About Biotechnology
Biotechnology questions in a recruitment paper cover the whole breadth of the subject - the structure of nucleic acids, the central dogma, the polymerase chain reaction, restriction enzymes and vectors, cloning and transformation, gel electrophoresis and sequencing, blotting, expression systems, biosafety and containment, and the industrial uses of the techniques - and most of them turn on a number that has to be worked out. This topic covers that breadth on one page, with the arithmetic that goes with it.
What you need to understand
- DNA is two antiparallel strands held together by hydrogen bonds between complementary bases: A pairs with T by two bonds and G pairs with C by three. A high G plus C content therefore means a more stable helix and a higher melting temperature.
- The two strands run in opposite directions, which is why one is written 5 prime to 3 prime and the other 3 prime to 5 prime. Every enzyme that copies nucleic acid works only in the 5 prime to 3 prime direction.
- Each base pair takes up about 0.34 nanometres of length, with ten pairs to a turn of the helix. Three thousand million base pairs, which is the human genome, therefore works out at about a metre of DNA in every cell.
- The central dogma runs DNA, then RNA, then protein. Replication copies DNA, transcription makes RNA from it, and translation makes protein from the RNA.
- The polymerase chain reaction doubles the target sequence at every cycle, so the number of copies is the starting number times two to the power of the cycles. Its three steps are denaturation, annealing of the primers and extension.
- A primer is a short single strand that gives the polymerase a place to start. The melting temperature can be estimated by the Wallace rule: two degrees for each A or T plus four for each G or C.
- A restriction enzyme cuts DNA at a specific recognition sequence, usually a palindrome. A linear molecule with n sites gives n + 1 fragments, because the two ends also count; a circular molecule gives exactly n.
- A vector is the carrier that takes a piece of foreign DNA into a cell. It needs an origin of replication, a selectable marker such as an antibiotic resistance gene, and a site where foreign DNA can be inserted.
- Transformation is the uptake of free DNA by a competent cell. Its efficiency is quoted as the number of transformants for each microgram of plasmid, and it depends on the plasmid as well as on the cells.
- Gel electrophoresis separates nucleic acids by size, because a charged molecule in an electric field moves faster the smaller it is. The distance travelled is roughly proportional to the logarithm of the length.
- A DNA solution is measured by its absorbance at 260 nanometres: an absorbance of one in a one centimetre cell is 50 micrograms per millilitre for double stranded DNA. The ratio of the readings at 260 and 280 nanometres shows how clean the sample is.
- Blotting transfers separated molecules to a membrane so a probe can find them: Southern for DNA, Northern for RNA and Western for protein.
- Sanger sequencing reads a sequence by interrupting the copying reaction with chain terminating nucleotides, so the fragments end at every possible position and the order of the lengths gives the sequence.
- Gel electrophoresis, blotting and sequencing all depend on the same idea: separate by one property, then identify by a specific probe so that the position means something.
- Industrial biotechnology runs the same biology at scale in a fermenter, and it is judged on titre, rate and yield - how much product, how fast, and how much of the substrate went into it.
- Containment is graded by risk: a disabled strain that cannot survive outside a fermenter can be handled in a lower containment level than one carrying a virulence gene.
How to work through these questions
- Decide which relation the question is about first: a doubling, a base composition, a melting temperature, a length, a fragment count or a concentration from an absorbance.
- For anything that doubles, write the power of two out and then multiply. Multiplying by the number of cycles instead of by two to that power is the commonest error in the amplification questions.
- For a base composition, count the letters in the sequence that is printed. A question gives the sequence so that it can be counted, and guessing from the look of it is where errors come from.
- For a fragment count, identify whether the molecule is linear or circular before counting. The same number of sites gives a different answer for the two, and that is usually the point of the question.
- For a concentration from an absorbance, remember both the factor for the molecule and the dilution the reading was taken at. The dilution is the part that is left out.
