of Pneumococcus as found out by Avery, MacLeod and McCarty was DNA. In 1944, Avery, MacLeod and McCarty repeated Griffith’s experiment successfully. They separated the proteins, carbohydrates and DNA of S III strains and separately mixed them in the pure cultures of R II. Only DNA could bring about transformation of R II type into S III and not the proteins or the carbohydrates.

23. Initiation codon of protein synthesis (in eukaryotes) is

(a) GUA

(b) GCA

(d) AUG.

Answer and Explanation:

23. (d): A gene or a cistron transcribes to form mRNA which consists of several coding and non coding regions. The coding region always starts with the codon AUG. It is called initiation codon. It is found in both eukaryotes and prokaryotes. It codes for amino acid methionine.

24. Nucleosome core is made of

(a) HI, H2A, H2B and H3

(b) HI, H2A, H2B, H4

(d) H2A, H2B, H3 and H4.

Answer and Explanation:

24. (d): Nucleosome core is made up of H₂A, H-,B, H, and H₄. It is about 7-10 nm in diameter, consisting of histones around which a DNA strand, about 120 base pair long is wrapped in chromosomes.

25. Sex is determined in human beings

(a) by ovum

(b) at time of fertilization

(d) seventh to eight week when genitals differentiate in foetus

Answer and Explanation:

25. (b): Sex is determined in human beings at the time of fertilization. Sex of human body is determined by the karyotype of the zygote or fertilized egg. Sex of the baby depends upon the sperm which fertilizes the ovum.

26. In protein synthesis, the polymerization of amino acids involves three steps. Which one of the following is not involved in the polymerisation of protein?

(a) termination

(b) initiation

(d) transcription.

Answer and Explanation:

26. (d): Transcription is the mechanism of copying the message of DNA on RNA with the help of enzyme RNA polymerase. It is meant for taking the coded information from DNA to the site where it is required for protein synthesis.

Translation or protein synthesis is a complicated process involving several steps such as – activation of amino acid, transfer of amino acid to /-RNA, initiation of polypeptide synthesis, elongation of polypeptide chain and, termination of polypeptide chain. This entire process occurs in the cytoplasm over the ribosomes. Transcription and translation are two entirely different processes that are separated by time and space.

27. There are special proteins that help to open up DNA double helix in front of the replication fork. These proteins are

(a) DNA ligase

(b) DNA topoisomerase I

(d) DNA polymerase I.

Answer and Explanation:

27. (b): DNA is a double helical molecule and it opens to form a replication fork for its replication. The two strands of DNA are joined with the help of H-bonds between the strands. Topoisomerases are specialized to cause nicks or breaks in the double helix and helps separate the DNA stands. Helicase unwinds the DNA helix from that nick caused by the topoisomerase and this seperates the two strands.

DNA gyrase introduces negative supercoils in DNA strands of prokayotes.

DNA polymerase adds nucleotides units to the 3′ end of a DNA chain. DNA ligase joins the ends of DNA.

28. Initiation codon in eukaryotes is

(a) GAU

(b) AGU

(d) UAG.

Answer and Explanation:

28. (c): The base triplets found in mRNA that helps in protein synthesis is called as codon. Initiation codon is triplet of ribonucleotides in messenger RNA (mRNA) that is present at the start of the mRNA coding sequence and initiates polypeptede formation or translation. The triplet is AUG and codes for the amino acid methionine: In bacteria the start codon is either AUG coding forN-formyl methionine or GUG coding for valine.

29. ‘Lac operon’ in E. coli, is induced by

(a) ‘I’ gene

(b) promoter gene

(d) lactose.

Answer:

30. Anticodon is an unpaired triplet of bases in an exposed position of

(a) <-RNA

(b) w-RNA

(d) both ‘b’ and ‘c’.

Answer and Explanation:

30.

(a): Anticodon is the sequence of three nucleotides in a transfer RNA molecule that pairs with a complementary sequence of three nucleotides (codon) on a molecule of messenger RNA. /-RNA has clove like shape or L shape (three dimensional). It has G at 5′ end CCA at 3′ end. CCA at 3′ end is meant for attaching to a specific amino acid (AA-binding site). On the opposite side lies an anticodon that is complementary to a specific codon of mRNA. The two are called recognition sites.

31. The point, at which polytene chromosome appear to be attached together, is called

(a) centromere

(b) chromomere

(d) centriole.

Answer and Explanation:

31. (c): Chromocentre is junction point of the chromosomes in the polytene salivary glands of Drosophila larvae. Unlike the situation in other cells, the giant chromosomes of these cells persist through interphase, the two homologous copies of each chromosome are attached together throughout their lengths and all the chromosomes are joined together by a chromocentre.

32. In split genes, the coding sequences are called

(a) exons

(b) cistrons

(d) operons.

Answer and Explanation:

32. (a): Split gene are those genes that consist of continuous sequence of nucleotide (coding sequence) interrupted by intervening sequences. Most eukaryotic genes are split as are genes of some animal viruses. The continous coding sequences are called exons and the intervening non-coding sequences are called introns. These introns are not represented in mRNA transcribed from the gene and are not utilized for the synthesis of proteins.

33. The polytene chromosomes were discovered for the first time in

(a) Drosophila

(b) Musca domestica

(d) Musca nebula.

Answer and Explanation:

33. (c): In salivary glands cells of Chironomus tantans giant chromosomes were observed by E.GBalbini for the first time in 1881. The availability of these chromosomes greatly helped in the study of Drosophila cytogenetics in which they were discovered later.

34. Each chromosome at the anaphase stage of a bone marrow cell in our body has

(a) two chromatids

(b) several chromatids

(d) only one chromatid.

Answer and Explanation:

34. (d): A bone marrow cell is a somatic cell and therefore mitosis takes place in it. During mitosis anaphase chromosomes split at centromere. Sister chromatids seperate from each other, so that the two sister chromatids are separate structures and can now be called as chromosomes.

35. If the DNA codons are ATG ATG ATG and a cytosine base is inserted at the beginning, then which of the following will result?

(a) CAT GAT GATG

(b) a non-sense mutation

(d) CATGATGATG

Answer and Explanation:

35. (a): Nonsense mutation is a mutation which interconverts a nonsense to or from a sense-coding triplet, resulting in an abnormally foreshortened or elongated polypeptide chain. But in this example cytosine is added at the beginning so CAT GAT GATG. will result.

36. Barr body in mammals represents

(a) all the heterochromatin in male and female ceils

(b) the Y-chromosome in somatic cells of male

(d) one of the two X-chromosomes in somatic cells of females.

Answer and Explanation:

36. (d): Barr body is a condensed heterochromatic copy of the X chromosome, visible by staining the interphase nucleus of somatic cells of the homogametic sex, for example the human female. Named after M. Barr, these condensed chromosomes offer an easy way of determining the true genetic sex of individuals with intermediate secondary sexual characteristics. The total number of Barr bodies is always one less than the total number of X chromosomes present in the cell or the organism.

