Answer D. Because the DNA is different
I. treating the substance with nucleases that destroy all nucleic acids and then determining whether it is still infectious
Explanation:
Analysis of the nature of the specimen like the infectious agents (bacterium, virus, viriod, or prion) is very important to diagnose the diseases caused by them, investigate disease outbreak, to determine the clinical course of the infectious disease etc.
There are many methods like direct detection, cultivation assays, serological assays to detect the pathogen type.
Molecular techniques like polymerase chain reaction (PCR) is done to further analyze or investigate the pathogen, like its microbial genome, the virulence factor etc. These are amplification methods which help to investigate the pathogenesis and disease progression. PCR method involves denaturing the DNA polymerase which is a nuclease type of enzyme. Nucleases are present with exonucleases and endonucleases, both of which acts at specific points on the DNA of the pathogen and will create new templates .
Answer: • Most prokaryotes lack well-defined organelles. They can move from one place to another using cilia.
• The presence of peptidoglycan in their cell walls allows them to be visible on staining.
Explanation:
A prokaryote lacks a nucleus that's enveloped enclosed. Also, most of them doesn't have well-defined organelles.
The presence of peptidoglycan in their cell walls allows them to be visible on staining and they can move by using cilia.
Answer:
= 
Explanation:
Given that the woman’s maternal grandfather is color-blind, then it is possible that the woman is a carrier of color-blind allele.
Then probability that the woman is a carrier of color-blind allele is = 
So if a carrier woman cross with a normal man
Let say the gene of the carrier woman is: 
and the normal man = 
so if we have a cross between and ; we have:
we can see the offspring to be :
=normal unaffected female
= normal unaffected male
= carrier female for color-blind
= carrier male for color-bind
Then probability that of inheriting defective allele = 
=
Let's not forget that the probability that the woman is a carrier of color-blind allele is =
∴ the probability that the son is colorblind = 
=
