So I don’t see answer choices here, but your answer is 50% of the offspring will be homozygous dominant with RR, and 100% of them will carry a homozygous dominant gene of Rr
If you take the two sets and put them into a punnett square, it would look like this (image attached):
When the two sets of alleles are crossed, you would end up with half of your pairs being fully dominant (RR), and the other half being dominant while containing a recessive gene (Rr). Since there’s only one recessive gene in these pairs, it gets overridden and the pair itself is dominant.
So your answer is 50% will be homozygous dominant with RR!
There is 50% probability that their daughter's son will have the disease, because daughter chromosomes are XX , and the X from father will have x- linked disease, now in next generation the son will have XY chromosomes, Y is from father and X will from mother, now mother have two "XX", out of these two one carry x- linked disease, that means 50 percent chances of having disease in son.
Answer:
700
Explanation:
If from the question the price per unit of fertilizer, p(x) = 300 - 0.1x
The cost of x units of fertilizer = C(x) = 15000 + 125x + 0.025x^2
And we know that Profit = Revenue - Cost
Revenue for x units of fertilizer R(X) = x*p(x)
= 300x - 0.1x^2
Hence Profit: P(x) = R(x) - C(x)
= [300x - 0.1x^2] - [15000 + 125x + 0.025x^2]
= -0.125x^2 + 175x - 15000
To determine critical values
P'(x) = -0.25x + 175 = 0
Thus x = 175/0.25
x = 700
To maximize profit, the manufactures must produce 700 units of fertilizers