RowQ
The Vault
RowQ
The Vault
CBSE Class 12 Biology · 11 questions · 26 marks
Here the abstract 'factor' of Mendel finally acquires a chemical identity: a long double-helical molecule of DNA. This chapter walks through the experiments that proved DNA is the genetic material, the mechanics of replication, transcription, and translation, and the elegant switch of the lac operon that lets a bacterium turn a gene on only when it is needed. It closes with the Human Genome Project and DNA fingerprinting, where all this molecular detail becomes practical technology.
A sample of double-stranded DNA is found to contain 22 percent adenine. What percentage of guanine does it contain?
Answer
28 percent is correct — by Chargaff's rule, adenine equals thymine, so T is also 22 percent and A + T together account for 44 percent. The remaining 56 percent is shared equally between guanine and cytosine, giving 28 percent guanine. The value 22 percent wrongly assumes G equals A, and 56 percent is the combined G + C figure.
The Meselson and Stahl experiment using heavy nitrogen ¹⁵N in Escherichia coli established that DNA replication is:
Answer
Semi-conservative is correct — after one generation in ¹⁴N medium every DNA molecule was of intermediate density, and after two generations the DNA was half intermediate and half light. This result is only possible if each new molecule keeps one parental strand and gains one newly made strand, ruling out both the conservative and the dispersive models.
Which statement about the genetic code is correct?
Answer
The code is degenerate, since one amino acid may be specified by more than one codon, is correct — for example, leucine has six codons. The code is unambiguous in the other direction: a given codon always specifies one and only one amino acid, so the first option is wrong. The code is non-overlapping and comma-less, and AUG is the initiator codon that also codes for methionine, not a stop signal.
In the lac operon of Escherichia coli, lactose functions as:
Answer
An inducer is correct — in the absence of lactose the repressor protein made by the regulatory i gene binds the operator and blocks transcription. When lactose is available, a small amount enters the cell and binds the repressor, changing its shape so that it can no longer sit on the operator, and RNA polymerase is then free to transcribe the z, y, and a genes. The repressor is a protein, while the operator and promoter are DNA sequences.
Assertion (A): During DNA replication one new strand is made continuously while the other is made as a series of short fragments. Reason (R): DNA polymerase can add nucleotides only in the 5' → 3' direction, and the two template strands are antiparallel.
Answer
Both A and R are true and R is the correct explanation of A — because the two template strands run in opposite directions, only one of them presents a template on which continuous 5' → 3' synthesis can follow the advancing replication fork. On the other template, synthesis must repeatedly start afresh behind the fork, producing short Okazaki fragments that DNA ligase later joins into a continuous strand.
Why is DNA a more suitable genetic material than RNA, even though RNA can also carry genetic information?
Answer
DNA is chemically far more stable than RNA. The 2' position of its sugar carries a hydrogen instead of a hydroxyl group, so it is not vulnerable to the alkaline hydrolysis that readily degrades RNA, and thymine in place of uracil gives it additional stability. Being double-stranded, it also has a complementary strand that acts as a template for accurate replication and allows damage on one strand to be repaired using the other. RNA, being single-stranded, is reactive, catalytically active, and mutates at a much faster rate, which suits it to short-lived roles in gene expression but makes it a poor store of information that must be transmitted unchanged across generations.
Explain the three main steps by which a eukaryotic primary transcript is processed before it can be translated.
Answer
Capping: a modified guanine nucleotide, methyl guanosine triphosphate, is added to the 5' end of the transcript. This cap protects the RNA from degradation by enzymes and is recognised by the ribosome during initiation of translation. Tailing: about 200 to 300 adenylate residues are added to the 3' end in a template-independent manner, forming the poly-A tail. This also protects the transcript and assists its export from the nucleus. Splicing: the non-coding intron sequences are precisely removed and the coding exons are joined together in the same order. Only after these three steps does the transcript become mature mRNA that can leave the nucleus and be translated in the cytoplasm.
What is DNA fingerprinting based on, and give one situation in which it is applied?
Answer
DNA fingerprinting is based on the fact that a large part of the genome consists of repetitive, non-coding satellite DNA containing short sequences repeated many times over — variable number tandem repeats. The number of repeats at a given site differs greatly between individuals but is inherited from the parents, so the overall pattern of repeat lengths is essentially unique to a person while being shared partly with each parent. It is applied in paternity disputes, where the child's bands must all be traceable to one or other parent, and in forensic investigation, where DNA from a small biological sample at a crime scene can be matched to a suspect.
Describe the process of translation in a bacterial cell, from the binding of mRNA to the release of the completed polypeptide.
