Calculations

Question 1

BIO1: What is osmosis? Be able to calculate osmotic pressure.

Answer 1

PI=c * R * T

Question 2

BIO1: What is the role of ATP? How much of it are we producing? Be able to perform simple stoichiometric calculations of energy conversion.

Question 3

BIO1: What is the Reynolds number? What is laminar flow? How does it scale with the pressure, length and diameter of the vessel?

Question 4

BIO2: You need to be able to understand the mathematics of the genetic code and perform simple calculations of alternative genetic codes, e.g. using doublets instead of triplets, or more or less nucleotides.

Question 5

BIO2: You need to be able to compare combinatorial diversity of DNA/RNA and protein sequences of different lengths.

Question 6

BIO2: What is an (open) reading frame? You need to be able to determine/estimate how many reading frames there are in a given piece of dsDNA or in a piece of DNA of given length (see exercises).

Question 7

BIO2: You need to be able to compute the quantity of information stored in a piece of DNA (see exercises).

Question 8

BIO3: Be able to interpret simple enzymatic energy landscapes

Answer 8

For the reaction S–> P

With enzyme E:
E+S–>ES—>EP—->E+P
(–>, dobbelt arrows since enxymes catalyze the reaction in both directions.)

The rate constants k_2 and k_-2:
k_2> k_-2 since the distance from ES to intermediate is shorter than EP to the intermediate

The rate limiting step without: S to P
with: ES to EP

Question 9

BIO3: Be able to perform simple stoichiometric calculations of antibody binding to target.

Answer 9

When someone is bitten or stung by a venomous animal, the venom contains toxins that can cause various harmful effects, including tissue damage, blood clotting disorders, paralysis, or even death. Antivenom works by binding to and neutralizing these toxins, thereby preventing or reducing the severity of the envenomation.

So, while antivenom is not produced by the human body like endogenous antibodies, it functions similarly by targeting specific toxins in venom, making it an important therapeutic tool in the treatment of envenomation.

Calculations:
Up to 290 mg of toxin per bite and LD50 (median lethal dose at this dose 50% of test animals/subjects die) of 0.389 mg/kg.

How much weight would you have to gain if you weighted 80 kg, in order to increase your change of survival to above 50%?
290 mg/0.389 mg/kg=745.5 kg and then 745.5 kg-80 kg=665 kg.

How many people weighing 80 kg could this bite theoretically kill with a
probability of 50%, and how many mice (25 g)?
745.5 kg/80 kg=9.32= 9 people could be killed:

745.5 kg/0.025 kg=29.820 mice

Question 10

BIO3: Be able to perform simple calculations of amino acid sequence diversity of peptides.

Answer 10

An amino acid sequence that consists of 5 peptides:
Peptide 1: ACGT
Peptide 2: AGCT
Peptide 3: ACGT
Peptide 4: ACGT
Peptide 5: AGCT

To calculate the amino acid sequence diversity of peptides, first the total number of peptides in the set is counted: 5
Then the amino acid sequence diversity is calculated:
Diversity=Number of unique peptides/total number of peptides=2/5=0.4

The amino acid sequence diversity for the given set of peptides is 0.4

The binding region on the antibody contains 25 amino acids.

How big would your phage library have to be if you want it to contain all possible single point mutations in this region?
First the number of single point mutations for each position in the binding region should be calculated. Since each position can be mutated to any of the 19 other amino acids (excluding it self), you multiply the number of positions (25) by the number of possible mutations at each position (19), resulting in 475 possible single point mutations.

What about if you want all possible double mutants?
There are then two positions. There are 19 possible aminoacids when excluding the other. Therefore the total number of combinations is 19x19. But in this number double counting is made. therefore divided by 2: (19x19)/2. Then:

191924*25/2=108300

Question 11

BIO4: Be able to perform simple calculation on numbers of ancestors as a function of time.

Question 12

BIO6: How is bacterial growth rate calculated

Question 13

BIO6: Hvornår vil bacteria C med en generation time på 20 min nå 10^9 celler/mL?

Answer 13

Antal generationer beregnes ved:
Antalcellerefterxgenerationer = 2^x;

hvor x er antallet af generationer

Det beregnes:
solve(10^9. = 2^x, x);
29.89735285

30 generationer for at nå fra 1 til 10^9 celler. 1 generation tager 20 min:
30*20min=10 timer

Question 14

BIO6: Beregn specifikke growth rate mu når generation time kendes

Answer 14

mu=(ln(2)-ln(1))/generation time=h^-1

Question 15

BIO6: Ved hvilke densiteter skal A ramme 10^9 celler/mL på samme tid som B?

Answer 15

FORMEL:
N(t)=N0exp(mu(t-t0))