CBSE Class 12 Biology Molecular Basis of Inheritance Worksheet Set C

MULTIPLE CHOICE QUESTIONS

Question. Balbiani rings are found in
(a) polysomes
(b) polytene chromosomes
(c) autosomes
(d) nonsense chromosomes
Answer. B

Question. In DNA helix, cytosine is paired with guanine by 
(a) covalent bond
(b) phosphate bond
(c) three hydrogen bonds
(d) two hydrogen bonds
Answer. C

Question. Which RNAs pick up specific amino acid from amino acid pool in the cytoplasm to ribosome during protein synthesis ? 
(a) tRNA
(b) mRNA
(c) rRNA
(d) hnRNA
Answer. A

Question. The structure of DNA is 
(a) linear
(b) double helix
(c) single helix
(d) triple helix
Answer. B

Question.Transposon was discovered by 
(a) Sutton
(b) Strassburger
(c) Fischer
(d) B.Mc Clintock
Answer. D

Question. Root cell of wheat has 42 chromosomes. What would be the number of chromosomes in the synergid cell ? 
(a) 7
(b) 14
(c) 21
(d) 28
Answer. C

Question. What is true about t-RNA? 
(a) It binds with an amino acid at it 3' end.
(b) It has five double stranded regions.
(c) It had a codon at one end which recognizes the anticodon on messenger RNA.
(d) It looks like clover leaf in the three dimensional structure.
Answer. D

Question. Which one of the following codons codes for the same information as UGC? 
(a) UGU
(b) UGA
(c) UAG
(d) UGG
Answer. A

Question. Which one of the following pairs is correctly matched with regard to the codon and the amino acid coded by it ? [2004, 2008]
(a) UUA-valine
(b) AM-lysine
(c) AUG-cysteine
(d) CCC-alanine
Answer. A

Question.

What is the error in above diagram? 
(a) Arrows are wrongly depicted.
(b) Polarity is incorrect.
(c) Both arrows and polarity are incorrect.
(d) None of the above.
Answer. B

Question. TATA box of eukaryotic promotor lies 
(a) about 25 bp upstream of the transcription start site.
(b) about 50 bp upstream of the transcription start site.
(c) about 75 bp upstream of the transcription start site.
(d) about 200 bp upstream of the transcription start site.
Answer. A

Question. Select the correct option: 
Direction of RNA Direction of reading of synthesis the template DNA strand
(a) 5´—3´ 3´—5´
(b) 3´—5´ 5´—3´
(c) 5´—3´ 5´—3´
(d) 3´—5´ 3´—5´
Answer. A

Question. There are three genes a, b, c. Percentage of crossing over between a and b is 20%, b and c is 28% and a and c is 8%. What is the sequence of genes on chromosome?
(a) b, a, c
(b) a, b, c
(c) a, c, b
(d) None of these
Answer. A

Question. Which one of the following group of codons is called as degenerate codons?
(a) UAA, UAG and UGA
(b) GUA, GUG, GCA, GCG and GAA
(c) UUC, UUG, CCU, CAA and CUG
(d) UUA, UUG, CUU, CUC, CUA and CUG
Answer. A

Question. The given figure shows the structure of nucleosome with their parts labelled as A, B & C. Identify A, B and C.

(a) A – DNA; B – H1 histone;C – Histone octamer
(b) A – H1 histone; B – DNA;C – Histone octamer
(c) A – Histone octamer; B – RNA;C – H1 histone
(d) A – RNA; B – H1 histone;C – Histone octamer
Answer. A

 

ASSERTION REASON QUESTIONS

Directions : These questions consist of two statements, each printed as Assertion and Reason. While answering these questions, you are required to choose any one of the following five responses.
(a) If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
(b) If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
(c) If the Assertion is correct but Reason is incorrect.
(d) If both the Assertion and Reason are incorrect.
(e) If the Assertion is incorrect but the Reason is correct.

