InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Do you think that the alternate splicing of exons may enable a structural gene to code for several isoproteins from one and the same gene? If yes, how? If not, why so ? |
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Answer» Fuctional mRNA of structural genes need not alwaus include all of its exons. This alternate splicing of exons is sex-specific, tissue-specific and even developmental stage-specific. By such alternate splicing of exons, a single gene may encode for several isoproteins and /or proteins of similar class. Protein/isoprotein. Such an extravagancy has been avoided in naturel phenomena by way of alternate splicing. |
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| 2. |
Would it be appropriate to use DNA probes such as VNTR in DNA fingerprinting of a bacteriophage? |
| Answer» Bacteriophage does not heve repetitive sequences such as VNTRs in its genome, as its genome is very small and have all the coding sequence. DNA finger printing is not done for phages. | |
| 3. |
Define a cistron. Giving examples differentiate between monocistronic and polycistronic unit. |
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Answer» A cistron is stretch of base sequences that codes for one polypeptide chain including adjacent control regions. It may also code for a tRNA, rRNA molecule or may perform other specific functions including regulating functions of other cistrons. This term has replaced the definition of a gene . Monocistronic transcription unit will have all the regulatory and coding sequences for a single polypeptide, Whereas polycistronic may have coding sequenes for more then one polypeptide. |
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| 4. |
Give any six features of the human genome. |
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Answer» Slient features of humen genome (i) The humen genome contains 3164. 7 million nucleotide bases. (ii) The average gene consists of 30000 the largest know humen gane being dystrophin at 2.4 Million bases . (iii) The total number of genes is estimated to be 30000 and 99.9% nuceotide bases are exactly the same in all people. (iv) The functions are unknown for over 50% of the discovered genes. (v) Less than 2% of the genome codes for proteins. (vi) The humen genome contains large repeted sequences . (vii) The repeated sequence is through to have no direct coding functions but they throw light on chromosomes struction , dynamics and evolution . (viii) Chromosome I has most genes (2968) and the Y has the fewest genes(231) (ix) Scientists have identified about 1.4 million location where single base DNA sequence differences called SNPs or Single Nucleotide Polymorphisms occur in humans. |
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| 5. |
Retroviruses do not follow central dogma. Comment. |
| Answer» Retroviruses do not follow central dogma of biology (DNA`to`RNA`to`Proein) because their genetic meterial is not DNA. Instead they have RNA that is converted to DNA by the enzyme reverse transcriptese. | |
| 6. |
During DNA replication, why is it that the entire molecule does not open in one go? Explain replication fork. What are the two functions that the monomers (dNTPs) play ? |
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Answer» While replicating, the entria DNA molecule to keep the whole molecule stabilised does not open in one go because it would be highly expens energetically. Actually unwiding creates tension in the molecule as uncoiled parts. Actually , unwinding creates tension in the molecule us cucoiled parts start forming super coils due to the interaction of exposed nucleotides . Instead, helicase enzyme acts on the double strand at ori site (origin of replication ) and a small stretch is unzipped. Immediately, it is held and stabilised by single strand binding proteins. Slowly with the help of enzymes, exposed strands are copied as a point of unwinding moves and ahead in both directins . It gives an appearance of Y-shaped structure which is called replication fork. The two functions that the monomer units of NTPs play are (i) They pair up with exposed nucleotides of the template strand and make phosphodiester linkages and release a pyrophosphate. (ii) Hydrolysis of this pyrophosphate by enzyme pyrophosphatase releases energy that will facillitate making hydrogen bonds between free nucleotides and bases of the template strand. |
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| 7. |
What would happen if histones were to be mutated and made rich in acidic amino acids such as aspertic acid and gultamic acid in place of basic amino acids such as lysine and arginine? |
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Answer» If histions were mutated and made rich in acidic amino acids. They will not be able to serve the purpose of keeping the DNA coilied around them. This is because DNA is negatively changed molecule and histones are positively changed because of basic amino acids . So , they are attracted to each other . If histiones become negatively charged, instead of bindding, they will rather repal DNA. The packaging of DNA in eukaryotes would not happen. Consequently, the chromatin fifre would not be formed. |
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| 8. |