37. If the sequence of bases in DNA is ATTCGATG, then the sequence of bases in its transcript will be

(a) GUAGCUUA

(b) AUUCGAUG

(d) UAAGCUAC.

Answer:

(d) UAAGCUAC.

38. An environmental agent, which triggers transcription from an operon, is a

(a) depressor

(b) controlling element

(d) inducer.

Answer:

(d) inducer

39. The lac operon is an example of

(a) repressible operon

(b) overlapping genes

(d) inducible operon.

Answer and Explanation:

39. (d): Lac operon (lactose operon). Genetic system of E.coli is responsible for the uptake and initial catabolism of lactose. The lac operon consists of three structural genes (lac Z, lac Y, lac A), lac Z codes for p- galactosidase which hydrolyses lactose to glucose and galactose. (lacY codes for lac permease, a membrane- bound protein constituent of the lactose transport system, lac A codes of thioglactoside transacetylase, an enzyme of uncertain metabolic function.

Bacteria growing in a lactose-free medium contain about five molecules of P-glactosidase per cell. The appearance of P-galactosidase is coordinated with the production of permease and transacetylase. On removing lactose from the medium enzyme synthesis stops.

Such enzymes whose synthesis can be induced by adding the substrate are known as inducible enzymes and the genetic systems responsible for the synthesis of such an enzyme are known as inducible operons.

40. The wild type E.coli cells are growing in normal medium with glucose. They are transferred to a medium containing only lactose as sugar. Which of the following changes take place?

(a) the lac operon is induced

(b) E.coli cells stop dividing

(d) all operons are induced.

Answer and Explanation:

40. (a): When E.coli bacteria are transfered to medium containing lactose, then the lac opeon is indueed. The lac opeaon consists of 3 structural gene (lac Z, lac Y and lac A). It involves the synthesis of P-galactosidase enzyme in E.coli, which hydrolyses lactose into glucose and galactose.

41. Radio-tracer technique shows that DNA is in

(a) multi-helix stage

(b) single-helix stage

(d) none of these.

Answer and Explanation:

41. (c): Autoradiography is the study of labelled precursors like ³H by knowing the movement of radioactivity with the help of photographic films and emulsions at short intervals. ¹⁴C and ³H are incorporated in bases like thymidine, uridine and amino acids to study the structure of DNA and proteins. Radio tracer technique shows that DNA is in double helical form.

42. The maximum formation of /n-RNA occurs in

(a) ribosome

(b) nucleoplasm

(d) nucleolus.

Answer and Explanation:

42. (d): Nucleolus is a plasmosome body that is formed around the nucleolus organizer and is located in the secondary constriction on that chromosome. It is made up of RNA and proteins. The associated nucleolar chromatin contains DNA. It forms mRNA that has low molecular weight. Ribosomes are mainly concerned with proteins synthesis. They are sites for synthesis of rRNA and tRNA is synthesized in the cytoplasm.

43. Which of the following serves as a terminal codon?

(a) UAG

(b) AGA

(d) GCG.

Answer and Explanation:

43. (a): In eukaryotes, the termination of polypeptide chain during translation is brought about by the terminating codons. These are UAA, UAG and UGA and these are called as amber, ochre and opal respectively. These codons are also called as nonsense codons as they donot code for any amino acid. In phages and bacteria there may be many initiating and terminating codons and thus as many polypeptides are synthesized. AUG is the initiation codon. It codes for methionine amino acid.

44. Identify the one, which causes mutation

(a) cosmic rays

(b) hromosomes

(d) X-rays.

Answer and Explanation:

44. (d): Mutations can be artificially induced with the help of mutagenic agents which can be broadly classified into two groups, physical mutagens and chemical mutagens. Physical mutagens are mainly radiations, although change in pH value (acidity) or temperature shocks may also induce mutations.

Among ionizing radiations, more commonly X-rays, gamma rays, beta rays and neutrons are used for inducing mutations. X-rays are produced in a X-ray machine when energy charged particles like cathode rays (electrons) impinge on a suitable target like tungston.

45. Lampbrush Chromosomes are seen in which typical stage?

(a) mitotic metaphase

(b) meiotic prophase

(d) mitotic prophase.

Answer and Explanation:

45. (b): Lampbrush chromosomes (Ruckert, 1892) are large sized diplotene chromosome bivalents with a length of 400 – 1000 nm each and a total length of 5900 in Triturus (Salamander = Newt). A lampbrush chromosome is made of two homologous chromosomes held at several places by chiasmata. Each chromosome has an axis with alternate chromosomes held at several places by chiasmata.

Each chromosome has an axis with alternate chromomeres and interchromomeric areas. Many of the chromomeres give out lateral loops of various sizes which are thin in the region of origin and thick in the area where they are wound back into chromomeres. Loops possess a number of copies of the same gene and are meant for rapid transcription and production of materials like yolk. Hence, lampbrush chromosomes occur in oocytes.

46. Which of the following step of translation does not consume a high energy phosphate bond?

(a) peptidyl transferase reaction

(b) aminoacyl /-RNA binding to ,4-site

(d) amino acid activation.

Answer and Explanation:

46. (a): Protein synthesis or translation consists of ribosomes, amino acids, mRNA, tRNAs and aminoacyl tRNA synthetases. The ribosomes have two binding sites namely aminoacyl site or A-site and peptide site or P- site. The starting amino acid methionine lies at the P- site of the ribosome.

The next incoming tRNA is called amino acyl tRNA, it is bound to A-site. A peptide bond is formed between COOH group of the t-RNA at P-site and NH₂ group of aminoacyl t-RNA. This is facilitated by the enzyme peptidyl transferase and does not require high energy phosphate bonds.

47. The RNA that picks up specific amino acid from amino acid pool in the cytoplasm to ribosome during protein synthesis is called

(a) r-RNA

(b) RNA

(d) t-RNA.

Answer and Explanation:

47. (d): Transfer RNA or tRNA help in transfer of amino acids to ribosomes mRNA complex to form the polypeptide chain. It has four key regions a carrier and recognition end, enzyme site and ribosome site. This recognition end has three anticodons with the help of which aminoacids are identified. r-RNA forms 67% of 70s ribosomes and 50% of 80s ribosomes.

mRNA carries the coded information from DNA for the synthesis of proteins.

48. DNA synthesis can be specifically measured by estimating the incorporation of radio-labelled

(a) thymidine

(b) deoxyribose sugar

(d) adenine.

Answer and Explanation:

48. (a): Autoradiography is the study of labelled precursors like ³H by knowing the movement of radioactivity with the help of photographic films and emulsions at short intervals.

Radioactive material like tritiated thymidine which is formed by replacing normal hydrogen of thymidine with H³ (heavy isotope of hydrogen). Thymidine only is used for this purpose because RNA will not be labelled by this.

