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A colleague determines that the B1 and B2 alleles of the B locus both occur with a frequency equal 0.45. Surprised, she redoes her work and confirms her results.

a. What could be the cause of your colleague's surprise? Please explain.

b. Because your colleague confirms her results she now needs to explain them. She turns to you for assistance. What do you suggest? Please be sure to explain how your explanation accounts for her observations.

Your colleague was probably surprised because she thought that B1 and B2 were the only two alleles that occurred at this locus in this population. Consequently, the discovery that their frequencies, p and q, summed to 0.9 as opposed to 1 was startling. Your suggestion to look for at least one additional allele to account for the 10% of the alleles unaccounted for in her study is well taken. She realizes that the existence of one or more additional alleles would explain the missing 10% and enable her to bring the summed allele frequencies for the B locus to 1.

Now that we have designated p to represent the freqeuncy of A allele and q, the a allele, we are ready to move forward with our efforts to construct the elements of the Hardy-Weinberg equation. Imagine that every individual in a population, in which both copies of both the A and a allele occur, is equally likely to survive and to reproduce.

What possible genotypes could occur in the offspring of this population?

To answer this question, determine all the possible genotypes that could be formed from a population of individuals whose loci collectively warehouse numerous copies of A and a alleles. Remember that, because these individuals are all equally likely to reproduce, all combinations of these two alleles have the potential to form. Visit this module if you have questions.

There are four possible genotypes:

  • A A
  • a a
  • A a
  • a A

Now that we know what genotypes could form, we can use the rule highlighted at the very beginning of this module to predict how frequently each of these genotypes will appear in the offspring generation.

What are these frequencies? Apply the highlighted (boxed) rule above to complete the phrases below using the symbols p and q.

If all individuals are equally likely to survive and to reproduce, then the

  • frequency of genotype AA will equal ____________________
  • frequency of genotype aa will equal _____________________
  • frequency of genotype Aa will equal ____________________
  • frequency of genotype aA will equal ____________________

Because the Aa and aA genotypes are genetically equivalent, we can summarize the relationships you articulated above as

  • frequency of genotype AA will equal p x p = p2
  • frequency of genotype aa aa will equal q x q = q2
  • frequency of genotype Aa will equal 2 x (p x q) = 2pq

And there you have it, the three fundamental elements of the Hardy-Weinberg equation that describe how frequently the three possible genotypes will appear in the offspring generation of a population that is not subject to an agent of evolution! Remember that only three genotypes are possible because we are only working with a gene for which only two alleles exist in a population.

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
I'm not sure why it wrote it the other way
I got X =-6
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
is it a question of log
Commplementary angles
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im all ears I need to learn
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what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
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a perfect square v²+2v+_
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
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Kristine 2*2*2=8
Bridget Reply
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Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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how do you translate this in Algebraic Expressions
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
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abeetha Reply
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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good afternoon madam
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what is the application of nanotechnology?
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
anybody can imagine what will be happen after 100 years from now in nano tech world
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
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silver nanoparticles could handle the job?
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I'm interested in Nanotube
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Understanding the hardy-weinberg equation. OpenStax CNX. Oct 22, 2007 Download for free at http://cnx.org/content/col10472/1.1
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