Question. Assertion : Histones are basic proteins of major importance in packaging of eukaryotic DNA. DNA and histones comprise chromatin forming the bulk of eukaryotic chromosome.
Reason : Histones are of five major types H1,H2A H2B,H3 andH4 .
Answer. A

Question. Assertion: mRNA attaches to ribosome through its 3' end.
Reason: The mRNA has F-capsular nucleotide and bases of lagging sequence.
Answer. D

Question. Assertion : Replication and transcription occur in the nucleus but translation in the cytoplasm.
Reason : m-RNA is transferred from the nucleus into the cytoplasm where ribosomes and amino acids are available for protein synthesis. 
Answer. A

Question. Assertion: An organism with lethal mutation may not even develop beyond the zygote.
Reason: All types of gene mutations are lethal.
Answer. C

Question. Assertion: Polytene chromosomes have a high amount of DNA.
Reason: Polytene chromosomes are formed by repeated replication. 
Answer. A   

 

Very Short Answer Questions

Question. In an experiment, DNA is treated with a compound which tends to place itself amongst the stacks of nitrogenous base pairs. As a result of which the distance between two consecutive base increases, from 0.34 nm to 0.44 nm. Calculate the length of DNA double helix (which has 2 × 109 bp) in the presence of saturating amount of this compound.
Answer. 2 × 109 × 0.44 nm.

Question. Calculate the length of the DNA of bacteriophage lambda that has 48502 base pairs.
Answer. Distance between two consecutive base pairs = 0.34 × 10–9 m
The length of DNA in bacteriophage lambda = 48502 × 0.34 × 10–9 m
= 16.49 × 10–6 m

Question. How does a degenerate code differ from an unambiguous one? 
Answer. Degenerate code meAnswer that one amino acid can be coded by more than one codon. Unambiguous code meAnswer that one codon codes for only one amino acid.

Question. Why is lactose considered an inducer in lac operon? 
Answer. Lactose binds to repressor and prevents it from binding with the operator, as a result RNA polymerase binds to promoter–operator region to transcribe the structural genes.

Question. What is cistron? 
Answer. A cistron is a segment of DNA coding for a polypeptide.

Question. State which human chromosome has
(i) the maximum number of genes and
(ii) the one which has the least number of genes? 
Answer. (i) Chromosome-1
(ii) Y-Chromosome

Question. Mention two functions of the codon AUG. 
Answer. Two functions of the codon AUG are:
(i) It acts as a start codon during protein synthesis.
(ii) It codes for the amino acid methionine.

Question. What is DNA fingerprinting? Mention its application.
Answer. DNA fingerprinting is the technique to determine the relationship between two DNA samples by studying the similarity and dissimilarity of VNTRs (variable number of tandem repeats). Its applications are:
(i) It is used as a tool in forensic tests to identify criminals.
(ii) To settle paternity disputes.
(iii) To identify racial groups to study biological evolution.

Question. List two essential roles of ribosome during translation.
Answer. Two essential roles of ribosome during translation are:
(i) One of the rRNA (23S in prokaryotes) acts as a peptidyl transferase ribozyme for formation of peptide bonds.
(ii) Ribosome provides sites for attachment of mRNA and charged tRNAs for polypeptide synthesis.

Question. In the medium where E. coli was growing, lactose was added, which induced the lac operon.
Then, why does lac operon shut down some time after addition of lactose in the medium?
Answer. As lac operon is an inducible operon therefore, it shuts down due to decrease in lactose substrate concentration.

Question. Explain (in one or two lines) the function of the following:
(a) Promoter (b) tRNA (c) Exons
Answer. (a) Promoter: It is the segment of DNA which lies adjacent to the operator and functions as the binding site for RNA polymerase to carry transcription if allowed by operator.
(b) tRNA: It acts as an adaptor molecule that picks up a particular amino acid from cellular pool and takes the same over to A site of mRNA for incorporation into polypeptide chain.
(c) Exons: These are the coding segments present in primary transcript which after splicing are joined to form functional mRNA.

