What is the meaning of Inspire Award?

The Department of Science & Technology (DST) is implementing a national program Innovation in Science Pursuit for Inspired Research (INSPIRE) for attracting talented young students to study Science and pursue career in Research.

Who is eligible for Inspire Award?

Students of class 6 to 10 are eligible for INSPIRE Awards. The students should be in the age group of 10 to 15 years. Around 2 Lakh students are awarded INR 5000 per child under INSPIRE Awards

What is the full form of Inspire?

Innovation in Science Pursuit for Inspired Research

Department of Science & Technology (DST), Government of India is successfully implementing ‘Innovation in Science Pursuit for Inspired Research (INSPIRE) scheme since 2010.

What is the Inspire Award?

For school students

INSPIRE Award MANAK ( Million Minds Augmenting National Aspiration and Knowledge) scheme, previously known as INSPIRE Award Scheme, is being implemented through State and UT governments with objectives (i) to attract young students to study science and pursue research career (ii) to promote creative thinking and foster .

What is Inspire Award scholarship?

For college students

INSPIRE Scholarship. This scheme offers 12,000 scholarships every year @ Rs.80,000/- each for undertaking Bachelor and Masters level education in the Natural & Basic sciences, possessing any of the following criteria

When to apply for Inspire Scholarship ?

The INSPIRE Scholarship  is for all the students who have opted 3-year B.Sc (Natural & Basic Science) or 5-year Integrated course and all the interested candidates can apply online on or before the last date

What is DST inspire fellowship?

There is a provision of 1000 INSPIRE Fellowships to be offered/given every year to eligible students and each Fellowship is tenable for a maximum period of five years or completion of their PhD degree, whichever is earlier. … or M. Tech. course while pursing Integrated M.Sc.-PhD/ M. Phil.-Ph.D./ M. Tech.-Ph.D. courses.

How much is inspire fellowship?

Each chosen INSPIRE Faculty shall be eligible to be given a consolidated amount of Rs.80,000/- pm. Moreover, a Research Grant of Rs 7 lakh annually for 5 years will also be provided to each successful candidate. The INSPIRE Faculty Award is for a maximum period of 5 (five) years.

How to apply-

If you have internet facility in the school or in the neighbourhood, you can fill the online form directly on INSPIRE website .one time registration for school authority.

Q 1) In corn, purple kernels are dominant to yellow. A random sample of 100 kernels is taken from a population in Hardy-Weinberg equilibrium. It is found that 9 kernels are yellow and 91 kernels are purple.What is the frequency of the yellow allele in this population?

ANS-0.3

Q 2) a) The frequency of two alleles in a gene pool is 0.19 (A) and 0.81(a). Assume that the population is in Hardy-Weinberg equilibrium. (a)Calculate the percentage of heterozygous individuals in the population.

ANS-According to the Hardy-Weinberg Equilibrium equation, heterozygotes are represented by the 2pq term. Therefore, the number of heterozygous individuals (Aa) is equal to2pq which equals 2 × 0.19 × 0.81 = 0.31 or 31%

b) Calculate the percentage of homozygous recessives in the population.

ANS- The homozygous recessive individuals (aa) are represented by the q2 term in the H-W equilibrium equation which equals 0.81 × 0.81 = 0.66 or 66%

Q 3) An allele W, for white wool, is dominant over allele w, for black wool. In a sample of 900 sheep, 891 are white and 9 are black. Calculate the allelic frequencies within this population, assuming that the population is in H-W equilibrium.

ANS- The allelic frequency of w is represented by the q term and the allelic frequency W is represented by the p term. To calculate the value of q, realize that qq or q2 represents the homozygous recessive individuals or the black sheep in this case. Since there are 9 black sheep, the frequency of black sheep = # individuals /total individuals=9/900=0.01 thus ww = q2 = 0.01 so q=square root of q square= 0.1 Additionally, p + q = 1 thus p = 1 – q or p = 1 – 0.1 or 0.9 ∴ p = W = 0.9 and q = w = 0.1

Q 4)In a population that is in Hardy-Weinberg equilibrium, the frequency of the recessive homozygote genotype of a certain trait is 0.09. Calculate the percentage of individuals homozygous for the dominant allele.

ANS- We know that the frequency of the recessive homozygote genotype is q2and equal to 0.09 and its square root is 0.30 AND we also know that p + q = 1 Thus, p = 1 – q ∴p = 1 – 0.30 = 0.70 ∴The homozygote dominants are represented by p2= (0.70)2 = 0.49 or 49%

Q 5.In a population that is in Hardy-Weinberg equilibrium, 38 % of the individuals are recessive homozygotes for a certain trait. In a population of 14,500, calculate the percentage of homozygous dominant individuals and heterozygous individuals.

