Population Genetics

Hardy-Weinberg Problem

Given: Population size which is 1000. 
       q^2 which is 0.38

Find: p=0.38
      q=0.62
      p^2=0.14
      2pq=0.47

How:

1.  To find q, square root the q^2. (√0.38=0.62)
2.  To find p, subtract one from q. (1-q=p, or 1-0.62=0.38)
3.  To find p^2, square p. (0.38^2=0,14)
4.  To find 2pq, multiply 2 to the factors of p and q. (2(0.38)(0.62)=0.47).

With these frequencies you can find the individual alleles for the population. 

Homozygous dominant individual:(p^2) 0.14*1000= 140 

Homozygous recessive individual: (q^2) 0.38*1000= 380

Heterozygous individual: (2pq) 0.47*1000=470 

The way to find the individuals of each allele is to multiply each of the frequencies to the population. 

In a total population of 1000 people we know that 14% of them are homozygous dominant. 38% of the population are homozygous recessive. 47% are heterozygous individuals. 62% have the recessive gene, and 38% have the dominant gene. 

The Effects on BlackWorms' Pulse from Stimulants and Depressants

In this experiment we had to figure out which three container of worms had the stimulant, the depressant, and normal water. Most of the data the class has collected was quite consistent, but there was the occasional outlier. To do my calculations I decided to not include them. To find the average amount of beats for each container, I had to find the mean for everyone's data. I have come to the conclusion that container A was the stimulant. The average pulses per minute were about 34 beats per minute. Container B was the depressant. The average pulses per minute were about 30 beats per minute. Container C was the container of worms with the normal water,so they had no affect on the worms. The average pulses per minute were about 32 beats per minute. Also with the data, I found out that the worms where not drastically affected by the stimulant and the depressant.