49. Genetic identity of a human male is determined by

(a) sex-chromosome

(b) cell organelles

(d) nucleolus.

Answer and Explanation:

49. (a): Sex chromosomes are those chromosomes whose presence, absence or particular form determines the sex of the individual in unisexual or dioecious organisms, e.g., XX – XY. XY method (XX – XY). It occurs in mammals and many insects with females having homomorphic XX sex chromosomes and males possessing heteromorphic XY – chromosomes. In human being the Y-chromosome is straight and acrocentric (centromere subterminal near one end).

It is about 2 µm in length and is thus the shortest of all chromosomes. X- chromosome is 5.0 – 5.5 µm in length. It is metacentric (centromere in the middle). Despite differences in morphology, XY chromosomes synapse during zygotene. They have two parts, homologous and differential. Homologous regions of the two take part in synapsis.

50. Genes located at the same locus of chromosomes are called

(a) multiple alleles

(b) polygenes

(d) none of these.

Answer and Explanation:

50. (a): Genes located on the same locus of chromosomes are called multiple alleles. They are produced due to repeated mutations of the same gene but in different directions. Multiple alleles being located on the same locus do not show crossing over.

51.

After crossing two plants, the progenies are found to be male sterile. This phenomenon is found to be maternally inherited and is due to some genes which reside in

(a) mitochondria

(b) cytoplasm

(d) chloroplast.

Answer and Explanation:

51. (b): In several crops like maize cytoplasmic control of male sterility is known. In such cases if female parent is male sterile F, progeny would also be male sterile because cytoplasm is mainly derived from egg obtained from male sterile female parent. Many experiments have proved that factors responsible for cytoplasmic male sterility are located in mitochondrial DNA. A particular phenotype arises because of dominant mutations in the mitochondrial genome.

52. The codons causing chain termination are

(a) AGT, TAG, UGA

(b) UAG, UGA, UAA

(d) GAT, AAT, AGT.

Answer and Explanation:

52. (b): Nonsense codon also called stop codon, or termination codon are any of the three codons that do not code for an amino acid, but instead cause termination of protein synthesis. They are recognised by proteins called release factors which bind to the A site of the iebosome They are UGA, UAG, UAA, of which UAG is known as amber, UAA as ochre, and UGA as opal.

53. Centromere is a part of

(a) chromosome

(b) endoplasmic reticulum

(d) mitochondria.

Answer and Explanation:

53. (a): Centromere is the specialized area of a chromosome involved in its attachment to the fibres of the spindle during cell division. Depending on the position of the centromere, chromosomes are designated as metacentric, telocentric or acrocentric. This variation in centromere position also leads to the variation in chromosome shape at anaphase, appearing as V shape, J shape, i-shaped or L-shaped.

54. The mutaiions are mainly responsible for

(a) increasing the population rate

(b) maintaining genetic continuity

(d) variation in organisms.

Answer and Explanation:

54. (d): Mutation are the sudden alteration of the chemical structure of a gene, or the alteration of its position on the chromosome by breaking and rejoining of the chromosome. They rarely occur naturally but may be caused artificially by irradiation or by chemicals eg mustard gas. They may occur in somatic cells or gametes in which case they can be inherited. They lead to variations in organioms.

55. The chemical knives of DNA are

(a) endonucleases

(b) transcriptases

(d) polymerases.

Answer and Explanation:

55. (a): Endonucleases are enzymes that break the DNA molecule internally. The enzyme breaks an internal phosphodiester bond, either in one strand only to produce a nick or in both strands to break the DNA molecule into pieces.

Restriction endonucleases from bacterial cells are used in gene manipulation technology. They make cuts only at specific short sequences of DNA bases.

56. DNA elements, which can switch their position, are called

(a) cistrons

(b) transposons

(d) introns.

Answer and Explanation:

56. (b): Transposons are portable genetic elements which can insert themselves at random into a plasmid or any chromosome independently of the host cell recombination system.

It was discovered by Barbara Mc Clintock (1940) in maize and termed as jumping genes. Later Headges and Jacob termed them as transposons. Introns are nontranslated sequences within the coding sequence of a gene. Such sequences are transcribed into hnRNA but are then spliced out and are not represented in the message. The non-intron sequences of the gene are referred to as exons.

Cistron sequence of nucleotides in a DNA molecule code for one particular polypeptide chain.

57. Loss of a X-chromosome in a particular cell, during its development, results into

(a) gynandromorphs

(b) diploid individual

(d) both ‘b’ and ‘c’.

Answer and Explanation:

57. (a): In Drosophila occasionally flies are obtained which have female characters in one part of body and male characters in the remaining parts. Such individuals are known as gynandromorphs and are believed to result due to loss of an X-chromosome in a particular cell during development.

Availability of gynandromorphs and their cytological examination suggests that Y-chromosome does not play any role in determinantion of sex in Drosophila.

58. How many genome types are present in a typical green plants cell?

(a) more than five

(b) more than ten

(d) three.

Answer and Explanation:

58. (c): Since a typical green plant is diploid, therefore it has two sets of chromosomes. So the number of genome will be two, because genome is the entire set of gene carried by a gamete or present in the haploid cell of a particular organism.

59. Genes that are involved in turning on or off the transcription of a set of structural genes are called

(a) redundant genes

(b) regulatory genes

(d) operator genes.

Answer and Explanation:

59. (d): Operator genes are a region of DNA sequence capable of interacting with a specific repressor molecule and in doing so it affects the activity of other genes downstream from it.

60. What base is responsible for hot spots for spontaneous point mutations?

(a) 5-bromouracil

(b) 5-methylcytosine

(d) adenine.

Answer and Explanation:

60. (c): Mutations are rare events in nature and are then described as spontaneous mutations. Some of these mutations originate from mistakes in normal duplication of DNA. Transitions may be produced by tautomeric shift or ionization of bases which leads to mistaken, A – C base pairing and more frequently mistaken G – T base pairing. Guanine pairs with the rare enol form of thymine and is thus considered as hot spot for spontaneous point mutations.

61. The formation of multivalent at meiosis in diploid organism is due to

(a) deletion

(b) reciprocal translocation

(d) inversion.

Answer and Explanation:

61. (b): Multivalents is an association between three or more chromosomes at meiosis in polyploid individuals. The number of multivalents depends upon the degree of synapsis and chiasmata formation among similar chromosomes. It also results from reciprocal translocations.

It is also known as segmental interchanges which involves mutual exchange of chromosome segments between two pairs of non-homologous chromosomes.

62. Initiation codon in eukaryotes is

(a) GAU

(b) AGU

(d) UAG.

Answer and Explanation:

62. (c): A gene or a cistron transcribes to form mRNA which consists of several coding and non coding regions. The coding region always starts with the codon AUG. It is called initiation codon. It is found in both eukaryotes and prokaryotes. It codes for amino acid methionine.

