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Population Genetics Problems (in Hindi)
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Hardy Weinberg equation and related problems

Manisha Malhotra
Msc in microbiology with gold medal, qualified GATE and Uttarakhand-set. Admin of the you tube channel Manisha Malhotra tutorials.

U
#Surgical Strike Historic Step taken by Government Of India. Salute to Indian Army on success of this strategic Mission.
I am sure we are all proud of it.. But try and criticize the point to get another dimension..say I am asking you this question.. INDIAN STRIKE WAS A MISTAKE, SUBSTANTIATE!!
Sn
if a wise person is pinched he will think and act while a rouge is pinched he wont think wise he thinks more horribly, hope our miltary is attentive hence forth and take good care of all the units along border
Indian strike was quite historic in the sense that we have for the first time engaged in an aggressive pursuit with a rogue neighbour and crossed the LOC for the first time to eliminate the terrorists who would have otherwise wrecked mayhem in India in the coming days.It is also a very tactical move as Pakistan can't admit that terrorists were present on their soil and neither can they digest this sudden move from India.Having said that, we need to focus now on maintaining extreme vigilance along the LOC in the coming days and can't afford a single mistake.
option A is the right one
Manisha Malhotra
4 months ago
All the solutions are given in the next course..pls go through that
Mam plz explain.. How to solve last question
Manisha Malhotra
7 months ago
I will provide u with the solutions in my next course
7 months ago
Okkk mam.. Thanku
SG
1- 0.4605 2- 64% 32% 4% ,option a 3- option a
Manisha Malhotra
7 months ago
Watch the lessons from doubt destroyer series to see whether your ans are correct or not😊
SG
Shubham Gupta
7 months ago
ok mam
1. 0.4605 3. p= 0.6 p= 0.4 mam plz solve second question
Manisha Malhotra
7 months ago
I have already provided the solutions under the course named doubt destroyer series
1. 0.4605 2. 64% of M, 32% of MN, 4 % of N 3. P=0.6, q=0.4
Manisha Malhotra
7 months ago
Gud attempt keep working smart and practice more
Pooja Luwang
7 months ago
yes maam
1. Hardy-weinberg equation For genotype frequency- p2+2pq+q2 1 For allele frequency- p+q-1 Freuency of homozygous dominant genotype is represented by 'p2 Freuency of homozygous recessive genotype is represented by 'q2 Freuency of hetrozygous genotype is represented by '2pq

2. Formula for calculating allele frequency Frequency of allele B - Number of allele B Total no. of alleles 2*Number of BB homozygous number of Bb heterozygote 2*Total number of individuals (p) -

3. Q In a population, 600 individuals have MM blood group, 300 have MN blood group and 100 have NN blood group. What will be the frequencies of M and N alleles in this population? (A) M 0.75 and N 0.25 (C) M 0.85 and N 0.15 (B) M 0.65 and N 0.35 (D) M 0.55 and N 0.45 Frequency of allele MN Number of allele M Total no. of alleles 2*Number of MM homozygous + number of MN heterozygote 2*Total number of individuals f(p)

4. 30 0 Total =1000 2 x Total 2 x 600 300 2 x 1o0o = 120 300-150 8.045 2o00

5. = 1-0-75 = 0-25

6. Frequency of allele B - Number of allele B Total no. of alleles 2*Number of BB homozygous number of Bb heterozygote 2*Total number of individuals f(p)

7. Fruit color of wild solamum nigrum is controlled by 2 alleles of a gene(A and a). The frequency of A, p-0.8 and a, q 0.2. in the neighboring field, a tetraploid genotype of S.nigrum was found. After critical examination 5 distinct genotypes were found which were AAAA, AAAa, Aaaa, AAaa, and aaaa.. Following Hardy-Weinberg principle and assuming the same allele frequency as that of diploid population, the no. of phenotypes calculated in each population of 1000 plants are close to one of the following- AAAA: AAAa: AAaa: Aaaa: aaaa 1. 409: 409: 154: 26: 2 2 420: 420: 140: 18: 2 3. 409: 409: 144: 36: 2 4. 409: 420: 144: 25: 2

8. Given -0.8, q-0.2 (p+q- (0.8)4+(0.2)4+4(0.8)3(0.2) 6(0.8)2(0.2)2+4(0.8)(0.2)3) 1000 (ptq)4- 409.6 1.6: 409.6:154:26 ANS-A

9. The frequency of M-N blood types in a popula- tion of 6129 individuals is as follows: Blood type Genotype Number of individuals LML LML 1787 3039 1303 MN The frequency of L allele in this population is 1. 0.4605 3. 0.5395 2. 0.2121 4. 0.2911

10. Among 205 individuals in a population, the frequency of the LM and LN alleles was 0.8 and 0.2, respectively. Calculate the percentage of individuals with M, MN and N blood type (A) 64% of M, 32% of MN and 4% of N (B) 64% of M, 30% of MN and 6% of N (C) 60% of M, 34% of MN and 6% of N (D) 60% ofM, 32% of MN and 800 ofN