Gene therapy I hope this helps
Answer:
why an athlete would need to be concerned about twitches or contractions is because, whenever they perform, and such thing happens, it will most likely distract the athlete, and cause the athlete doing his/her performance to not be as good, and may lead to failure, for example, when someone is running very fast for a sprinting race, and he has a contraction in his leg it will cause him to react by showing signs of pain by slowing down or even tripping and falling, causing him to lose the race.
Hope this Helped!
The question is incomplete as it does not have the options which are:
a. The woman with stretch marks has a greater collagen-to-elastic fiber ratio.
b. The woman may have different gender babies.
c. The woman with less stretch marks has larger skin cells.
d. The woman without stretch marks has more cholesterol in her diet, which affects her cell membranes.
Answer:
The woman with stretch marks has a greater collagen-to-elastic fiber ratio.
Explanation:
The stretch marks on the hips, stomach and breast of pregnant women is a very common problem. These stretch marks appear pinkish in color which is very difficult to treat.
These stretch marks are the result of the collagen to elastin ratio which increases and stretches. This stretching happens due to the sudden or rapid increase in the weight of the body or the affected portion.
Thus, the selected option is correct.
Answer:
a. the environmental variance (VE) = 3.5 g ²
b. 17.5 g ² for population
c. the heritability of broad sense (H2) = 0.83
Explanation:
a.With the information we have we can infer that environmental factors influence A, which is considered isogenic, thus, the environmental variance (VE) = 3.5 g ²
b.When studying population B, comparing it without environmental changes with respect to population A, we found that its total variance (VT) = 21.0g ²
We generate the following formula with the data obtained previously to find the genetic variation
VT = VE + VG
then VG = VT-VE
replacing data:
VG = 21.0 - 3.5 = 17.5 g ² for population
c. Regarding the heritability of broad sense (H2) in population B, we can reach the result with the data previously obtained like this:
H2 = VG / VT = 17.5 / 21.0 = 0.83