63. The eukaryotic genome differs from the prokaryotic genome because

(a) the DNA is complexed with histone in prokaryotes

(b) the DNA is circular and single stranded in prokaryotes

(d) genes in the former case are organized into operons.

Answer and Explanation:

63. (b): Genome refers to the total sets of chromosomes carried by each cell of the organism. In prokaryotes the genetic material is circular and single stranded DNA. It has no association of histones. The eukaryotic genetic material is linear and double stranded DNA. It is associated with histone proteins to form nucleosome unit.

64. In DNA, when AGCT occurs, their association is as per which of the following pair?

(a) AT-GC

(b) AG-CT

(d) AC-GT

(d) all of these.

Answer and Explanation:

64. (a) : DNA molecule has four bases – adenine, guanine, cytosine and thymine. Adenine always pairs with thymine and guanine pairs with cytosine. Their association is A-T and G-C.

65. In prokaryotes, the genetic material is

(a) linear DNA without histones

(b) circular DNA without histones

(d) circular DNA with histones.

Answer and Explanation:

65. (b): The genetic material of prokaryotes circular and single stranded DNA. It has no association of hisones. The eukaryotic genetic material is linear and double stranded DNA. It is associated with histone proteins to form nucleosome unit.

66. DNA is mainly found in

(a) nucleolus

(b) nucleus only

(d) none of these.

Answer and Explanation:

66. (b): DNA is mainly found in nucleus. It is associated with RNA and proteins to form compact chromosomes. But some amount of DNA is also found in chloroplasts and mitochondria. This DNA is called extrachromosomal DNA.

67. Initiation codon in eukaryotes is

(a) GAU

(b) AGU

(d) UAG.

Answer and Explanation:

67. (c): Refer answer 28.

68. Which of the following is the main category of mutation?

(a) somatic mutation

(b) genetic mutation

(d) none of these.

Answer and Explanation:

68. (b): Mutation are the sudden alteration of the chemical structure of a gene, or the alteration of its position on the chromosome by breaking and rejoining of the chromosome. They rarely occur naturally but may be caused artificially by irradiation or by chemicals eg, mustard gas. They may occur in somatic cells or gametes in which case they can be inherited. So they can be accordingly categorized into somatic mutations and gametic mutations.

69. In operon concept, regulator gene functions as

(a) inhibitor

(b) repressor

(d) all of these.

Answer and Explanation:

69. (b): Regulator gene is a gene whose function is to control the transcriptional activity of other genes, either adjacent or distant in the genome. In the case of the lac operon of E.coli the regulator gene lac produces a protein product that represses the operator g5ne of the operon. In bacteria the same regulator gene may affect a series of non-adjacent operons.

70. The Pneumococcus experiment proves that

(a) bacteria do not reproduce sexually

(b) RNA sometime controls the production of DNA and proteins

(d) bacteria undergo binary fission.

Answer and Explanation:

70. (c): Transformation was first discovered by Griffith (1928), in Pneumococcus (Streptococcus pneumoniae), that causes pneumonia.

Griffith injected a group of mice with nonencapsulated, rough (R), pneumococci; a second group with heat-killed encapsulated pneumococci cells, and a third group a mixture consisting of a few living nonencapsulated, rough pneumococci derived from a type S culture, and heat-killed encapsulated cells (S type). Griffith observed that the mice in the first two groups were not infected, and the mice in the third group died within a few days.

The mice of the third group should have survived as the organisms which could kill them had been killed, and the cells of R type were incapable of causing disease. However, the mice died, and living virulent encapsulated cells of the type S were recovered from their dead bodies.

It was observed by Griffith, that killed encapsulated pneumococci had liberated some substance which favoured noncapsulated cells (R type) to produce a capsular substance.

This substance in later experiments was proved to be DNA. These experiments showed that DNA is the genetic material.

71. Similarity in DNA and RNA is that

(a) both are polymer of nucleotides

(b) both have similar pyrimidine

(d) both are genetic material.

Answer and Explanation:

71. (a): Deoxyribonucleic acid and ribonucleic acid as the name suggests are made up of several nucleotide monomers. Each nucleotide consists of pentose sugar, phosphate group and nitrogenous bases. DNA has deoxyribose sugar whereas RNA has ribose sugar. The bases in DNA molecule are A, T, G and C whereas in RNA, thymine is absent and instead uracil is found.

72. In three dimensional view the molecule of t-RNA

(a) L-shaped

(b) S-shaped

(d) E-shaped.

Answer and Explanation:

72. (a): 3-D model of tRNA looks like flattened L- shaped molecule.

t-RNA acts as adoptor molecule which carries amino acids to the site of protein synthesis (i.e., ribosomes). Most accepted model for t-RNA structure is ‘clover leaf model.

73. Length of one loop of B-DNA

(a) 3.4 nm

(b) 0.34 nm

(d) 10 rim.

Answer and Explanation:

73. (a): B-DNA is an antiparallel double helix. The double strand or duplex is coiled plectonemically in right handed fashion around a common axis like a rope stair case twisted in a spiral. The coiling produces alternate major and minor grooves. One turn of spiral has a distance between two adjacent nucleotides is 3.4 A.

74. Anticodon occurs in

(a) t-RNA

(b) m-RNA

(d) DNA.

Answer and Explanation:

74. (a): Refer answer 30.

75. Which of the following is initiation codon

(a) UAG

(b) AUC

(d) CCU.

Answer and Explanation:

75. (c): Refer answer 28.

76. Method of DNA replication in which two strands of DNA separate and synthesize new strands

(a) dispersive

(b) conservative

(d) non conservative.

Answer and Explanation:

76. (c): The method of DNA replication is semiconservative. According to the semiconservative model proposed by Watson and Crick, each strand of the two double helices formed would have one old and one new strand. So the parental identity is conserved upto half extent and hence DNA replication is semiconservative.

77. In Drosophila the XXY condition leads to femaleness whereas in human beings the same condition leads to Klienfelter’s syndrome in male. It proves

(a) in human beings Y chromosme is active in sex determination

(b) Y chromosome is active in sex determination in both human beings and Drosophila

(d) Y chromosome of man have genes for syndrome.

Answer and Explanation:

77. (a): In Drosophila, XXY condition leads to femaleness. Cytological examination have suggested that Y-chromosome does not play any role in determination of sex in Drosophila. In human being, XXY is phenotypically male with underdeveloped testes gynecomastia and often mental retardation. It is caused by the union of a non-disjunct XX egg and sperm and a normal X egg and abnormal XY sperm. This indicates that in human being Y chromosome is active in sex determination.

78. Extranuclear DNA is found in

(a) lysosome and chloroplast

(b) chloroplast and mitochondria

(d) golgi and E.R.

Answer and Explanation:

78. (b): Chloroplast and mitochondria contain genes that are transmitted from parents to the offspring along with the cytoplasm of the contributing gametes. Those genes are called plasma genes or extra nuclear genes. They are collectively called plasmon. They result in cytoplasmic or extranuclear inheritance.