Short Answer Questions

Question. (a) Name the enzyme responsible for the transcription of tRNA and the amino acid the initiator tRNA gets linked with.
(b) Explain the role of initiator tRNA in initiation of protein synthesis.
Answer. (a) RNA polymerase III is responsible for transcription of tRNA and the initiator tRNA gets linked with the amino acid methionine.
(b) The initiator tRNA, which is charged with amino acid methionine, reaches the smaller subunit of ribosome. Its anticodon UAC recognises the codon AUG on mRNA and binds by
forming complementary base pairs. The large subunit of ribosome joins the smaller subunit and initiates translation.

Question. Given below is a schematic representation of lac operon:

(a) Identify i and p.
(b) Name the ‘inducer’ for this operon and explain its role.
Answer. (a) i is the regulatory gene and p is the promoter gene.
(b) Lactose is the inducer. It is the substrate for the enzyme beta-galactosidase and it regulates switching on and off of the operon.

Question. How are the structural genes inactivated in lac operon in E. coli? Explain.
Answer. The regulator gene produces repressor which when free, binds to the operator region of the operon and prevents RNA polymerase from transcribing the structural genes.

Question. How are the structural genes activated in the lac operon in E. coli?
Answer. Lactose acts as the inducer that binds with repressor protein that cannot bind to operator and hence frees the operator gene. RNA polymerase freely moves over the structural genes, transcribing lac mRNA, which in turn produces the enzymes responsible for the digestion of lactose.

Question. “Genes contain the information that is required to express a particular trait.” Explain.
Answer. The genes present in an organism show a particular trait by way of forming certain product.
This is facilitated by the process of transcription and translation (according to central dogma of Biology)

Question. 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 be synthesised which is essential for transport of lactose from medium into the cells. And if lactose cannot be transported into the cell, then it cannot act as inducer. Hence, cannot relieve the lac operon from its repressed state. Therefore, lac operon is always expressed.

Question. Where is an ‘operator’ located in a prokaryote DNA? How does an operator regulate gene expression at transcriptional level in a prokaryote? Explain.
Answer. The operator region is located adjacent to promoter elements or prior to structural gene.
The operator regulates switching on and off the operon when the repressor binds to the operator region it is switched off and prevents transcription.
In the presence of inducer the repressor is inactivated and operator allows RNA polymerase to access the promoter. The operon is switched on and transcription proceeds.

Question. Would it be appropriate to use DNA probes such as VNTR in DNA fingerprinting of a bacteriophage? 
Answer. Bacteriophage does not have repetitive sequence such as VNTR in its genome as its genome is very small and have all the codon sequenced. Therefore, DNA fingerprinting is not done for bacteriophages.

Question. What would happen if histones were to be mutated and made rich in amino acids aspartic acid and glutamic acid in place of basic amino acids such as lysine and arginine?
Answer. If histone proteins were rich in acidic amino acids instead of basic amino acids then they may not have any role in DNA packaging in eukaryotes as DNA is also negatively charged molecule. The packaging of DNA around the nucleosome would not happen. Consequently, the chromatin fibre would not be formed.

Question. Mention two applications of DNA polymorphism.
Answer. DNA polymorphism is applicable in genetic mapping and DNA finger printing.

Question. (a) Construct a complete transcription unit with promoter and terminator on the basis of the hypothetical template strand given below: 

(b) Write the RNA strand transcribed from the above transcription unit along with its polarity.

Answer. (a)

Question. (i) Name the enzyme that catalyses the transcription of hnRNA.
(ii) Why does the hnRNA need to undergo changes? List the changes that hnRNA undergoes and where in the cell such changes take place.
Answer. (i) RNA polymerase II.
(ii) hnRNA has non-functional introns in between the functional exons. To remove these, it undergoes changes. The changes that hnRNA undergoes include capping, i.e., methyl guanosine triphosphate is added to 5′ end; tailing in which poly A tail is added at 3′ end; and splicing by which introns are removed and exons are joined.