In a nucleus, the number of ribonucleoside triphophates is 10 times the number of deoxy `xx10` ribonucleoside triphosphaes, but only deoxy ribonucleotides are added during the DNA replication. Suggest a mechanism. |
| Answer» DNA polymerase enzyme is highly specific to recognise only deoxy ribonuceoside triphosphates. Therefore, it connot hold RNA`beta`-nucleotides. | |
| 9. |
Sometimes cattle or even humen beings give birth to their young ones that are having extremely different sets of organs like limba/position of eye(s) etc. Comment . |
| Answer» Sometimes cattle or even humen being give birth young ones that are having extremely different of organs like limbs/position of eye etc. It happens due to the disturbance in coordinated regulation of expression in sets of genes, which are associated with orgen development. | |
| 10. |
What is DNA polymorphism? Why is it important to study it? |
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Answer» DNA polymorphism refers ro the variation in DNA arising through mutation at non-coding sequences. A special type of polymorphism, called VNTR (Variable Number of Tendem Repeats), is composed of repeated copies of a DNA sequence that lie adjacent to one another on the chromosome. Since, polymorphism is the basis of genetic mapping of humen genome, therefore, it forms the basis of DNA fingerprinting too. history as well as in cese of paternity testing . |
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| 11. |
Based on your understanding of genetic code, explain the formation of any abnormal heamoglobin molecule. What are the known consequences of such a change? |
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Answer» Due to point mutation in `beta`-globin chain of haemoglobin molecule, glutamic acid (Glu) is replaced by valine (Val) at the sixth position. Under stress condition erythrocytes lose their circular shape and become sickle -shaped . As a result, the cells cannot pass through narrow capillaries. Blood capillaries are clogged and thus affect blood supply to different organs. |
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| 12. |
What is the function of histones in DNA packaging ? |
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Answer» Functions of histones in DNA packaging are (i) Histones as units of octomer participate in primary packaging of DNA . (ii) Basic histone proteins neutralise the acidic DNA molecule. |
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| 13. |
In some viruses, DNA is synthesised by using RNA as template . Such a DNA is calledA. A-DNAB. B-DNAC. cDNAD. rDNA |
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Answer» Correct Answer - C In some viruses, like retroviruses (e.g., HIV), an enzyme called reverse transcriptase is used to generate complementary DNA(cDNA) from an RNA template. This process is termed reverse transcription . |
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| 14. |
The total number of genes in humans is far less(lt25000) the than the pravious estimate (up to 140000 gene). Comment. |
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Answer» The total number of genes is estimated at 25000 much lower than provious estimates of 14000 that had been based on extrapolations from gene-rich areas as opposed to composite of gene-rich rich and gene-poor areas. This information promises to revolutionise the processes of finding chromosomal locations for diseases- associated sequence and tracing humen history . |
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| 15. |
The net electric charge on DNA and histones isA. both positiveB. both negativeC. Both (a) and (b)D. zero |
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Answer» Correct Answer - C DNA consists of a nirtogenous base, pentose suger and a phosphate group. DNA has negative change due to the presence of phosphate group `(PO_(4)^(3-))` Histones are rich in the basic amino acid residues lysines and arginines , which carry positive charges in their side chains. Therefore, histones are positively charged. |
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| 16. |
Which of the following statements is the most appropriate for sickle cell anaemiaA. It connot be treated with iron supplementsB. it is a molecular diseaseC. It confers resistance to acquiring malariaD. All of the above |
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Answer» Correct Answer - D (Sickle-cell anaemia is an autosome linked recessive trait. In this genatic disorder point mutation in `beta`-globin chain results in change of glutamate (glutamic acid) to valine at sixth position . Olny the homozyhous individuals for `Hb_(2)^(s)`,i.e., `Hb^(s) Hb^(s)` show the diseased phenotype. The heterozygous `(Hb^(s)// Hb^(A))` are carriers. It is also known that heterozygotes, having both types of heamoglobin, show resistance to malaria infection because the body targets the P. falsiparum (protozoan) infected cells for destruction of RBC. |
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| 17. |
With regard to mature mRNA in eukaryotesA. exons and introns do not appear in the mature RNAB. exons appear but introns do not apper in the mature RNAC. introns appear but exons do not apper in the mature RNAD. both exons. And introns appear in the mature RNA |
