People with GENIUS IDEAS

go subscribe to azzyland

Hardy-Weinberg

 Step #1.) I was given the homozygous recessive (q2) which was .61. 

Step #2.) In order to get the recessive allele (q) I had to square-root .61 and the answer I got was .78 so applied it into the list below.

Step #3.) Now that I have the recessive allele (q) I can get the dominant allele (p) by subtracting it by 1. My answer for the dominant allele (p) was .22, *1-.78= .22*

Step #4.) The next step is the find the Homozygous dominant (p2) I got this by squaring the p (dominant allele) ((.22*.22= .05))

Step #5.) The last step is to get the heterozygous (2pq). To do that I multiplied q (recessive allele) and the p (dominant allele) ((2 (.22)(.78) = .34))

q2 (Homozygous recessive)- .61

q (Recessive allele)- .78

p (Dominant allele)- .22

p2 (Homozygous dominant)- .05

2pq (Heterozygous)- .34 
population: 1,000

~The genotypes are q2, p2, and 2pq.~ 

In the total of 1,000 individuals, I was asked to find the individuals of each genotype. Soooo the next thing I did was multiply each genotype by 1,000.

 For the q2 (homozygous recessive) I did this > .61*1,000 = 610
              p2 (homozygous dominant) > .05*1,000 = 50
             2pq (heterozygous) > .34*1,000 = 340 
                All together the individuals should add up to be 1,000. I checked, they do (:

 In conclusion there will be 610 that have the homozygous recessive allele in which they are not affected, 50 individuals who have the homozygous dominant allele, and 340 individuals who are heterozygous and carry it. 

Okay okay so I found this picture and I noticed that dry tree and bushy bushes in the back and said, well this can be my first picture for the scavenger hunt, and yes I know, I'm very late haha I was confused on how to make this blog thing. well there you have it folks, my first ecology selfie :) . Have a blessed day.

The California worm was a very interesting organism, its skin was transparent which helped us all/apeezy students find the pulse. although i'm pretty sure there was lots of mistakes, some may have counted the blood flow instead of the heart pulse, maybe some of us didn't get an exact number due to the worms squirming all around making it very difficult to keep track of the pulse. less likely would there have been a precise number through out the lab. Although it was definitely a frustrating lab due to knowing if there was a precise number, we did our best to do our calculations. For solution A we got an average number of 41.6, solution B we calculated 57 and lastly solution C had the mean of 37. With that being said, my answer would be that solution A is the neutral water as for solution B would be the stimulate and, solution C would have had to be the stimulate solution.