The Hardy-Weinberg Equilibrium!
In preparation for
the Upcoming Genetics TestThe Hardy-Weinberg Equilibrium illustrates how the % of each of the three genotypes remains constant in a population. This also helps to illustrate the stable portions of the theory of punctuated equilibrium. It is important to remember, however, that the theory of punctuated equilibrium also includes the punctuated sections during which there is a great deal of natural selection, ultimately leading to speciation. The factors that affect the evolution of populations, both in terms of Macro-evolution and Micro-evolution (which occurs in small populations, and on a small scale) , are the same as the those that cannot happen for the Hardy-Weinberg Equilibrium to be maintained. These factors, in turn, illustrate many of the finely tuned details of modern evolutionary theory:
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Conditions Necessary for the |
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This relates to the following aspects of Evolution (Macro & Micro): |
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| 1. | There is no mutation occurring | 1. | Remember how mutation provides the variation in a population, which is the raw material for natural selection. |
| 2. | There is no natural selection occurring | 2. | This is more likely in an extremely stable environment. |
| 3. | The population is infinitely large | 3. | The Founder Effect, the Bottleneck Effect, and the importance of a large sample size! |
| 4. | All members of the population breed | 4. | The effect of Sexual Selection (e.g., think of the peacock's tail feathers!). |
| 5. | Mating is entirely random | 5. | This, as with the step above, relates to Sexual Selection. |
| 6. | Every organism produces the same number of offspring | 6. | Remember the Guano Ring, and why the Blue Footed Booby always lays two eggs? |
| 7. | There is no migration in/out of the population (no immigration or emigration) | 7. | Migration not only affects the overall percentages, but is also related to the Bottleneck Effect. |
For those students who prefer to approach these problems in a linear manner, I have outlined a few steps to follow . . .
That’s all there is to it!1. Be able to write out the formula from memory!
Remember:p2 + 2pq + q2 = 1
% + % + % = 100%
BB + Bb + bb = 100%
The % will be expressed as a decimal (e.g. 14% = 0.14)2. I will give you the % of the population that has the recessive trait!
Since % will be expressed as a decimal, be sure to convert it! (e.g. 14% = 0.14)
The % = q23. Figure out q.
q = the Square Root of the % given above (i.e. the Square Root of 0.14)
Example q = .37 (round it to 2 decimal places)4. Figure out p.
p + q = 1
therefore
p = 1 - q
Example p = 1 - .37 = .63 (p = .63)5. Figure out:
A) p2
p2 = p x p
Example p2 = .63 x .63 = .40 (p2 = .40)
(round it to 2 decimal places)2pq = 2(p x q)B) 2pq
2(.63 x . 37) = 2(.23) (round it to 2 decimal places) = .466. Write the numbers out underneath the formula!
p2 + 2pq + q2 = 1
.40 + .46 + .14 = 1
Note: the numbers in RED below refer to the
portion of the population with the recessive phenotype . . .
Taken from 1999 A Period Biology II Honors: Answers are available BELOW! |
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Group 1: 3/85
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Group 2: 7/112
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Group 3: 11/96
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Group 4: 19/77
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Taken from 1999 D Period Biology II: Answers are available BELOW! |
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Group 1: 7/85
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Group 2: 11/112
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Group 3: 19/96
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Group 4: 13/77
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Taken from 1999 F Period Biology II: Answers are available BELOW! |
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Group 1: 13/85
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Group 2: 19/112
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Group 3: 17/96
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Group 4: 11/77
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Taken from 2000 5th Period Biology II: Answers are available BELOW! |
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Group 1: 87/179
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Group 2: 43/213
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Group 3: 67/187
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Group 4: 59/203
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Taken from 2000 6th Period Biology II: Answers are available BELOW! |
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Group 1: 89/323
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Group 2: 209/492
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Group 3: 113/288
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Group 4: 407/517
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Taken from Fall 2001 1st Period Biology II Honors: Answers are available BELOW! |
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Group 1: 431 / 918
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Group 2: 398 / 741
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Group 3: 643 / 895
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Group 4: 587 / 959
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