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Answer» Correct Answer - D In eukaryotes, the monocistronic structural genes have interrupted coding sequences i.e., the genes in eukaryotes are split. The coding sequences or expressed sequences are defined as exons. These sequences (exons ) appear in mature or processed RNA. THe exons are interrupted by introns or intervening sequences which do not appear in mature or processed RNA. |
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| 18. |
One of the following is true with respect to AUGA. It codes for methionine onlyB. It is also an initation codenC. It codes for methionine in both prokaryotes and eukaryotesD. All of the above |
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Answer» Correct Answer - D Polypeptide synthesis is signalled by two initiation codons commonly AUG or methionine codon and rarely GUG or valine codon. Since there are 64 triplet codons and only 20 amino acids, the incorporation of same amino acids must be influenced by more then one codon. Only trytophan (UGG) or methionine (AUG) are specified by single codons. AUG codes for methionine in both prokaryotes and eukaryotes. |
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| 19. |
A low level of expression of lac operon occurs at all the time. Can you explain the logic behind this phenomena. |
| Answer» In the complete absence of expression of lac operon, permease will not ve synthesised which is essential for transport of lactose from medium into the cells. And if lactose connot be transported into the cell, then it cannot act as inducers. Hence , cannot relieve the lac operon from its repressed state. | |
| 20. |
The first genetic material could beA. ProteinB. carbohydratesC. DNAD. RNA |
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Answer» Correct Answer - D RNA was the first genetic material. There is now enough evidence to suggest that essential life processes (such as metabolism, translation, splicing, etc.) evolved around RNA. RNA used to act as a genetic maerial as well as a catalyst (there are some important biochemical reactions in living systems that are catalysed by RNA catalysts and not by protein enzymes ). But , RNA being a catalyst was reactive and hence unstable . Therefore, DNA has evoled from RNA with chemical modifications that make it more stable. DNA being double-stranded and having complementary strand further resists changes by evolving a prosess of repair . |
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| 21. |
Give an account of the methods used in sequencing the human genome. |
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Answer» Sequencing of humen genome has made it possible to understend the link between various genes and their functions . If there are any gene defacts that defect that express as disorders or that increase the susceptibility of an individual to a disease then specific gene therepies can be worked out Methodologies of humen genome sequencing The methods involve two major approaches (i) Expressed Sequence Tags (ESTs) This method focusses on identifying all the genes that are expressed as RNA . (ii) Sequence annotation It is an apperoach of simply sequencing the whole set of genome that contains all the coding and non-coding sequence, and later assigning different regions in the sequence with functions . For sequencing, first the total DNA from cell is i.e., solated and broken down in relatively small sizes as fragments. There DNA fragments are cloned in suitable host using suitable vectors. When becteria is used as vactor , they are called Bacterial Artificial Chromosomes (BAC) and when yeast is used as vector, they are called Yeast Artifical Chromosomes (YACs). Frederick Sanger developed a principle according to which the fragments of DNA are sequenced by automated DNA sequences. On the basis of overlapping regions on DNA fragment, these sequences are arranged accordingly . For alignment of these sequences, specialised computer-based programmes were developed. Finally, the genetic and physical maps of the genome were constructed by collecting information about cartian repetitive DNA sequences and DNA polymorphism, based on endonuclease recognition sites. |
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| 22. |
Comment on the utility of variability in number of tansem repeats during DNA fingerprinting . |
| Answer» Tendemness in repets provides many copies of the sequence for finger-priniting and variability in nitrogen base sequences present in them. Being individual-specific, this proves to be useful in the process of DNA fingerprinting. | |
| 23. |
In an experiment, DNA is treated with the compound which tends to place itself amongst the stacks of nitrogenous base pairs. As a result of this, the distance between two consecutive base increases.From 0.34-0.44nm, calculate the length of DNA double helix (Which has `2x10^(9)bp) in the presence of saturating of this compound. |
| Answer» The length of DNA double helix `=2xx10^(9)xx0.44xx10^(-9)/ bp` | |
| 24. |
List the various markers that are used in DNA fingerprinting |
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Answer» Dr. Alec Jeffreys developed tht technique of DNA fingerprinting in an attempt to identify DNA marker for inherited diseases. DNA fingerprinting uses short nucleotide repeats called Variable Number Tendem Repeats (VNTRs) as markes . VNTRs very from persom and are inherited from one genaeration to the next. Only closely indiviuals have similar VNTRs. |
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