- Check the answer against the technique: a fragment count of zero, a percentage above a hundred, or a melting temperature below the annealing temperature all mean something has been inverted.
Mistakes that cost marks
- Multiplying the starting number of molecules by the number of cycles rather than by two to that power, which turns exponential amplification into a straight line.
- Counting the sites of a restriction enzyme and forgetting that a linear molecule has two ends as well, so the fragment count is one too low.
- Giving a circular molecule the linear answer, or the other way round, when the question has said which one it is.
- Leaving the dilution out of a concentration worked out from an absorbance, which understates the sample by the whole dilution factor.
- Using five degrees for a G plus C pair instead of four, or two for an A plus T pair instead of four, in the melting temperature rule.
- Reading a 260 to 280 ratio the wrong way round: a pure DNA sample is close to 1.8 and a pure RNA sample close to 2.0, and protein contamination lowers the figure.
- Confusing the distance moved on a gel with the size of the fragment. The small fragments travel furthest, so the relationship is the inverse of what it looks like at first.
- Treating 0.34 nanometres per base pair as 0.34 micrometres, which makes every length a thousand times too large.
Worked example
A polymerase chain reaction starts with 100 copies of the target and is run for 20 cycles. How many copies are there at the end? A linear DNA molecule of 4000 base pairs has 3 sites for a restriction enzyme: how many fragments does it give?
- For the amplification, every cycle doubles what is there, so the count after n cycles is the starting number times two to the power of n.
- Two to the twentieth is 1 048 576, so the copies are 100 x 1 048 576 = 104 857 600.
- Check the size against the method: twenty cycles from a hundred copies giving a hundred million is the whole point of the technique, and it is why a single molecule can be detected. Adding 100 per cycle would give only 2100, which is out by five orders of magnitude.
- For the fragments, the molecule is linear, so its two ends count as boundaries as well as the three sites.
- The number of fragments is therefore 3 + 1 = 4. The same molecule in circular form would give exactly 3, and that difference is what the question is testing.
Answer: 104 857 600 copies; 4 fragments from the linear molecule
Practice questions with answers
A few Biotechnology questions with the full solution shown, so you can see
how the method is applied before you attempt the timed set.
Question 1
A polymerase chain reaction starts with 10 copies of the target sequence and is run for 16 cycles. How many copies are there at the end?
-
A
1310720
-
B
160
-
C
327680
-
D
655360
Answer: Option D — with explanation
Every cycle doubles the number of copies of the target, because each new strand becomes a template for the next cycle. The number after n cycles is the starting number times two to the power of n.
That is 10 x 2 to the 16, which is 10 x 65536 = 655360 copies.
The doubling is exponential, which is why twenty cycles gives over a million copies from a single starting molecule, and why the reaction is normally followed by a detection step rather than run for many more cycles - the reagents run out and the reaction plateaus.
Common mistakes
- 1310720 is not the answer: 1310720
- 160 is not the answer: 160
- 327680 is not the answer: 327680
Question 2
The sequence of a DNA fragment is ATTTTTATCATAGATCGAAAAGATC. What is its guanine and cytosine content, as a percentage?
Answer: Option B — with explanation
The guanine and cytosine content is the number of G and C bases divided by the number of bases altogether, written as a percentage. It is usually called the GC content and it matters because a G with a C is held by three hydrogen bonds while an A with a T is held by two, so a high GC content means a more stable double helix.
Counting the bases of ATTTTTATCATAGATCGAAAAGATC: there are 6 of them that are G or C out of 25 in all, so the content is 6 / 25 x 100 = 24 per cent.
The remaining bases are A and T, so the AT content is the complement - the two always add to a hundred.
Common mistakes
- 12 is not the answer: 12
- 76 is not the answer: 76
- 6 is not the answer: 6
Question 3
The sequence of a primer is 5'-AGTAAGGTCCAATCGT-3'. Taking 2 degrees Celsius for each A and T and 4 for each G and C, what melting temperature does the primer have, in degrees Celsius?