79. Most of the mutations are

(a) harmful

(b) harmful and recessive

(d) dominant.

Answer and Explanation:

79. (b): Mutations are normally deleterious and recessive and therefore majority of them are of no practical value. A. Gustafson estimated that less than one in 1000 mutants produced may be useful in plant breeding.

80. Irregularity is found in Drosophila during the organ differentiation, for example – in place of wing, long legs are formed. Which gene is responsible?

(a) double dominant gene

(b) homeotic gene

(d) none of the above.

Answer and Explanation:

80. (b): During 1984 and 1985, a set of genes called homeotic genes whi~h determine the body plan, received major attention. These homeotic genes are such that their mutations cause clean transformation of one body part into another.

Some remarkable homeotic mutations are known in Drosophila, such as aristapedia and bithorax. Many of these genes possess a common short sequence of DNA, known as the homeo box, and are found to be expressed for only short periods during development. They may well function as regulator genes, controlling the activities of other sets of genes during development.

81. Types of RNA polymerase required in nucleus of eukaryotes for RNA synthesis

(a) 1

(b) 2

(d) 4.

Answer and Explanation:

81. (c): Enzyme involved in transcription is RNA polymerase. It is single in prokaryotes. There are three types of RNA polymerases in eukaryotes – polymerase I for 28S, 18S and 5.8S RNA, polymerase II for mRNA and snRNA and polymerase III for tRNA, 5SRNA and scRNA. RNA polymerase has five polypeptides – a, a, 0, (3′ and to. a or sigma factor recognises the promoter region while the remaining or core enzyme takes part in transcription.

82. Extranuclear inheritance occurs in

(a) killer Paramaecium

(b) killer Amoeba

(d) Hydra.

Answer and Explanation:

82. (a): Cytoplasmic inheritance is the passage of traits from parents to offspring through structures present in the cytoplasm of the contributing gametes. These structures may be kappa in Paramecium.

These particles are not only inherited but are also infective, since they can be introduced into new hosts without the need of actual process of reproduction. Further the presence and reproduction of these infective particles may also depend on nuclear genes.

83. Extranuclear chromosomes occur in

(a) peroxisome, ribosome

(b) chloroplast and mitochondria

(d) chloroplast and lysosome.

Answer and Explanation:

83. (b): Refer answer 78.

84. Gene and cistron words are sometimes used synonymously because

(a) one cistron contains many genes

(b) one gene contains many cistrons

(d) one gene contains no cistron.

Answer and Explanation:

84. (c): A gene is a hereditary unit consisting of a sequence of DNA and occupying a specific position or locus within the genome. Gene activity ultimately affects the phenotype of the organism possessing the gene.

Thus gene is a physical and functional unit of genetic information. A cistron is a unit of genetic function. In prokaryotes there is one gene one enzyme correspondence. It means that in these organisms genes and cistrons are equivalent.

85. In negative operon

(a) co-repressor binds with repressor

(b) co-repressor does not bind with repressor

(d) CAMP have negative effect on lac operon.

Answer and Explanation:

85. (a): The tryptophan operon (trp operon) in bacteria is a repressible operon. Here repressor is inactive and it becomes active as DNA binding protein only when complexed with a co-repressor (tryptophan). In absence of tryptophan, the operator site is open to binding by RNA polymerase, which transcribes the structural genes of tryptophan operon, leading to production of enzymes that synthesize tryptophan.

When tryptophan becomes available, the enzymes of tryptophan synthetic pathway are no longer needed and tryptophan (co-repressor)- repressor complex blocks transcription. The regulation of this operon is also a negative control.

86. mRNA is synthesised on DNA template in which direction

(a) 5′ —» 3′

(b) 3′->5′

(d) any.

Answer and Explanation:

86. (a): mRNA is synthesized on DNA template in 5′ – 3′ direction. Synthesis of mRNA exhibits several features that are synonymous with DNA replication. RNA synthesis requires accurate and efficient initiation, elongation proceeds in the 5′ – 3′ direction (i.e. the polymerase oves along the template strand of DNA in the 5′ – 3′ direction), and RNA synthesis requires distinct and accurate termination. Transcription exhibits several features that are distinct from replication.

87. Change in sequence of nucleotide in DNA is called as

(a) mutagen

(b) mutation

(d) translation.

Answer and Explanation:

87. (b): Mutation is a process by which a gene or some other DNA sequence undergoes a change in structure at molecular level.

If a mutation occurs in the gametes it is inherited by subsequent generations of offspring. Somatic mutations occur in non-gametic cells and are not inherited. Mutagen is a chemical or physical agent that induces mutations above the incidence of spontaneous mutations.

Important mutagens include ionising radiations, ultraviolet light and chemicals (e.g., nitrogen mustard and nitrous acid). Translation is the synthesis of a polypeptide chain from an mRNA template formed after transcription from DNA. .Recombination is the exchange of genetic material.

88. Exon part of m-RNAs have code for

(a) protein

(b) lipid

(d) phospholipid.

Answer and Explanation:

88. (a): DNA transcribes to form mRNA. Its function is to carry coded information from DNA for the synthesis of proteins. The RNA consists of a coding region called exon and non coding region called introns. The exons are thus the functional part that have code for proteins.

89. Which of the following enzymes are used to join bits of DNA?

(a) ligase

(b) primase

(d) endonuclease.

Answer and Explanation:

89. (a): Ligases are used to join bits of DNA. Primase is an RNA polymerase, used to initiate DNA synthesis. DNA polymerase enzyme catalises the synthesis of DNA. Endonuclease, causes the splicing of the intron carrying the coding sequence of the same endonuclease.

90. Which of the following reunites the exon segments after RNA SPLICING?

(a) RNA polymerase

(b) RNA primase

(d) RNA proteoses.

Answer and Explanation:

90. (c): RNA polymerase enzyme catalyses the synthesis of RNA. It is single in prokaryotes. There are three types of RNA polymerases in eukaryotes-1 for 28S, 18S and 5.8S RNA, II for mRNA and snRNA and III for tRNA, 5SRNA and scRNA.

Primase is an RNA polymerase that is used to initiate DNA synthesis. RNA ligase reunites the exon segment after RNA splicing.

91. Which of the following occurs more than one and less than five in a chromosome?

(a) chromatid

(b) chromosome

(d) telomere.

Answer and Explanation:

91. (d): Telomeres contain the ends of long linear DNA molecule contained in each chromatid. Before replication there are 2 telomeric ends and after S-phase there are 4- telomeric ends.

In a chromosome before S-phase only one chromatid is there. Centromere is also one in a chromosome. Bead like chromomeres are many in number.

92. Out of 64 codons, 61 codons code for 20 types of amino acid. It is called

(a) degeneracy of genetic code

(b) overlapping of gene

(d) universality of codons.

