In this video, we will be talking about how selection violates the Hardy Weinberg equilibrium using a population of turtle ducks
living in a medium-sized pond.
Before selection occurred, there were yellow, brown, and green-shelled turtle ducks living in this pond.
There were a total of 85 turtle ducks.
With 13 brown, 25 green, and 47 yellow turtle ducks.
To calculate the allele frequency for Y and B we use the following equation.
Number of Y alleles over the total number of alleles
To do 2 times 47 yellow, plus 25 green, divided by 2 times the total of 85 to get .7
To get the frequency of B, you can do 1 minus the frequency of Y to get .3
or you can use the equation from above to do 2 times 13 brown plus 25 green divided by 2 times the total of 85 to get .3 as well.
To verify that you did this calculation correctly you can use the equation p+q=1
 
The frequency of Y plus the frequency of B should equal 1.
 
.7 + .3 does, in fact, equal, so this calculation has been done correctly.
To calculate the genotypic frequencies for each of the genotypes we use the following equations.
For yellow [turtle ducks], we did [the number of] yellow turtle ducks over total [number of] turtle ducks to get .5529
For green [turtle ducks], we did [number of] green turtle ducks over total [number of] turtle ducks to get .2941
For brown [turtle ducks], we did brown turtle ducks over the total [ number of] turtle ducks] to get .1529
To check if the population is in Hardy Weinberg Equilibrium, we can use the p squared, plus 2 times p times q, plus q squared equals 1.
Using the values we got earlier [for p and q], we can plug them into the equation to get a sum of 1
proving that the population of turtle ducks is in Hardy Weinberg equilibrium.
In this population, brown turtle ducks are more susceptible to getting eaten by predators due to their shell color standing out in the water, unlike yellow or green turtle ducks.
The fitness for yellow turtle ducks is 1.
For brown turtle ducks, it's .54
and for green turtle ducks, it's .875
After directional selection occurs, the frequencies for each genotype will change.
To calculate this, first, we have to find the proportion of unadjusted survivors by multiplying the initial frequency by fitness.
For yellow turtle ducks, it would be .49
For green turtle ducks, it would be .3675
and for brown turtle ducks, it'd be .0486.
Then to find the frequency after selection for each of the genotypes, we have to divide the adjusted survivors by the sum of the row.
After selection for yellow turtle ducks, the frequency would be .5408.
For green turtle ducks, it would be .4056.
and for brown turtle ducks, it would be .0536.
To calculate the frequency of Y [after selection], you do the frequency of YY plus the frequency of YB times .5 to get .7436.
To get the frequency of B [ after selection], you do the frequency of YB times .5 plus the frequency of BB to get .2546.
If 100 offspring were produced in the next generation,
49 would be yellow shelled, 42 would be green-shelled and 9 would be brown-shelled.
Directional selection would play against brown-shelled turtle ducks in each generation, decreasing the number of brown turtle ducks
and increasing [the number of] yellow and green turtle ducks in each future generation.
 