Question. One of the salient features of the genetic code is that it is nearly universal from bacteria to humAnswer. Mention two exceptions to this rule. Why are some codes said to be degenerate?
Answer. The genetic code is universal except in mitochondria and few protozo Answer.
Some codes are said to be degenerate because some amino acids are coded by more than one code.

Question. (a) Draw a clover leaf structure of tRNA showing the following:
(i) Tyrosine attached to its amino acid site.
(ii) Anticodon for this amino acid in its correct site (codon for tyrosine is UCA).
(b) What does the actual structure of tRNA look like?
Answer. 

(b) The actual structure of tRNA looks like inverted L.

Question. Explain the significance of satellite DNA in DNA fingerprinting technique.
Answer. Satellite DNA are very specific in each individual, vary in number from person to person and are inherited. These sequences show high degree of polymorphism. Each individual inherits the satellite DNA from, his/her parents which are used as genetic markers in DNA finger printing.
200. DNA fingerprinting is a technique of determining nucleotide sequences of certain areas of DNA which are unique to each individual. DNA fingerprints can be prepared from extremely minute amounts of DNA sample from blood, semen, hair bulb or any other cells of the body.

Question. (a) Explain DNA polymorphism on the basis of genetic mapping of human genome.
(b) State the role of VNTR in DNA fingerprinting.
Answer. (a) Polymorphism is variation at genetic level which arises due to mutations. The polymorphism in DNA sequences is the basis of genetic mapping of human genome as well as DNA finger printing. If an inheritable mutation is observed in a population at high frequency, it is referred to as DNA polymorphism.
(b) Short nucleotide repeats in the DNA are very specific in each individual and vary in number from person to person but are inherited. These are the ‘Variable Number Tandem Repeats’ (VNTRs). These are also called “minisatellites”. Each individual inherits these repeats from his/her parents which are used as genetic markers in a personal identity test. For example, a child might inherit a chromosome with six tandem repeats from the mother and the same tandem repeated four time in the homologous chromosome inherited from the father. One half of VNTR alleles of the child resemble that of the mother and other half with that of the father.

Question. The following is the flow chart highlighting the steps in DNA fingerprinting technique. Identify a, b, c, d, e and f.
Isolation of DNA from blood cells
         ↓
Cutting of DNA by ‘a’
         ↓
Separation of DNA fragments by
electrophoresis using ‘b’
         ↓
Transfer (blotting) of fragments to ‘c’ gel
         ↓
DNA splits into single strand
         ↓
Introduction of labelled ‘d’ probe
         ↓
‘e’ of single strands with ‘d’
         ↓
Detection of banding pattern by ‘f ’
Answer. a → Restriction endonuclease b → Agarose gel
c → Nitro cellulose membrane d → VNTR
e → Hybridisation
f → Autoradiography

Question. What are satellite DNA in a genome? Explain their role in DNA fingerprinting. 
Answer. A small stretch of DNA sequences repeats many a time, shows a high degree of polymorphism are called as satellite DNA. Short nucleotide repeats in DNA are very specific in each individual and vary in number from person to person but are inherited. Each individual inherits these repeats from his/her parents which are used as genetic markers in DNA fingerprinting.

Question. (a) Differentiate between repetitive and satellite DNA.
(b) How can satellite DNA be isolated? Explain.
(c) List two forensic applications of DNA fingerprinting. 
Answer. a) Difference between repetitive and satellite
DNA are as follows :

Repetitive DNA

Satellite DNA

Repetitive DNA consist of short identical sequences which are repeated several hundred or thousand times. It is of three types. Terminal repeats, tandem repeats and interspersed repeats. e disposition of repeatitive element consist either in arrays of tandemly repeated sequences or in repeats dispersed throughout the genome.

