Answer:
4m/s2
Explanation:
took test and got it right
...require a living host cell to survive.
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:
they have cell walls as an outer boundry
Explanation:
the cells of all living things do not have cell wall. animal cells do not have cell wall. protozoans also do not have cell wall. while cells of plants algae fungi and bacteria have cell wall.