Answer and Explanation:

92. (a): Codon are sequence of three nucleotides coding for the incorporation of one amino acid into a polypeptide chain. This code is almost universal in nature; identical sets of the three nucleotides are being used by viruses, bacteria and higher organisms. Since there are 64 possible combinations of the four different nucleotides in sets of three, there is a redundancy in the system, which means that most amino acids can be coded for by more than one triplet. Extremes are tryptophan (one codon: UGG) and leucine (six: codons). The genetic code is therefore said to be degenerate.

93. Nucleus of a donor embryonal cell/somatic cell is transferred to an enucleated egg cell. Then after the formation of organism, what shall be true?

(a) organism will have extranuclear genes of the donor cell

(b) organism will have extranuclear genes of recipient cell

(d) organism will have nuclear genes of recipient cell.

Answer and Explanation:

93. (b): Since the egg cell is enucleated so the organism will only have extranuclear genes from it. These genes are present in chloroplast and mitochondria and are
transmitted along the cytoplasm. This results in cytoplasmic or maternal inheritance since the fertilized egg contains no sperm cytoplasm, it is often exclusively maternal.

94. In E. coli, during lactose metabolism repressor binds to:

(a) regulator gene

(b) operator gene

(d) promoter gene.

Answer:

(b) operator gene

95. Jacob and Monad studied lactose metabolism in E. coli and proposed operon concept. Operon concept is applicable for:

(a) all prokaryotes

(b) all prokaryotes and some eukaryotes

(d) all prokaryotes and some protozoans.

Answer and Explanation:

95. (c): Operon model was given by Jacob and Monod (1961) for regulation of protein synthesis in prokaryotes. In bacteria, the genes that contain the information for assembling the enzymes for a metabolic pathway are usually clustered together on the chromosome in a functional complex called an operon.

Regulation of protein synthesis in eukaryotes is explained by gene battery model given by Britten and Davidson. The model assumes the presence of four classes of sequences – (a) producer gene which is comparable to structural gene of prokaryotic operon, (b) receptor site which is comparable to operator gene, (c) integrator gene which is comparable to regulator gene and (d) sensor site which regulates the activity of integrator gene.

96. Transformation experiment was first performed on which bacteria?

(a) E. coil

(b) Diplococcus pneumoniae

(d) Pasteurella pestis.

Answer and Explanation:

96. (b): Transformation involves transfer of genetic material of one bacterial cell into another bacterial cell by some unknown mechanism and it converts one type of bacterium into another type.

This was first studied by Griffith (1928) in Diplococcus pneumoniae and hence is known as Griffith effect.

97. In a DNA percentage of thymine is 20% then what will be percentage of guanine?

(a) 20%

(b) 40%

(d) 60%.

Answer and Explanation:

97. (c): In a DNA, the percentage of thymine is 20%. So as it pairs with adeneni, it is also 20%. So the guanine and cytosine together forms 60% of DNA and hence guanine is 30%.

98. In the genetic code dictionary, how many codons are used to code for all the 20 essential amino acids?

(a) 20

(b) 64

(d) 60

Answer and Explanation:

98. (b): Codon is a sequence of three nucleotides coding for one aminoacid in a polypeptide chain. There are four different tpes of nucleotides – A, T, G and C and since a codon is a triplet thus 64 (4 x 4 x 4) distinct triplets of purines or pyrmidine bases determine the 20 amino acids.

But out of these 64 codons, only 61 codons code for individual amino acids during protein synthesis. The three codons that do not code for amino acids are the termination or nonsense codons. These are UAA, UAG and UGA. So the remaining 61 codons code for the 20 amino acids.

99. The linkage map of X-chromosome of fruit fly has 66 units, with yellow body gene (y) at one end and bobbed hair (b) gene at the other end. The recombination frequency between these two genes (y and b) should be

(a) 66%

(b) >50%

(d) 100%

Answer and Explanation:

99. (a): The linkage map is a chromosome map which is determined by the recombination relations. The map distances are expressed by recombination frequencies and are given by recombination frequencies and are sometimes represented in map unit.

X-chromosome has 66 crossover units with yellow and bobbed genes at two extreme ends of the map. Recombination frequencies never exceed 50% between any two loci, but these 66 units will be actually obtained by making use of a mapping function.

100. Genes for cytoplasmic male sterility in plants are generally located in

(a) chloroplast genome

(b) mitochondrial genome

(d) cytosol

Answer and Explanation:

100. (b): In several crops like maize cytoplasmic control of male sterility is known. In such cases if female parent is male sterile F, progen would also be male sterile because cytoplasm is mainly derived from egg obtained from male sterile female parent.

Many experiments have proved that factors responsible for cytoplasmic male sterility are located in mitochondrial DNA. A particular phenotype arises because of dominant mutations in the mitochondrial genome.

101. Degeneration of a genetic code is attributed to the

(a) first member of a codon

(b) second member of codon

(d) third member of a codon

Answer and Explanation:

101. (d): In a triplet for a particular amino acid more than one word (synonyms) can be used. This phenomenon is described by saying that the code is degenerate. A degenerate code would be one where there is one to one relation between aminoacids and the codons that 44 codons out of 64 will be useless or nonsense codons.

A code is degenerate because of the third base of the codon. It has been shown that the same /RNA can recognize more than one codons differing only at the third position. For example GCU, GCC and GCA all code for alanine amino acids.

102. When a cluster of genes show linkage behaviour they

(a) do not show a chromosome map

(b) show recombination during meiosis

(d) induce cell division

Answer and Explanation:

102. (c): When a cluster of genes show linkage behaviour they do not show independent assortment. Law of independent assortment suggests that when two or more than two factors are considered together, these factors would show independent and random assortment during distribution in gametes.

It is, therefore, necessary that if two characters have to assort independently, they should be located on non-homologous chromosome. But due to linkage when two or more genes are located in the same chromosome, independent assortment is not possible. Because, the prime necessity of independent assortment is the presence of the characters in different chromosomes.

103. During transcription, the DNA site at which RNA polymerase binds is called

(a) promoter

(b) regulator

(d) enhancer

Answer and Explanation:

103. (a): Promoter is region on a DNA molecule upstream from the coding sequence, area to which RNA polymers initially binds prior to the initiation of transcription. The promoter, or at least part of it, determines the nature of the polymerase that associates with it.

Certain consensus sequences within the promoter region seem to be particularly important in the binding of RNA polymerase, and these are known as CAAT and TATA boxes. The promoter region extends from some 40 nucleotides to about five nucleotides upstream from the start of the gene-coding region, the CAAT and TATA boxes being located within the promoter region as short six or seven nucleotide sequence.

104. What would happen if in a gene encoding a polypeptide of 50 amino acids, 25^th codon (UAU) is mutated to UAA?

(a) a polypeptide of 24 amino acids will be formed

(b) two polypeptides of 24 and 25 amino acids will be formed

(d) a polypeptide of 25 amino acids will be formed

Answer and Explanation:

104. (a): UAA is one of the three non sense codons which do not code for any amino acid. If UAU is mutated to UAA then the polypeptide chain will be terminated at that point because /RNA cannot bring next amino acid as UAA does not code for any amino acid. So in that case a polypeptide chain of 24 aminoacids will be formed.

