Answer:
a.The phenotypic proportions obtained after having the genotypes are 50% marbled seeds, 25% spotted and dotted seeds since they are codominant, 25% spotted seeds.
b. Taking into account the F1 genotypes in the previous point, the expected phenotypes for the first crossing are 100% marbled seeds and for the second crossing 100% dotted seeds.
Explanation:
Let's suppose:
Marbled allele: M
Spotted allele: S
Dotted allele: D
Allele for Clear: C
a. Because both crosses were between homozygous parents, the entire F1 genotype is the same.
For the first crossing the descendants have the MS genotype, and for the second crossing the descendants have the DC genotype. It is enough to make a Punnett square to obtain the different combinations of genotypes between the crossing of MS and DC.
Answer:
The correct answer is C) 1 L of 1.0 M NaCl
Explanation:
NaCl is a ionic compound so it dissociates in water into Na⁺ and Cl⁻ ions. Glucose is a covalent solute so it does not dissociates into ions. So, when we dissolve NaCl we have twice the amount of particles in solution in comparison with glucose. According to this:
A) and B) are solutions with the same concentration (0.5 M) but NaCl solution will have more solute particles than glucose.
C) and D) are solutions with more solute amount because they are more concentrated (1 M), but NaCl solution will have more solute particles than glucose solution ( 1 mol of Na⁺ ions + 1 mol of Cl⁻ ions).
The solution with the greatest solute particle number is C).
Answer:
B
Explanation:
first it attaches to the host cell membrane
Answer:
Test Variable (independent variable)
Explanation:
X is always the independent variable/time. Test grades is dependent because it's the outcome. Trust me I took the test.