The proportion of the DNA of a eukaryotic cell that consists of very large numbers of copies of a tandem repeatedly short nucleotide sequence. It occurs mainly around the centromeres and telomeres of the chromosomes. e highly repetitive nature of this DNA fraction gives it a distinctive base composition, and consequently when samples of DNA are centrifuged, it forms so-called ‘satellite bands’ quite separate from the band representing the bulk of the cell’s DNA.

 

Question. Following the collision of two trains a large number of passengers are killed. A majority of them are beyond recognition. Authorities want to hand over the dead to their relatives. Name a modern scientific method and write the procedure that would help in the identification of kinship. 
Answer. By using DNA fingerprinting technique, the kinships can be identified.
DNA fingerprinting is a technique of determining nucleotide sequences of certain areas of DNA which are unique to each individual. It identifies a person on the basis of his/her DNA specificity.
The major steps in DNA fingerprinting are : (i) DNA is extracted from the cells.
(ii) DNA is amplified by making many copies of it using polymerase chain reaction.
(iii) Digestion of DNA by restriction endonucleases. (iv) Separation of DNA fragments by electrophoresis. (v) Transferring of separated DNA fragments to synthetic membranes, such as nitrocellulose or nylon.
(vi) Hybridisation using VNTR probe.
(vii) Detection of hybridised DNA fragments by autoradiography.

Question. “A very small sample of tissue or even a drop of blood can help determine paternity”. Provide a scientific explanation to substantiate the statement. 
Answer. DNA fingerprinting helps in determining the paternity from a small sample of tissue or a drop of blood. DNA fingerprinting is a technique for identifying individuals, generally using repeated sequences in the human genome that produce a pattern of bands which is unique for every individual. Important for DNA fingerprinting are short nucleotide repeats that vary in number from person to person, but are inherited. These are the Variable Number of Tandem Repeats or VNTRs. The VNTRs of two persons may be of the same length and sequence at certain sites, but vary at others.
DNA fingerprints can be prepared from extremely minute amounts of blood, semen, hair bulb or any other cells of the body

Question. How would lac operon operate in E. coli growing in a culture medium where lactose is present as source of sugar?
Answer. When lactose is present in the culture medium, then the lac operon in E. coli is switched on. It is because the inducer (lactose) binds to the repressor protein thereby inactivating it. It prevents binding of repressor to the operator. Consequently, RNA polymerase gets access to the promoter and transcription of structural genes proceeds. 

Question. Explain the role of regulatory gene in a lac operon. Why is regulation of lac operon called negative regulation?
Answer. Regulatory gene (i gene) produces a repressor. In the absence of an inducer (i.e. lactose), the repressor binds to the operator gene making it non- functional. RNA polymerase enzyme cannot move over it to reach the structural genes. Thus, structural genes are inactivated and transcription cannot take place.
As regulatory gene exerts a negative control over the working of structural genes, therefore regulation of lac operon is called negative regulation.

Question. A considerable amount of lactose is added to the growth medium of E. coli. How is the lac operon switched on in the bacteria? Mention the state of the operon when lactose is digested?
Answer. Inducer is a chemical (substrate, hormone or some other metabolite) which after coming in contact with the repressor, changes the latter into non-DNA binding state so as to free the operator gene.
The inducer for lac-operon of Escherichia coli is lactose (actually allolactose, or metabolite of lactose). In the presence of inducer (lactose), the repressor gets inactivated due to its interaction with it. This allows RNA polymerase to access the promoter and tranciption proceeds. Hence, the lac operon is switched on.
When lactose is digested, glucose and galactose are formed. Then, the lac operon will stop due to the accumulation of glucose and galactose in the cell as they cannot be used as an inducer for lac operon.