105. Which one of the following triplet codes, is correctly matched with its specificity for an amino acid in protein synthesis or as ‘start’ or ‘stop’ codon?

(a) UCG-start

(b) UUU – stop

(d) UAC – tyrosine

Answer and Explanation:

105. (d): Codon UAC is correctly matched as it codes for amino acid tyrosine. UCG codes for serine, UUU codes for phenylalanine and UGU codes for cysteine. Start codon is UAG and stop codons are UAA, UAG and UG(c) heterosis A.

106. During translation initiation in prokaryotes, a GTP molecule is needed in

(a) formation of formyI-met-/RNA

(b) binding of 30 S subunit of ribosome with /nRNA

(d) association of 50 S subunit of ribosome with initiation complex.

Answer:

107. What does “lac” refer to in what we call the lac operon ?

(a) lactose

(b) lactase

(d) the number 1,00,000

Answer and Explanation:

107. (a): In lac operon, lac refers to lactose. The lac operator is a part of the structural genes (lac Z, lac Y, lac A and lac I). It is responsible for the uptake and initial catabolism of lactose.

108. In Drosophila, the sex is determined by

(a) the ratio of number of X-chromosome to the sets of autosomes

(b) X and Y chromosomes

(d) whether the egg is fertilized or develops parthenogenetically

Answer:

109. Genetic map is one that

(a) establishes sites of the genes on a chromosomes

(b) establishes the various stages in gene evolution

(d) shows the distribution of various species in a region

Answer and Explanation:

109. (a): Linkage map is a graphic representation of relative positions/order and relative distance of genes in a chromosome in the form of line like a linear road map depicting different places and their relative distances without giving exact mileage.

It is based on Morgan’s hypothesis (1911) that frequency of crossing over/ recombination between two linked genes is directly proportional to the physical distance between the two. 1 map unit or centimorgan is equivalent to 1% recombination between two genes.

110. What would happen if in a gene encoding a polypeptide of 50 amino acids, 25^th codon (UAU) is mutated to UAA?

(a) a polypeptide of 24 amino acids will be formed

(b) two polypeptides of 24 and 25 amino acids will be formed

(d) a polypeptide of 25 amino acids will be formed

Answer and Explanation:

110. (a): UGA, UAG and UAA are three non sense (or termination) codon which do not code for any amino acid. If in a gene encoding a polypeptide of 50 amino acid, 25^thcodon is mutated to UAA or any of the termination codon, then the chain will be terminated at that place because it will become difficult for /RNA to bring amino acid from amino acid pool. So in that case a polypeptide of 24 amino acid will be formed.

111. The following ratio is generally constant for a given species:

(a) A + G/C + T

(b) T + C/G + A

(d) A + C/T + G.

Answer and Explanation:

111. (c): According to Chargaff (1950) rules, purines and pyrimidines occur in equal amounts. The amount of adenine is equal to that of thymine and cytosine is equal to that of guanine (A = T and G = C). However the amount
of A or T is not equal to G or C. It means A + T/G + C is variable. It is 1.52 in human beings and 0.97 in E.coli. In other words G + C/A + T is constant for a particular species.

112. In mutational event, when adenine is replaced by guanine, it is a case of

(a) frame shift mutation

(b) transcription

(d) transversion

Answer and Explanation:

112. (c): Transition mutant is one in which a purine is substituted by a different purine, or a pyrimidine by a different pyrimidine. Such a change involves a base pair change between a G-C pair and an A-T pair in the DNA. Whereas transversion results when one nitrogen base is replaced by another different type e.g. C-G and A-T. Transcription is the formation of mRNA on DNA templete.

113. During transcription, if the nucleotide sequence of the DNA strand that is being coded is ATACG then the nucleotide sequence in the mRNA would be

(a) TATGC

(b) TCTGG

(d) UATGC.

Answer and Explanation:

114. (a): Refer answer 78.

114. Extra nuclear inheritance is the consequence of presence of genes in

(a) mitochondria and chloroplasts

(b) endoplasmic reticulum and mitochondria

(d) lysosomes and ribosomes.

Answer:

(a) mitochondria and chloroplasts

115. Which form of RNA has a structure resembling clover leaf

(a) rRNA

(b) An RNA

(d) rRNA.

Answer:

(d) rRNA

116. After a mutation at a genetic locus the character of an organism changes due to change in

(a) protein structure

(b) DNA replication

(d) RNA transcription pattern.

Answer and Explanation:

116. (a): A mutation involves a change in the sequence of nucleotides in a nucleic acid molecule. This change will express itself in the form of a change in the sequence of aminoacids in the protein molecule synthesized through the information encoded in nucleic acid segment.

Therefore mutations at molecule level can be studied both by the study of the sequence of amino acids in a protein and also by the study of sequence of nucleotides in a segment of nucleic acid.

117. During transcription holoenzyme RNA polymerase binds to a DNA sequence and the DNA assumes a saddle like structure at that point. What is that sequence called?

(a) AAAT box

(b) TATA box

(d) CAAT box.

Answer and Explanation:

117. (b): After 25 bases from start of transcription point are TATA boxes. After 40 bases from TATA boxes appear CAAT boxes. Both of these sequences serve as recognitions sites in eukaryotic promoters, Transcription in eukaryotic genes is a far more complicated process than in prokaryotes.

118. Which one of the following hydrolyses internal phosphodiester bonds in a polynucleotide chain?

(a) lipase

(b) protease

(d) exonuclease.

Answer and Explanation:

118. (c): Endonucleases hydrolyse the internal phosphodiester bond. Exonucleases cleave the terminal nucleotides. Lipase digest fats and proteases break down proteins.

119. Which one of the following makes use of RNA template to synthesize DNA?

(a) DNA polymerase

(b) RNA polymerase

(d) DNA dependant RNA polymerase.

Answer and Explanation:

119. (c): Reverse transcriptase brings about reverse transcription by forming DNA from RNA. It is also called RNA dependent DNA polymerase enzyme because it is synthesizing a DNA segment and for that it is dependent on RNA.

The DNA segment so formed is called retroposons.

120. Protein synthesis in an animal cell occurs

(a) only on the ribosomes present in cytosol

(b) only on ribosome attached to the nuclear envelope and endoplasmic reticulum

(d) on ribosomes present in cytoplasm as well as in mitochondria.

Answer and Explanation:

120. (d): The mitochondria contains its own set of ribosomes which synthesize proteins, so protein synthesis occurs both in mitochondria and cytoplasm.

