A pairs with T
C pairs with G
and vice versa so
TTTTTAACG
CCGAAATGT
Answer:
"U"-shaped, with peaks at the low and high extremes of beak size; disruptive selection
1.What is the probability that a sperm from the father will contain the PKU allele?
Probability = ½
2.What is the probability that an egg from the mother will contain the PKU allele?
Probability = ½
3.What is the probability that their next child will have PKU?
Probability = ¼ (because each parent has ½ chance ½ X ½ = ¼)
4.What is the probability that their next child will be heterozygous for the PKU gene?
Probability = ½ (because each parent has ½ chance of donating the ‘P’ allele and ½ chance of donating the ‘p’ allele (½ x ½) + (½ x ½) = ½
The simplest answer would be to look at a punnet square. it gives you the random chance that a trait gets passed on to offspring. it's made off probability of dominant and recessive genes. But it's random in the end.
Answer:The ratio of the concentrations of 
and 
when the buffer has a pH of 7.02 is 0.69
Explanation:
The dissociation constant for formic acid = 
Concentration of HA= 0.5 mM
Concentration of = 0.1 mM
pH = 6.16
First we have to calculate the value of 
.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[A^-}{[HA]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BA%5E-%7D%7B%5BHA%5D%7D)
Now put all the given values in this expression, we get:
![6.16=pK_a+\log (\frac{[0.1]}{0.5})](https://tex.z-dn.net/?f=6.16%3DpK_a%2B%5Clog%20%28%5Cfrac%7B%5B0.1%5D%7D%7B0.5%7D%29)
To calculate the ratio of the concentrations of and when the buffer has a pH of 7.02.
Using Henderson Hesselbach equation :
![7.02=6.86+\log \frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=7.02%3D6.86%2B%5Clog%20%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
![\frac{[A^-]}{[HA]}=1.44](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D%3D1.44)
![\frac{[HA]}{[A^-]}=0.69](https://tex.z-dn.net/?f=%5Cfrac%7B%5BHA%5D%7D%7B%5BA%5E-%5D%7D%3D0.69)
Thus the ratio of the concentrations of and when the buffer has a pH of 7.02 is 0.69