Question. Explain the role of lactose as an inducer in a lac operon.
Answer. An operon is a part of genetic material (or DNA) which acts as a single regulated unit having one or more structural genes, an operator gene, a promoter gene, a regulator gene, a repressor and an inducer or corepressor (from outside).
Lactose acts as an inducer in lac operon. The repressor molecule coded by i gene is inactivated by interaction with the lactose. This allows RNA polymerase to access the promoter and transcription proceeds. The operon gets switched ‘off ’ in the absence of lactose as the repressor molecule binds with the operator region of the operon and prevents RNA polymerase from transcribing the operon. 

Question. Write the scientific importance of single nucleotide polymorphism identified in human genome. 
Answer. Single nucleotide polymorphism (SNPs or snips) help in finding chromosomal locations for disease associated sequences and tracing human history.

Question. Mention the contribution of genetic maps in human genome project. 
Answer. Genetic maps have helped in gene sequencing, DNA fingerprinting, tracing human history, etc.

Question. Mention any two ways in which SingleNucleotide Polymorphism (SNPs) identified in human genome can bring revolutionary change in biological and medical sciences. 
Answer. (a) In finding chromosomal locations for disease associated sequences
(b) In tracing human history

Question. Which human chromosome has (a) maximum number of genes, and which one has (b) fewest genes?
Answer. (a) Chromosome 1 (b) Chromosome Y 

Question. Expand ‘BAC’ and ‘YAC’. Explain how they are used in sequencing human genome.
Answer. BAC = Bacterial Artificial Chromosome. YAC = Yeast Artificial Chromosome.
BAC and YAC are the vectors into which DNA fragments are inserted to form rDNA (recombinant DNA) using recombinant DNA technology and are then multiplied in suitable host. 

 

Very Short Answer

Question. Who developed the technique DNA fingerprinting?
Answer. DNA fingerprinting was developed by Alec Jeffreys.

Question. Define genetic codes?
Answer. The genetic code is the set of rules by which information encoded in genetic material is translated into proteins by living cells. .

Question. What are exons?
Answer. The coding sequences or expressed sequences are known as exons.

Question. What are the structural genes in a transcription unit?
Answer. The promoter and terminator flank is the structural gene in a transcription unit.

Question. What are the types of RNA?
Answer. There are three types of RNA: mRNA, tRNA, and rRNA.

 

Short Answer

Question. What do you understand by the term bioinformatics?
Answer. Bioinformatics is a science field that is similar to but distinct from biological computation, while it is often considered synonymous to computational biology. Bioinformatics and computational biology involve the analysis of biological data, particularly DNA, RNA, and protein sequences.

Question. What are the some important goals of the human genome project?
Answer. Some important goals of HGP are:
1. Identify all the approximately 20,000 to 25,000genes in human DNA.
2. Store the information in databases.
3. Improve tools for data analysis.
4. Transfer related technologies to other sectors.
5. Determine the sequences of 3 billion chemical base plants that make up human DNA.

Question. Distinguish between coding strand and template strand?
Answer. 
1. Template strand contains the same nucleotide sequence as the tRNA whereas coding strand contains the complementary nucleotide sequence as the tRNA.
2. The template strand runs in a 3' to 5' direction whereas the coding strand runs in a 5' to 3' direction

Question. What do you mean by regulation of gene expression?
Answer. Regulation of gene expression, or gene regulation, includes a wide range of mechanisms that are used by cells to increase or decrease the production of specific gene products. Gene expression is regulated during transcription and RNA processing, which take place in the nucleus, and during protein translation, which takes place in the cytoplasm.

Question. What do you meant by Lac operon?
Answer. The lac operon is an operon, or group of genes with a single promoter. The genes in the operon encode proteins that allow the bacteria to use lactose as an energy source.The operon contains genes coding for proteins in charge of transporting lactose into the cytosol and digesting it into glucose. This glucose is then used to make energy.

