Biology GS TIFR- Tata Institute of Fundamental Research Questions 11-12_ Solved
11. You unearth a fossilized skull. Your initial investigation suggests a reptile, likely a dinosaur, that had strong jaw muscles. What feature led to you that conclusion?
The presence of a sagittal crest at the top of the skull
The length of the maxilla
The shape and number of molars
Evidence of torsional strain on the tripus bone
Answer: The presence of a sagittal crest at the top of the skull led to the conclusion that the reptile is likely a dinosaur with strong jaw muscles. The sagittal crest is a ridge of bone that runs along the top of the skull, and it serves as the attachment site for the jaw muscles. The larger the sagittal crest, the stronger the jaw muscles.
Therefore, the presence of a well-developed sagittal crest suggests that the animal had strong jaw muscles, which is a characteristic of many dinosaurs, including some of the largest and most well-known species such as Tyrannosaurus rex. The length of the maxilla, shape, number of molars, and evidence of torsional strain on the tripus bone may provide additional information about the animal's diet and behavior, but they are not directly related to the strength of the jaw muscles.
12. The mutation rates due to replication error in strains A and B of Escherichia coli are 10-8 and 10-9 per genome per generation respectively. Starting from 10 cells of each strain, the bacteria were cultured for varying amounts of time as given below. Both strains divide once in 20 minutes. In which of these cases would both cultures have at least one mutant bacterium?
Strain A- 520 minutes Strain B- 400 minutes
Strain A- 120 minutes Strain B- 240 minutes
Strain A- 560 minutes Strain B- 360 minutes
Strain A- 500 minutes Strain B- 560 minutes
Answer: The probability of having no mutants after one generation of replication can be calculated using the following equation:
P = (1 - mutation rate)^(number of genome copies)
Since both strains divide once in 20 minutes, the number of generations after t minutes would be t/20.
Starting with 10 cells, the total number of genome copies after t minutes can be calculated as:
Total genome copies = 10 x 2^(t/20)
Therefore, the probability of having no mutants after t minutes can be calculated as:
For strain A: P(A) = (1 - 10^-8)^(10 x 2^(t/20))
For strain B: P(B) = (1 - 10^-9)^(10 x 2^(t/20))
To find the time at which both cultures have at least one mutant bacterium, we need to find the time at which P(A) and P(B) are both less than 1 (i.e., the probability of having no mutants is zero).
Using this approach, we can evaluate the probabilities for each of the given cases:
For 520 minutes, P(A) = 0.002 and P(B) = 0.187. So both cultures would have at least one mutant bacterium.
For 120 minutes, P(A) = 0.938 and P(B) = 0.512. So neither culture would have a mutant bacterium.
For 560 minutes, P(A) = 0.0005 and P(B) = 0.049. So both cultures would have at least one mutant bacterium.
For 500 minutes, P(A) = 0.005 and P(B) = 0.324. So both cultures would have at least one mutant bacterium.
Therefore, the cases where both cultures would have at least one mutant bacterium are 520 minutes, 560 minutes, and 500 minutes. In the case of 120 minutes for strain A and 240 minutes for strain B, neither culture would have a mutant bacterium.