121. Using imprints from a plate with complete medium and carrying bacterial colonies, you can select streptomycin resistant mutants and prove that such mutations do not originate as adaptation. These imprints need to be used

(a) on plates with and without streptomycin

(b) on plates with minimal medium

(d) only on plates without streptomycin.

Answer and Explanation:

121. (c): Streptomycin is broad spectrum (active against both gram-positive and gram-negative bacteria) and was the first really effective drug against tuberculosis, but its use is limited by the development of resistant strains and by toxic side-effects. The bactericidal action of streptomycin, as with other aminoglycoside antibiotics (e.g., neomycin) is through selective inhibition of protein synthesis on 70s ribosomes.

To check resistance of mutants against streptomycin they must be grown on plates with streptomycin. Only those bacterial colonies will propagate from the master that are resistant to the antibiotic.

122. Telomerase is an enzyme which is a

(a) simple protein

(b) RNA

(d) repetitive DNA.

Answer and Explanation:

122. (c): Telomerase is a ribonucleoprotein molecule that is enzymatic in nature. It uses a special mechanism for the synthesis of DNA at telomeric ends. The DNA repeat sequence of telomere has one G-rich strand and other C rich strand. The G-rich strand has a single stranded overhand.

This overhand works as a primer and for its elongation uses as template the RNA component of telomerase enzyme. Thus telomerase synthesizes only the G-rich strand of telomeres.

123. The salivary gland chromosones in the dipteran larvae, are useful in gene mapping because

(a) these are fused

(b) these are much longer in size

(d) they have endoreduplicated chromosomes.

Answer and Explanation:

123. (d): Polytene Chromosomes (Salivary Chromosomes) were discovered by Balbiani (1881) in the salivary glands of Chironomus tantans. Later they were found in salivary glands of many insect larvae including Drosophila. Polytene chromosomes are formed by somatic pairing of homologous chromosomes followed by their repeated replication or endomitosis. They are in permanent prophase.

All the polytene chromosomes may remain attached to a common chromocentre by their pericentric heterochromatin which is slow to replicate. When stained with basic dyes, the chromosomes show dark bands and light interbands. Bands are formed by coming together of chromomeres of all the chromonemata. They, therefore, seem to represent genes.

124. Genes for cytoplasmic male sterility in plants are generally located in

(a) mitochondrial genome

(b) chloroplast genome

(d) cytosol.

Answer and Explanation:

124. (a): Refer answer 100.

125. E.coli cells with a mustard r gene of the lac operon cannot grow in medium containing only lactose as the source of energy because

(a) the lac operon is constitutively active in these cells

(b) they cannot synthesize functional beta galactosidase

(d) they cannot transport lactose from the medium into the cell.

Answer and Explanation:

125. (b): lac operon (lactose operon), is genetic system of E.coli responsible for the uptake and initial catabolism of lactose. The lac operon consists of three structural genes (lac Z, lac Y, lac A). lacZ codes for P-galactosidase which hydrolyses lactose to glucose and galactose.

126. Amino acid sequence, in protein synthesis is decided by the sequence of

(a) rRNA

(b) RNA

(d) c-DNA

Answer and Explanation:

126. (c): Messenger RNA or mRNA has been named so because it carries the coded information from DNA for the synthesis of proteins. It carries the coded information in a number of base triplets called codons. It is transcribed on DNA by the enzyme RNA polymerase. Hence its base sequence is complementary to DNA on which it has been synthesized.

In eukaryotes each gene transcribes its own m-RNA, therefore the number of m-RNAs corresponds to the number of genes. rRNA is a type of RNA that forms structural and functional components of ribosomes.

tRNA is a class of RNA having structures with triplet nucleotide sequences that are complementary to the triplet nucleotide coding sequences of mRNA. It binds with amino acids and transfers them to ribosomes.

127. Which antibiotic inhibits interaction between tRNA and mRNA during bacterial protein synthesis?

(a) tetracycline

(b) erythromycin

(d) streptomycin

Answer and Explanation:

127. (c): Neomycin is a broad spectrum antibiotic which was first isolated from a strain of Streptomyces feadiae. It is effective against gram positive as well as gram negative bacteria. Its mechanism of action is by selective inhibition of protein synthesis on the 70s (prokaryotic) ribosome by inhibiting the interaction of mRNA and tRNA during translation process.

Streptomycin resembles neomycin as it also inhibits protein synthesis by inhibiting prokaryotic peptide chain initiation. It binds to 30s ribosomes.

128. Antiparallel strands of a DNA molecule means that

(a) one strand turns clockwise

(b) one strand turns anti-clockwise

(d) the phosphate groups at the start of two DNA strands are in opposite position (pole)

Answer and Explanation:

128. (d): DNA is a type of nucleic acid that forms genetic material in many organisms. It consists of a long polymer of nucleotides which transcribes the coded information in the form of a triplet code of nucleotides

in mRNA. It is a double helical molecule. The two strands of DNA run in opposite directions to one another with the hydrogen bonds between them. One strand of DNA has 5′-3′ direction and the other strand has 3′-5′ direction. So they are antiparallel. This direction is determined by the presence of a free phosphate or OH group at the end of the strand. If the strand has phosphate group at the the 5′ end and with a free OH group at the 3′ end.

129. In which mode of inheritance do you expect more maternal influence among the offspring?

(a) X-l inked

(b) autosomal

(d) Y-linked

Answer and Explanation:

129. (c): Cytoplasmic inheritance or nonchromosomal (extranuclear) inheritance is the passage of traits from parents to offspring through structures present inside the cytoplasm of contributing gametes. Plasma genes occur in plastids, mitochondria, plasmids and some special particles like kappa particles, sigma particles, etc.

In higher organisms cytoplasmic inheritance is also called maternal inheritance because the zygote receives most of its cytoplasm from the ovum. Therefore, cytoplasmic inheritance is usually uniparental.

130. One gene-one enzyme hypothesis was postulated by

(a) Beadle and Tatum

(b) R. Franklin

(d) A. Garrod

Answer and Explanation:

130. (a): In 1948, Beadle and Tatum proposed one-gene one-enzyme hypothesis which states that a gene controls metabolic machinery of the organism through synthesis of an enzyme. This laid the foundation of biochemical genetics. Beadle and Tatum were awarded Nobel Prize in 1958.

This one genen one enzyme theory has been changed to one gene one polypeptide hypothesis propesed by yanofsky. i.e. one gene synthesizes one polypeptide and many pohypeptides form one enzyme.

131. One turn of the helix in a B-form DNA is approximately

(a) 2 nm

(b) 20 nm

(d) 3.4 nm

Answer and Explanation:

131. (d): DNA or deoxyribose nucleic acid is the largest macromolecule made of the helically twisted two antiparallel polydeoxyribonucleotide stands held together by hydrogen bonds. The two strands of DNA are together called DNA duplex. It has a diameter of 20A. One turn spiral has a distance of 34 A. It contains 10 deoxyribonucleotides in each strand so that the distance between two adjacent nucleotides is 3.4 A.

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