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Biotechnology and its Applications - Molecular Diagnosis

Grade 12CBSEBiology

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Conventional methods of diagnosis, such as serum and urine analysis, generally do not allow for early detection of pathogens because the concentration of the pathogen must be high enough to be measured.

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Recombinant DNA technology, Polymerase Chain Reaction (PCR), and Enzyme-Linked Immuno-Sorbent Assay (ELISA) are some of the techniques that serve the purpose of early diagnosis.

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PCR (Polymerase Chain Reaction) is used to detect pathogens like bacteria or viruses even when their concentration is very low by amplifying their nucleic acids. It is used to detect HIV in suspected AIDS patients and mutations in genes in suspected cancer patients.

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The total number of DNA copies produced after nn cycles of PCR starting from a single template is given by 2n2^n.

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In Molecular Diagnosis, a single-stranded DNA or RNA, tagged with a radioactive molecule called a probe, is allowed to hybridize to its complementary DNA in a clone of cells followed by detection using autoradiography.

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During autoradiography, the clone having the mutated gene will not appear on the photographic film because the probe will not have complementarity with the mutated gene.

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ELISA is based on the principle of antigen-antibody interaction. Infection by a pathogen can be detected by the presence of antigens (proteins, glycoproteins, etc.) or by detecting the antibodies synthesized against the pathogen.

📐Formulae

Nn=N0×2nN_n = N_0 \times 2^n

Tm=2(A+T)+4(G+C) ∘CT_m = 2(A + T) + 4(G + C) \text{ } ^\circ\text{C}

Sensitivity=True PositivesTrue Positives+False Negatives×100\text{Sensitivity} = \frac{\text{True Positives}}{\text{True Positives} + \text{False Negatives}} \times 100

💡Examples

Problem 1:

If a scientist starts with a single molecule of double-stranded DNA (N0=1N_0 = 1) in a PCR reaction, how many DNA molecules will be present after n=20n = 20 cycles?

Solution:

Using the formula Nn=N0×2nN_n = N_0 \times 2^n, we substitute the values: N20=1×220N_{20} = 1 \times 2^{20} N20=1,048,576N_{20} = 1,048,576

Explanation:

PCR results in the exponential amplification of the target DNA sequence. Each cycle doubles the amount of DNA present in the previous cycle.

Problem 2:

A probe with the sequence 5′−ATGC−3′5'-ATGC-3' is used to detect a gene. What is the complementary sequence it will bind to on the target DNA, and why would a mutation in the target sequence 3′−TACG−5′3'-TACG-5' prevent detection?

Solution:

The complementary sequence is 3′−TACG−5′3'-TACG-5'. If the target DNA undergoes a mutation (e.g., 3′−TTCG−5′3'-TTCG-5'), the hydrogen bonding between the probe and the DNA is disrupted.

Explanation:

Molecular diagnosis using probes relies on the principle of base pairing (AA with TT and GG with CC). Any mutation changes the sequence, leading to a lack of complementarity, which prevents the radioactive probe from binding. Consequently, the mutated gene does not show up on the autoradiograph.

Problem 3:

Calculate the approximate melting temperature (TmT_m) for a short oligonucleotide probe with the sequence 5′−AAGCTTCG−3′5'-AAGCTTCG-3'.

Solution:

Count the bases: A=2A=2, T=2T=2, G=2G=2, C=2C=2. Using the formula Tm=2(A+T)+4(G+C)T_m = 2(A + T) + 4(G + C) Tm=2(2+2)+4(2+2)T_m = 2(2 + 2) + 4(2 + 2) Tm=2(4)+4(4)T_m = 2(4) + 4(4) Tm=8+16=24∘CT_m = 8 + 16 = 24 ^\circ\text{C}

Explanation:

The melting temperature determines the stability of the probe-target hybrid. G−CG-C pairs have three hydrogen bonds and contribute more to stability than A−TA-T pairs (two hydrogen bonds).