 

Long Answer

Question. Explain the salient features of the genetic code?
Answer. 
The salient features of genetic code are:
1. The codon is triplet. 61 codons code for amino acids and 3 codons do not code for any amino acids, hence they function as stop codons.
2. Some amino acids are coded by more than one codon hence the code is degenerate.
3. The codon is read in mRNA in a contiguous fashion. There are no punctuations.
4. The code is nearly universal, for example: from bacteria to human UUU would code for Phenylalanine (phe). Some exceptions to this rule have been found in mitochondrial codons and in some protozoans.
5. AUG has dual functions. It codes for Methionine (met), and it also act as initiator codon.
6. UAA, UAG, UGA are stop terminator codons.

Question. Write short note on DNA fingerprints?
Answer. 
Digital Fingerprinting Technology enables the content owner to exercise control on their copyrighted content by effectively identifying, tracking, monitoring and monetising it across distribution channels (web, broadcast, radio, streaming, etc.) by converting their content into a compact digital asset or impression by a known fingerprint algorithm. In a fingerprinting algorithm, a large data item (audio or video or any files) maps to a much shorter bit string, i.e. fingerprint of the file which uniquely identifies the original data for all practical purposes. Digital Fingerprints have contents’ characteristics and enough details to identify a content variant upon comparison. Fingerprints can be used in all sorts of ways: Providing biometric security, for example: to control access to secure areas or systems. Identifying amnesia victims and unknown deceased such as victims of major disasters, if their fingerprints are on. Fingerprints have been developed on porous surfaces (papers, etc.) forty years and later after their deposition. On non-porous surfaces, they can also last a very long time. The nature of the matrix of the latent print will often determine whether it will survive environmental conditions

Question. Explain the salient features of the human genomeuman human genome?
Answer. 
The salient features of human genome are:
1. The human genome contains 3164.7 million nucleotide bases.
2. The average gene consists of 3000 bases but sizes vary greatly with the largest known human gene being dystrophin at 2.4 million bases.
3. The total number of genes is estimated at 30,000 much lower than the previous estimates of 80,000 to 1, 40,000 genes. Almost all nucleotide bases are exactly the same in all people.
4. The functions are unknown for over 50 percent of the discovered genes.
5. Less than 2 % of the genome codes for proteins.
6. Repeated sequences make up very large portion of the human genome.
7. Chromosome 1 has most genes and the Y has the fewest.
8. Repetitive sequences are stretches of DNA sequence that are repeated many times, sometimes the hundred to the thousand times.

Question. State the difference between transcription and translation?
Answer. 
1. In transcription synthesis of RNA copies of the genetic instructions written in the genome is the main purpose whereas in translation its main purpose is the synthesis of proteins from RNA which are copied from genes.
2. In transcription template is the genes in the genome whereas in translation template is the mRNA.
3. Transcription occurs in the nucleus whereas translation occurs in the cytoplasm.
4. In transcription RNA polymerase are the enzymes whereas in translation ribosomes are enzymes.
5. In transcription transcript is released, the enzyme detaches and DNA rewinds whereas in translation ribosome dissembles by encountering into one of the three stop codons, and polypeptide chain is detached.

Question. State the difference between mutation and polymorphism?
Answer. 
1. A mutation is a physical event whereas polymorphism is a population attribute.
2. A single base pair change in the nucleotide sequence of a gene is called a point mutation whereas a single base pair change in the nucleotide sequence is called a single nucleotide polymorphism.
3. A permanent alteration of a nucleotide sequence of a gene is referred to as a mutation whereas the presence of more than one allele at a particular locus in a particular population is referred to as polymorphism.
4. Natural selection selects the mutations that are best suited for the environment whereas natural selection does not affect alleles that bring polymorphism.
5. Sickle cell anemia, hemophilia, cystic fibrosis, Klinefelter syndrome, and Turner syndrome occur due to mutations whereas ABO blood group and the gender of humans are the examples of polymorphism.

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