The best and most correct answer among the choices provided by the first question is the third choice or letter C "<span>principle of uniformitarianism."</span>
On the other hand, the best and most correct answer among the choices provided by the second question is the third choice or letter "Worms don't have hard parts."
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
C: The polar bear will probably not survive.
Multicellular organisms have various levels of organization within them. Individual cells may perform specific functions and also work together for the good of the entire organism. The cells become dependent on one another. I hope that makes sense
Answer:
Allele frequency
Normal allele 
Mutant r allele 
Genotype frequency
Homozygous normal bugs 
Homozygous mutant bug 
Heterozygous normal bug with one mutant r allele and one normal allele 
Explanation:
It is given that 99% of the bugs were killed after the spray of pyrethrum. This suggests that 1% of the bugs that were not killed must be homozygous for the mutant type allele "r"
Thus, the frequency of homozygous "rr" species i.e 
From this we can evaluate the frequency of mutant "r" allele.
Thus, 

As per Hardy-Weinberg first equilibrium equation, 
Substituting the value of q in above equation, we get

Thus, the frequency of homozygous normal bug is equal to

As per Hardy-Weinberg second equilibrium equation-

Substituting all the available values we get -

Allele frequency
Normal allele 
Mutant r allele 
Genotype frequency
Homozygous normal bugs 
Homozygous mutant bug 
Heterozygous normal bug with one mutant r allele and one normal allele 