ANS- Always start with the homozygous recessive percentage if given, which is equal to q2. ∴ q =Square root of q∴ q =square root of 0.38 ∴q = 0.616 . Solving for p is now straightforward: p + q = 1, thus p = 1 – q = 1 – 0.616 = 0.384 The homozygous dominant individuals are represented by p square=0.384×0.384=0.147 so,14500×0.0147=2,132 homozygous dominant individual.The heterozygotes are represented by the 2pq term and 2×0.616×0.384=0.473 so,14500×0.473=6,859

Q 6)A rare disease which is due to a recessive allele (a) that is lethal when homozygous, occurs within a specific population at a frequency of one in a million. How many individuals in a town with a population of 14,000 can be expected to carry this allele? ANS-Always start with the homozygous recessive percentage if given, which is equal to q2. The frequency of the recessive genotype, q2 is 1/1, 000, 000 = 0.000001 ∴q = square root of q = 0.000001 = 0.001 = the frequency of allele a and p + q = 1 thus, p = 1 – q = 1 – 0.001 = 0.999 Therefore, the frequency of allele A = p = 0.999 and the frequency of allele a = q = 0.001 Carriers are heterozygous and are equal to 2pq. So, 2x 0.999x 0.001 x 14, 000=27.97

Q 7) In a certain African population, 4 % of the population is born with sickle cell anemia (aa). Calculate the percentage of individuals who enjoy the selective advantage of the sickle-cell gene (increased resistance to malaria)?

ANS-The individuals that are heterozygous enjoy the selective advantage of increased resistance to malaria and are carriers of the sickle cell trait. They have a genotype of Aa and are represented by the 2pq term in the H-W equilibrium equations. We are told that 4% of the population is affected with sickle cell anemia, thus q2 = 4% = 0.04 when expressed as a decimal. So, q = square root of q or we can say aquare root of 0.04=0.20 and p + q = 1, thus p = 1 – q = 1 – 0.20 = 0.80 Thus, A allele has a frequency of 0.80 and the a allele has a frequency of 0.20. Therefore, the percentage of the population that is heterozygous (Aa) and are carriers is = 2x 0.20x 0.80=0.32 or 32%.

Q 8) In the United States, approximately one child in 10,000 is born with PKU (phenylketonuria), a syndrome that affects individuals homozygous for the recessive allele (aa). (a)Calculate the frequency of this allele in the population.

ANS-Always start with the homozygous recessive data if given, which is equal to aa and to q2. The frequency of the recessive genotype, aa or q2 is 1/10, 000= 0.0001 Thus, q =sqaure root of q = 0.01

(b)Calculate the frequency of the normal allele.

ANS-A represents the “normal” allele and is also represented by p in the H-W equilibrium equations. p + q = 1 thus, p = 1 – q = 1 – 0.01 = 0.99

(c)Calculate the percentage of carriers of the trait within the population.

ANS-The heterozygotes are the carriers of the trait and are represented by the 2pq term in the H-W equilibrium equations. 2X 0.01 X 0.99=0.0198 which is then multiplied by 10,000 to obtain the number of heterozygous carriers in the population. So, 0.0198 X 10, 000 = 198 individuals in the population that carry the trait.

Q 1) In corn, purple kernels are dominant to yellow. A random sample of 100 kernels is taken from a population in Hardy-Weinberg equilibrium. It is found that 9 kernels are yellow and 91 kernels are purple.What is the frequency of the yellow allele in this population?

ANS-0.3

Q 2) a) The frequency of two alleles in a gene pool is 0.19 (A) and 0.81(a). Assume that the population is in Hardy-Weinberg equilibrium. (a)Calculate the percentage of heterozygous individuals in the population.

ANS-According to the Hardy-Weinberg Equilibrium equation, heterozygotes are represented by the 2pq term. Therefore, the number of heterozygous individuals (Aa) is equal to2pq which equals 2 × 0.19 × 0.81 = 0.31 or 31%

b) Calculate the percentage of homozygous recessives in the population.

ANS- The homozygous recessive individuals (aa) are represented by the q2 term in the H-W equilibrium equation which equals 0.81 × 0.81 = 0.66 or 66%

Q 3) An allele W, for white wool, is dominant over allele w, for black wool. In a sample of 900 sheep, 891 are white and 9 are black. Calculate the allelic frequencies within this population, assuming that the population is in H-W equilibrium.