Answer
Activation of amino acids: before translation begins, each amino acid is joined to its specific tRNA by an aminoacyl-tRNA synthetase in an ATP-requiring reaction. The charged tRNA carries an anticodon triplet at one end that will read the mRNA codon, and the amino acid at the other end — this is why tRNA is called the adaptor molecule. Initiation: the mRNA binds to the smaller subunit of the ribosome, which scans to the AUG initiator codon. The initiator tRNA carrying methionine base-pairs with this codon, and the larger subunit then joins, creating a functional ribosome with the initiator tRNA in the P site. The untranslated regions on either side of the coding sequence are required for this step but are not translated. Elongation: a charged tRNA whose anticodon matches the next codon enters the A site. The ribosome, acting as a ribozyme through its 23S rRNA, catalyses formation of a peptide bond between the amino acid in the P site and the one in the A site. The ribosome then translocates one codon along the mRNA in the 5' → 3' direction, the now-empty tRNA leaves, and the next charged tRNA enters the vacated A site. This cycle repeats codon by codon, lengthening the polypeptide. Termination: when a stop codon — UAA, UAG, or UGA — enters the A site, no tRNA can read it. A release factor binds instead, the completed polypeptide is freed from the last tRNA, and the ribosomal subunits, mRNA, and tRNA dissociate. The polypeptide then folds into its functional shape. In bacteria several ribosomes may translate the same mRNA simultaneously as a polysome, and because there is no nucleus, translation can begin while transcription is still in progress.
Describe the experiments of Griffith and of Hershey and Chase, and explain how together they identified the genetic material.
Answer
Griffith's experiment: working with Streptococcus pneumoniae, Griffith used two strains — the smooth S strain, which has a mucous coat and is virulent, and the rough R strain, which lacks the coat and is non-virulent. Mice injected with live S bacteria died; mice injected with live R bacteria survived; mice injected with heat-killed S bacteria survived. However, when heat-killed S bacteria were mixed with live R bacteria and injected, the mice died, and living S bacteria were recovered from them. Griffith concluded that some substance from the dead S cells had transformed the harmless R cells into the virulent form, and he called it the transforming principle. He could not, however, say what that substance chemically was. Avery, MacLeod, and McCarty followed this up by purifying the cell-free extract and treating it separately with protein-digesting, RNA-digesting, and DNA-digesting enzymes. Transformation still occurred when proteases and RNase were used, but was abolished by DNase, pointing strongly to DNA. Hershey and Chase's experiment: they used bacteriophages, which consist only of a DNA core and a protein coat. One batch of phages was grown in medium containing radioactive ³²P, which labelled only the DNA since phosphorus is present in DNA but not in these proteins. Another batch was grown with radioactive ³⁵S, which labelled only the protein coat since sulphur occurs in certain amino acids but not in DNA. Each batch was allowed to infect Escherichia coli, and the cultures were then agitated in a blender to shake off the empty phage coats and centrifuged. The bacteria infected by ³²P-labelled phages were radioactive, showing that DNA had entered the cell; the bacteria infected by ³⁵S-labelled phages were not radioactive, and the radioactivity remained in the discarded coats outside. Since only the material that entered the bacterium could direct the production of new phage particles, DNA and not protein must be the genetic material. Griffith's work had established that a chemical could carry heredity between cells, and the Hershey-Chase experiment settled unambiguously which chemical it was.
A forensic laboratory receives a bloodstain recovered from a burgled house along with reference samples from three suspects. DNA is extracted from each, cut with a restriction enzyme, separated by gel electrophoresis, transferred to a membrane, and probed with a labelled repetitive sequence. The band pattern from the bloodstain matches suspect 2 at every position, while suspects 1 and 3 differ at several positions. In a separate case at the same laboratory, a disputed-paternity sample from a child shows some bands matching the mother and the remaining bands matching the alleged father. (a) Name the technique being used and the type of DNA sequence the probe detects. (b) State the purpose of the electrophoresis step. (c) What conclusion can be drawn about suspect 2, and what does the mismatch of suspects 1 and 3 show? (d) Explain why, in the paternity case, some of the child's bands match the mother and the rest match the father.
Answer
(a) The technique is DNA fingerprinting. The probe detects repetitive, non-coding satellite DNA — specifically variable number tandem repeats, short sequences repeated a highly variable number of times at particular sites in the genome. (b) Electrophoresis separates the restriction fragments according to size. Because the fragments carry a negative charge from their phosphate groups, they migrate through the gel towards the positive electrode, with smaller fragments travelling further, so a person's fragments are resolved into a characteristic ladder of bands that can be compared with another sample. (c) The complete match at every position indicates that the bloodstain almost certainly came from suspect 2, since the probability of two unrelated people sharing the number of repeats at all these highly variable sites is extremely small. Suspects 1 and 3 differ at several positions, which excludes them definitively — a single genuine mismatch is enough to rule a person out. (d) A child inherits one chromosome of each pair from the mother and the other from the father, so every repeat site in the child has one maternal and one paternal copy. The bands that match the mother correspond to the copies inherited from her, and the remaining bands must have come from the biological father. Any band in the child that could not have come from either claimed parent would disprove the relationship.
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