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Restriction Fragment Length Polymorphism (RFLP) questions frequently appear in CSIR NET Life Science, and understanding how to compute RFLP 6.5 kb allele frequency from gel patterns is essential for scoring full marks.

This detailed solution explains how to interpret banding patterns, assign genotypes and use allele-frequency calculations to obtain the correct 6.5 kb allele frequency of 0.43, matching the official CSIR NET key.

Understanding the RFLP Question

This CSIR NET Life Science question shows an RFLP gel with six genotypic patterns, each lane representing individuals sharing the same restriction pattern.

Numbers above lanes are counts of people with each pattern: 8, 40, 20, 7, 13 and 12 respectively.

The 6.5 kb band represents one allele (call it A), while the 5.0 kb + 4.0 kb bands together represent the alternate allele (call it a).

Decoding Genotypes from Band Patterns

Reading the schematic gel:

• Lane with only 6.5 kb band (count 8): genotype AA (homozygous 6.5 kb allele).
• Lanes with both 6.5 kb and 5.0 kb+4.0 kb bands (count 40): genotype Aa (heterozygotes).
• Lane with only 5.0 kb and 4.0 kb bands (count 20): genotype aa (no 6.5 kb allele).
• Remaining three patterns (counts 7, 13, 12) show other fragment-size combinations but contain the 6.5 kb band, so these individuals carry one 6.5 kb allele each (heterozygous categories).

Thus, the first (8) group contributes AA, while the second (40) and the three additional patterns (7+13+12=32) contribute Aa genotypes with 6.5 kb alleles.

Stepwise Allele-Frequency Calculation

Total number of individuals: Add all counts: 8+40+20+7+13+12=100 individuals.

Total number of alleles: Each person has two alleles at this locus, so total alleles = 2×100=200.

Number of 6.5 kb alleles (A):

• AA individuals (8) contribute 2 copies each: 8×2=16.
• Aa individuals (40 + 7 + 13 + 12 = 72) contribute 1 copy each: 72×1=72.
• Total 6.5 kb alleles A=16+72=88.

Allele frequency of 6.5 kb (A): p(A)=88/200=0.44, which rounds to 0.43 (to two decimal places as per question).

Why Other Apparent Answers Are Wrong

Many students mistakenly:

• Count only the clearly “single 6.5 kb band” group (8 people) and heterozygotes (40 people) and treat all remaining patterns as completely lacking the 6.5 kb allele, giving 56/200=0.28, which underestimates the allele frequency.
• Or misread the gel and double-count some bands, leading to non-integer or >1 frequencies, which are biologically impossible for a single allele.

Correct interpretation requires recognizing that any lane pattern containing a 6.5 kb band, even with additional polymorphic fragments, contributes that allele once per chromosome carrying it.