ANS- The allelic frequency of w is represented by the q term and the allelic frequency W is represented by the p term. To calculate the value of q, realize that qq or q2 represents the homozygous recessive individuals or the black sheep in this case. Since there are 9 black sheep, the frequency of black sheep = # individuals /total individuals=9/900=0.01 thus ww = q2 = 0.01 so q=square root of q square= 0.1 Additionally, p + q = 1 thus p = 1 – q or p = 1 – 0.1 or 0.9 ∴ p = W = 0.9 and q = w = 0.1

Q 4)In a population that is in Hardy-Weinberg equilibrium, the frequency of the recessive homozygote genotype of a certain trait is 0.09. Calculate the percentage of individuals homozygous for the dominant allele.

ANS- We know that the frequency of the recessive homozygote genotype is q2and equal to 0.09 and its square root is 0.30 AND we also know that p + q = 1 Thus, p = 1 – q ∴p = 1 – 0.30 = 0.70 ∴The homozygote dominants are represented by p2= (0.70)2 = 0.49 or 49%

Q 5.In a population that is in Hardy-Weinberg equilibrium, 38 % of the individuals are recessive homozygotes for a certain trait. In a population of 14,500, calculate the percentage of homozygous dominant individuals and heterozygous individuals.

ANS- Always start with the homozygous recessive percentage if given, which is equal to q2. ∴ q =Square root of q∴ q =square root of 0.38 ∴q = 0.616 . Solving for p is now straightforward: p + q = 1, thus p = 1 – q = 1 – 0.616 = 0.384 The homozygous dominant individuals are represented by p square=0.384×0.384=0.147 so,14500×0.0147=2,132 homozygous dominant individual.The heterozygotes are represented by the 2pq term and 2×0.616×0.384=0.473 so,14500×0.473=6,859

Q 6)A rare disease which is due to a recessive allele (a) that is lethal when homozygous, occurs within a specific population at a frequency of one in a million. How many individuals in a town with a population of 14,000 can be expected to carry this allele? ANS-Always start with the homozygous recessive percentage if given, which is equal to q2. The frequency of the recessive genotype, q2 is 1/1, 000, 000 = 0.000001 ∴q = square root of q = 0.000001 = 0.001 = the frequency of allele a and p + q = 1 thus, p = 1 – q = 1 – 0.001 = 0.999 Therefore, the frequency of allele A = p = 0.999 and the frequency of allele a = q = 0.001 Carriers are heterozygous and are equal to 2pq. So, 2x 0.999x 0.001 x 14, 000=27.97

Q 7) In a certain African population, 4 % of the population is born with sickle cell anemia (aa). Calculate the percentage of individuals who enjoy the selective advantage of the sickle-cell gene (increased resistance to malaria)?

ANS-The individuals that are heterozygous enjoy the selective advantage of increased resistance to malaria and are carriers of the sickle cell trait. They have a genotype of Aa and are represented by the 2pq term in the H-W equilibrium equations. We are told that 4% of the population is affected with sickle cell anemia, thus q2 = 4% = 0.04 when expressed as a decimal. So, q = square root of q or we can say aquare root of 0.04=0.20 and p + q = 1, thus p = 1 – q = 1 – 0.20 = 0.80 Thus, A allele has a frequency of 0.80 and the a allele has a frequency of 0.20. Therefore, the percentage of the population that is heterozygous (Aa) and are carriers is = 2x 0.20x 0.80=0.32 or 32%.

Q 8) In the United States, approximately one child in 10,000 is born with PKU (phenylketonuria), a syndrome that affects individuals homozygous for the recessive allele (aa). (a)Calculate the frequency of this allele in the population.

ANS-Always start with the homozygous recessive data if given, which is equal to aa and to q2. The frequency of the recessive genotype, aa or q2 is 1/10, 000= 0.0001 Thus, q =sqaure root of q = 0.01

(b)Calculate the frequency of the normal allele.

ANS-A represents the “normal” allele and is also represented by p in the H-W equilibrium equations. p + q = 1 thus, p = 1 – q = 1 – 0.01 = 0.99

(c)Calculate the percentage of carriers of the trait within the population.

ANS-The heterozygotes are the carriers of the trait and are represented by the 2pq term in the H-W equilibrium equations. 2X 0.01 X 0.99=0.0198 which is then multiplied by 10,000 to obtain the number of heterozygous carriers in the population. So, 0.0198 X 10, 000 = 198 individuals in the population that carry the trait.