<span>D. earthworm would be classified in the same kingdom as the hawk (both animals). </span>
Answer:
The mentioned parental types are c+m- and c-m+. Thus, the recombinants will be c+m+ and c-m-.
Now, the given distance between c and m is 8 map units. Thus, the recombinant frequency is 8% or 0.08.
The total recombinants from 1000 plaques will come out to be 80,
Thus, the recombinants of each type will be 40.
Total parental type will be 920, and therefore, each parental type count will be 460.
Thus, expected c+m- = 460, expected c-m+ = 460, expected c+m+ = 40 and expected c-m- = 40.
Answer:
C. 1/4
Explanation:
Let's assume that the allele for the spotted coat is "S" and the one for the even coat is "s". The allele "L" gives short horns while the recessive allele "l" imparts long horns. The genotype of the cattle heterozygous for both traits would be SsLl. A cross between two heterozygous cattle would produce progeny in following phenotype ratio=
9 spotted coat and short horns: 3 even coat and short horns: 3 spotted coat and long horns: 1 even coat and long horns.
Therefore, the proportion of the progeny with long horns = 4/16= 1/4
The answer is D its tightly packed cells allow for proctection against harmful substances.