Answer:
17
Explanation:
In corn,
Smooth kernel (S) is dominant to rough kernel (s)
Purple kernel (P) is dominant to yellow kernel (p)
Since no information has been given about the cross here let us assume it to be a standard dihybrid cross.
The phenotypic ratio of dihybrid cross is 9:3:3:1. 9/16 will show dominant phenotype for both traits, 3/16 will show dominant phenotype for one trait, other 3/16 will show dominant phenotype for the second trait, 1/16 will have recessive phenotype for both the traits.
Here,
Purple rough (P_ss) = 52
Yellow smooth (ppS_) = 50
Both of them show dominant phenotype for one trait so they each form 3/16 of the progeny.
If the total progeny was z, 3/16*z = 50
z = (50*16)/3 = 267
Total progeny = 267 kernels
Yellow rough kernels (ppss) = (1/16*267) = 17
So approximately 17 corn kernels will be yellow and rough.
An amylase is an enzyme that catalyses the hydrolysis of starch into a sugar.
1 mol 2 mol
84g of N2(g) = 84/28 = 3 mol of N2(g)
From the above reaction 1 mol of N2(g) gives 2 mol 0f NH3(g). So 3 mol of N2(g) should give 6 mol or (6x17) 102 g of NH3(g). But we are getting only 85.0 g of NH3.
Hence %yield of NH3 = 100x85.0/102 = 83.3%
Answer D. is correct.
2. 2HgO (s) --------------------- 2Hg (l) + O2 (g)
2 mol 2 mol 1 mol
4 mol 4 mol 2 mol
As only 1.5 mol of O2 is obtained, so % yield = 1.5x100/2 = 75.0%
Moreover 603 g of Hg = 603/200.59 = 3.00 molof Hg. Considering this also
% yield = 3x100/4 = 75%
The answer would be 36. They would be ATP Molecules formed by glycolysis, link reaction, and the Krebs cycle.
