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In a population of 5000 individual, 450 show recessive trait. what will be the number of heterozygous individual in this population.PLEASE REPLY FAST

In a population of 5000 individual, 450 show recessive trait. what will be the number of heterozygous individual in this population.PLEASE REPLY FAST

Grade:12th pass

5 Answers

Mayajyothi C J
126 Points
6 years ago
This can be solved using Hardy-Weinberg Equation;
p = AA (dominant)
q = aa  (recessive)
2pq = Aa (Heterozygous)
p2  + 2pq + q2  = 1
p  +  q = 1
p= no.of dominant individuals/total population
q= no. of recessive individuals/total population
q = 450/5000
= 0.09
p+q = 1
p = 1 – q = 1 – 0.09 = 0.91
2pq = 2 x 0.91 x 0.09
= 0.1638
Total no. of heterozygous individuals = 0.1638 x 5000
= 819 
This is the correct answer. If you have any doubt please ask. Otherwise please approve ^_^
Parul Rajput
25 Points
6 years ago
But ans is 2100 given...please give correct answer..i think first we have to divide 450 by 100 ...but not getting the ans
Mayajyothi C J
126 Points
6 years ago
The answer could only be 819 anyway not 2100. You divide 450 by 100 only when the allele frequency % percentage is given. 
Eg Q: Chicken with rose combs made up of 36% of a population..….........
Here you take q=36/100 = 0.36 (allele should be less than 1 always)
But here if you divide 450/100 it will be greater than 1  (4.5) 
The answer is 819 I’m sure. Just type hardy-Weinberg population problems in youtube and you will get videos to how to solve it.
Mayajyothi C J
126 Points
6 years ago
Just add “www.” before this and you will get into my source of this problem “germanna.edu/documents/Hardy-WeinbergEquilibriumSept2012_002.pdf”
Amit Kumar Chand
11 Points
6 years ago
Total individual = 5000400= recessive i.e. q2So q2 = 5000/400= 0.09And q= 0.3Hence we know p+q = 1So p - 0.3 = 1=> P= 0.7And 2pq = 2× 0.7×0.3= 0.21×2= 0.42Hence 2pq = 0.42× 5000= 2100.Ok.. got it??

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