Biology GS TIFR- Tata Institute of Fundamental Research Questions 6-10_ Solved
6. Human DNA contains 18% C on a molar basis. What is the mole percentage of A, G, and T respectively?
(i) 32, 18, 32
(ii) 18, 32, 32
(iii) 32, 32, 18
Answer: If human DNA contains 18% C on a molar basis, then the total percentage of nucleotides in the DNA must be 100%, and the remaining nucleotides must be A, G, and T.
Since DNA is double-stranded, the percentage of A must be equal to the percentage of T, and the percentage of G must be equal to the percentage of C.
Therefore, the mole percentage of A and T together must be 100% - 18% - 18% = 64%.
Since A and T occur in equal amounts, the mole percentage of each is 64% / 2 = 32%.
Similarly, the mole percentage of G and C together is 18% + 18% = 36%.
Since G and C occur in equal amounts, the mole percentage of each is 36% / 2 = 18%.
Therefore, the mole percentages of A, G, and T in human DNA are:
7. Which of the following statements explains why net primary productivity is higher near the tropics?
More predators in the tropics kill herbivores, increasing plant growth in the tropics
Increased species diversity results in increased plant growth
Availability of newly weathered inorganic nutrients
Water and warm temperatures in the tropics promote photosynthesis
Answer: The statement that explains why net primary productivity is higher near the tropics is "Water and warm temperatures in the tropics promote photosynthesis."
Net primary productivity (NPP) is the rate at which primary producers (usually plants) convert solar energy into biomass through photosynthesis, minus the rate at which they use some of that biomass through respiration. NPP is an important measure of the amount of energy available to support the rest of the ecosystem, including herbivores and carnivores.
Water and warm temperatures are two key factors that influence photosynthesis, as they are necessary for the chemical reactions that occur during the process. The tropics have high levels of both, with abundant rainfall and high levels of solar radiation throughout the year. These conditions provide optimal growing conditions for plants, which in turn leads to higher NPP.
While increased species diversity may contribute to increased plant growth in some cases, and the availability of newly weathered inorganic nutrients may also play a role in some ecosystems, these factors are not as consistently linked to higher NPP as water and warm temperatures are in the tropics. The presence of more predators in the tropics killing herbivores does not increase plant growth; rather, it may actually decrease it if herbivores are important in maintaining the balance of the ecosystem.
8. The stages of the cell cycle for a cultured mammalian cell line require the following periods of time: G1 = 8 hr, S = 5 hr, G2 = 1 hr, M = 1 hr. An asynchronous culture of these cells is exposed to radioactive thymidine for five minutes and then allowed to continue to grow in a nonradioactive medium. The figure below shows the results.
If the length of G2 were increased by one hour, which of the following statements would be true regarding the slope of the line from the first appearance of labeled mitotic cells until all cells are labeled?
It would increase.
It would decrease.
It would stay the same.
It would switch from a positive to a negative slope
Answer: Based on the given information, the graph represents the results of a pulse-chase experiment, where a population of cells is exposed to radioactive thymidine for 5 minutes and then allowed to continue growing in a non-radioactive medium. During this time, the cells incorporate thymidine into their DNA, and as they divide, the radioactive label is distributed among the daughter cells.
The graph shows that initially, there is a lag phase (latent period) where no cells are labeled, followed by a period of exponential growth in the number of labeled cells, and finally, a plateau where all cells are labeled.
If the length of G2 were increased by one hour, it would delay the onset of mitosis and therefore also delay the appearance of labeled mitotic cells. This delay would result in a shallower slope in the exponential phase of the curve since the same number of labeled cells would be spread out over a longer time.
Therefore, the correct statement regarding the slope of the line from the first appearance of labeled mitotic cells until all cells are labeled would be: The slope of the line would be shallower or less steep.
9. X-linked recessive disorders are most commonly exhibited by - because
females, because they have two X-chromosomes and are therefore more likely to inherit one with a disease
males, because they have only one X-chromosome females,
because one of their X-chromosomes is deactivated both males and females,
because of equal penetrance
Which of the following statement is correct, given this phylogeny of four species?
A has gone extinct
A is more closely related to C than D D evolved from C
B, C and D shared a common ancestor
Answer: X-linked recessive disorders are most commonly exhibited by males because they have only one X-chromosome. If the X-chromosome they inherit carries the disease-causing allele, they will express the disorder since there is no corresponding dominant allele on the Y-chromosome to mask it.
Based on the given phylogeny, the correct statement is: B, C and D shared a common ancestor. The phylogeny shows a common ancestor of B, C, and D that diverged from the lineage that led to species A. It does not provide information on whether A has gone extinct, whether A is more closely related to C than D, or whether D evolved from C.
10. A trihybrid test cross (order of loci unknown) produced progeny classes as follows: 35 AbC/abc, 37 aBc/abc, 8 ABc/abc, 10 abC/abc, 3 ABC/abc, 5 abc/ abc, 1 Abc/abc, 1 aBC/abc. The gene order is
CBA BAC BCA CAB
A newly discovered extra-terrestrial organism was found to contain proteins that are made up of 75 different amino acids. Assuming that one codon codes for a unique amino acid, which of the following is possibly correct about its DNA and codons?
The DNA could be composed of 4 bases and the organism has a triplet codon.
The DNA could be composed of 3 bases and the organism has a triplet codon.
The DNA could be composed of 3 bases and the organism has a quartet codon.
The DNA could be composed of 2 bases and the organism has a quartet codon.
Answer: Based on the progeny classes provided, we can determine the gene order by looking for the progeny class that has the least frequent recombination events. In this case, the progeny class with the least frequent recombination event is abc/abc, which indicates that this progeny class is the parental type. Therefore, we can determine that the order of the loci is BAC, as follows:
35 AbC/abc and 5 abc/abc indicate that A is linked to C
37 aBc/abc and 10 abC/abc indicate that B is linked to C
8 ABc/abc, 3 ABC/abc, 1 Abc/abc, and 1 aBC/abc all indicate double crossovers, which occur between the middle gene (B) and the other two genes (A and C). These double crossovers allow us to determine the relative order of A and C.
The double crossovers can be used to infer the gene order as follows:
8 ABc/abc: this progeny class indicates that there was a crossover between A and B, and another crossover between B and C. Therefore, the order is BAC.
3 ABC/abc: this progeny class indicates that there was a crossover between B and C, and another crossover between A and C. Therefore, the order is BCA.
1 Abc/abc and 1 aBC/abc: these progeny classes indicate that there was a crossover between A and B, and another crossover between B and C. Therefore, the order is CAB.
Since the order of the loci is BAC, we can determine the frequency of recombination events between adjacent loci. The recombination frequency between B and A is (8 + 3 + 1 + 1) / 100 = 0.13, the recombination frequency between B and C is (8 + 3 + 1 + 1) / 100 = 0.13, and the recombination frequency between A and C is (35 + 5 + 1 + 1) / 100 = 0.42.
Therefore, the gene order is BAC and the recombination frequencies between adjacent loci are 0.13 and 0.42.
Regarding the extra-terrestrial organism, if there are 75 different amino acids and each codon codes for a unique amino acid, then there must be at least 3 codons for each amino acid. Since there are 64 possible codons (4^3), the organism must use at least 75/64 = 1.17 codons per amino acid on average. This means that the organism could use codons with either 3 or 4 bases.