H - Weezy Problem

q ^2	(given) .41
q	.64
p	.36
p ^2	.13
2pq	(2)(.36)(.64) = .46

p^2 + 2pq + q^2	(.13) + (.46) + (.41) = 1
# of homo dominant	q^2 = .41 = 41% (q^2 x 1000) = (.41 x 1000) = 410
# of homo recessive	p^2 = .13 = 13% (p^2 x 1000 = (.13 x 1000) = 130
# of heterozygous	2pq = .46 = 46% (2pq x 1000) = (.46 x 1000) = 460

            We are given the value q^2 (homozygous dominant) of .41 and an initial population of 1000 individuals. 

Step 1: Find q by taking the square root of q^2 which equals .64

Step 2: Find p by knowing q + p = 1, therefore 1 - .64 = .36

Step 3: Find p^2 by taking p and squaring it, or multiplying it by itself. .36 x .36 = .13

Step 4: Find 2pq by plugging in the values p and q. (2)(.36)(.64) = .46

After finding all values, I then found the # of homo dom. individuals, # of homo rec. individuals, and # of heterozygotes in the population. 

In order to do so, I multiplied the value of q^2 (.41) ,which also equals 41% of the population, by the total population (1000) and got 410 individuals.

The number of individuals that contain the homozygous recessive alleles are 13% of the population which is 130 individuals, which is found by multiplying p^2 by the total population as well.

Lastly, the number of heterozygous individuals takes up 46% of the population which is 460 individuals. This is found by multiplying 2pq by 1000, the total population as well.

Worm Lab

Solution A has the depressant. Solution B has the stimulant. Solution C is normal

According to the data my group and I have collected, the results were very obvious. For solution A, the pulse rates we collected were 19, 16, 12, 16, and 17 per minute. For solution B, the pulse rates resulted in 35, 28, 30, 29, and 31 per minute. Lastly in solution C, the rates per minute collected were 18, 25, 25, 20, and 19. A method I used for analyzing the data would be finding the averages of every solution. The average for solution A was 16, B was 25.2, and C was 21.4. The lowest average was solution A, therefore leading to the conclusion that it contained the depressant. Whereas solution B had the highest average creating the assumption that it contained the stimulant. Lastly, leaving C to being the middle number, which meant that it was normal.

This is a picture of the California Blackworms that were observed in this lab.

This a a picture of the worms through the microscope.