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# Biology GS TIFR- Tata Institute of Fundamental Research Questions 1-5_ Solved 1. In a population of a remote Pacific island, the frequency of people with green eyes is 3%. In that same population, the frequency of people having dimples is 0.5%. Assuming independent assortment, what is the expected number of people who will have both dimples and green eyes, given that the population of the island is 3400 individuals?

(i)51

(ii)102

(iii)119 (iv)150

Solution: We can start by using the multiplication rule of probability to find the probability of an individual having both green eyes and dimples:

P(green eyes and dimples) = P(green eyes) x P(dimples)

P(green eyes and dimples) = 0.03 x 0.005

P(green eyes and dimples) = 0.00015

This means that the probability of a randomly selected individual having both green eyes and dimples is 0.00015.

To find the expected number of individuals with both green eyes and dimples in a population of 3400 individuals, we can multiply the probability by the total population size:

Expected number = Probability x Population size

Expected number = 0.00015 x 3400

Expected number = 0.51

Therefore, we can expect approximately 0.51 individuals in the population to have both green eyes and dimples. Since we cannot have a fraction of a person, we can say that we expect 1 individual to have both traits with a probability of 0.51.

2. In a skewed distribution with a long right tail like the distribution of annual incomes in most developing countries, which of the following statements is true?

(i) mean<mode<median

(ii) mean<median<mode

(iii) mode<mean<median

(iv) mode<median<mean

Answer: In a skewed distribution with a long right tail, like the distribution of annual incomes in most developing countries, the following statement is true:

The median is typically smaller than the mean. This is because the right tail of the distribution has a few very high values that pull the mean upward, making it larger than the median. The median, on the other hand, is less sensitive to extreme values and is therefore a better measure of central tendency in skewed distributions.

The mode may not be a very informative statistic. Skewed distributions often have multiple modes or no clear mode, especially if the distribution is heavily skewed.

The distribution is positively skewed. This means that the tail of the distribution extends to the right, and most of the data is concentrated on the left side of the distribution.

There are more low-income earners than high-income earners. This is because the majority of the population typically falls within the lower income brackets, and there are relatively fewer people who earn very high incomes.

Outliers are more common in the right tail of the distribution. This is because the right tail of the distribution contains the extreme values that pull the mean upward and make the distribution skewed. These extreme values are often referred to as outliers and can have a significant impact on statistical analysis.

3. A bat echolocates and receives an echo 10 milliseconds later. Assuming a speed of sound in air of 340 m/s, how far away is the object?

(i) 3.4 m

(ii) 1.7 m

(iii) 6.8 m

(iv) 5.1 m

Solution: The time taken for sound to travel from the bat to the object and back to the bat (i.e., the round-trip time) is 10 milliseconds or 0.01 seconds.

The speed of sound in air is 340 m/s, which means that sound travels 340 meters in one second. Therefore, the distance traveled by sound during the round-trip time is:

distance = speed × time

distance = 340 m/s × 0.01 s

distance = 3.4 meters

However, this distance is the total distance traveled by sound, which includes the distance from the bat to the object and the distance from the object back to the bat. To find the distance from the bat to the object, we need to divide the total distance by two:

distance from bat to object = 3.4 meters ÷ 2

distance from bat to object = 1.7 meters

Therefore, the object is located 1.7 meters away from the bat.

4. A badminton tournament has 128 participants and needs to produce just one winner. In each match, only the winner goes to the next round and plays with one of the other winners. This proceeds till ultimately there is only one winner. How many matches will a person need to reach the semi-finals?

(i)8 (ii)7 (iii)6

(iv)5

Solution: To reach the semi-finals, a player must win 4 matches in total: one in the round of 128, one in the round of 64, one in the round of 32, and one in the round of 16.

To determine how many matches will be played in the entire tournament, we can use the fact that in each round, half of the remaining players are eliminated. Therefore, the number of matches played in each round is equal to the number of remaining players divided by 2.

In the first round, there are 128 players, so 64 matches will be played (128/2 = 64). In the second round, there are 64 players remaining, so 32 matches will be played. In the third round, there are 32 players remaining, so 16 matches will be played. And in the fourth round, there are 16 players remaining, so 8 matches will be played.

Thus, a player would need to win 4 matches to reach the semi-finals, and a total of 120 matches will be played in the entire tournament (64 + 32 + 16 + 8 = 120).

5. Kleiber’s law states that for a number of animal species, the metabolic rate, scales as the 3/4th power of its mass. Thus, a cat that weighs about 80 times more than a mouse:

(i)consumes about 27 times the energy of the mouse.

(ii) consumes about 1/27 times the energy of the mouse.

(iii)consumes as much energy as the mouse.

(iv)consumes about 60 times the energy of the mouse.

Answer: According to Kleiber's law, if a mouse weighs x, and a cat weighs 80x, then the cat's metabolic rate would be approximate:

Metabolic rate of cat = (Mass of cat)^0.75

= (80x)^0.75

= 80^0.75 * x^0.75

= 31.7 * x^0.75

So, if the metabolic rate of a mouse is m, then the metabolic rate of the cat would be:

Metabolic rate of cat = 31.7 * m

Therefore, based on Kleiber's law, the metabolic rate of a cat that weighs about 80 times more than a mouse would be approximately 31.7 times greater than the metabolic rate of the mouse.

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