Answer: Option C — with explanation
The rule used here is the Wallace rule, which counts two degrees for each A or T and four for each G or C. It works because a G-C pair is held by three hydrogen bonds and an A-T pair by two, so the stronger pair contributes about twice as much to the temperature at which the two strands come apart.
The primer has 7 G or C bases and 9 A or T bases, so the melting temperature is 2 x 9 + 4 x 7 = 46 degrees Celsius.
The rule is only a good guide for primers up to about twenty five bases. Beyond that the nearest neighbour interactions between stacked bases matter more than the base composition, and the Wallace rule overestimates the temperature.
Common mistakes
- 32 is not the answer: 32
- 48 is not the answer: 48
- 50 is not the answer: 50
Question 4
A double stranded DNA molecule is 20000 base pairs long. Taking the rise per base pair as 0.34 nanometres, what is the length of the molecule, in micrometres?
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A
6800
-
B
6.80
-
C
13.60
-
D
68
Answer: Option B — with explanation
The two strands of DNA wind round each other so that each base pair takes up a fixed amount of the length - about 0.34 nanometres, which is the rise of the B form of the double helix. Measuring the length of a molecule therefore comes down to multiplying the number of pairs by that rise.
Here that is 20000 x 0.34 = 6800 nanometres, and a micrometre is a thousand nanometres, so the molecule is 6.80 micrometres long.
The human genome is about three thousand million base pairs, which at this rise works out at about a metre of DNA in every cell - a figure worth remembering when the question asks about the length of a molecule.
Common mistakes
- 13.60 is not the answer: 13.60
- 6800 is not the answer: 6800
- 68 is not the answer: 68
Question 5
A restriction enzyme has 6 recognition sites in a CIRCULAR DNA molecule of 5000 base pairs. How many fragments are produced?
Answer: Option A — with explanation
A restriction enzyme cuts at each recognition site, and the number of pieces depends on whether the molecule has ends. A linear molecule with n sites has n + 1 pieces, because the two ends count as boundaries as well as the sites. A circular molecule has no ends, so n sites give exactly n pieces.
This molecule is circular and has 6 sites, so the number of fragments is 6.
The lane on a gel tells you which it was: a circular molecule cut once gives one fragment of the whole length, while a linear molecule cut once gives two.
Common mistakes
- 8 is not the answer: 8
- 7 is not the answer: 7
- 5 is not the answer: 5
Frequently asked questions
Why does a G plus C pair count for more than an A plus T pair?
Because it is held by three hydrogen bonds instead of two, so more energy is needed to pull the strands apart. That is why a primer with a high G plus C content needs a higher annealing temperature, and why such sequences are harder to melt.
Why is the reaction stopped after twenty or thirty cycles?
Because the reagents run out and the enzyme loses activity, so the number of copies stops doubling and the curve flattens. The useful range of the reaction is the exponential part, which is why the cycle count is chosen rather than left open.
Why does a restriction enzyme give a different number of fragments from a circle and a line?
Because a line has two ends, and each end is a boundary of a fragment as well as each cut. A circle has no ends, so the cuts alone bound the fragments and the count is exactly the number of sites.
What does the 260 to 280 ratio tell me?
It shows how much protein is in the sample, because protein absorbs strongly at 280 nanometres and DNA hardly at all. A ratio near 1.8 is clean DNA; a lower figure means protein contamination, and the sample should be purified again before it is used.
Why is the absorbance reading taken on a diluted sample?
Because the relation between absorbance and concentration only holds up to about one absorbance unit. A concentrated sample is diluted until it reads in that range, and the dilution is then multiplied back in - which is the step people forget.
Which travels furthest on a gel, a large fragment or a small one?
The small one. The gel is a mesh and the smaller molecule threads through it more easily, so it travels further in the same time. That is why the largest fragments are found nearest the wells and the smallest nearest the far end.
