Explanation:
The universal wave equation states that the velocity of a wave is the product of its wavelength and its frequency.
Mathematically speaking,
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
v
=
λ
f
a
a
∣
∣
−−−−−−−−−−−
where
−−−−−
:
v
=
velocity of wave (metres per second)
λ
=
wavelength (metres)
f
=
frequency (hertz or reciprocal seconds)
Step 1
Convert the given wavelength from centimetres into metres.
3.0
i
c
m
×
m
100
i
c
m
=
3.0
c
m
×
m
100
c
m
=
0.030
i
m
Step 2
Rearrange for frequency,
f
, in the universal wave equation.
v
=
λ
f
Becomes,
f
=
v
λ
Step 3
Substitute your known values to solve for the frequency of the pulse.
f
=
1500
i
m
s
0.030
i
m
f
=
50000
i
H
z
Rounding off the value to two significant figures,
f
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
5.0
×
10
4
i
H
z
a
a
∣
∣
−−−−−−−−−−−−−−−−
Before DNA can be replicated, the double stranded molecule must be “unzipped” into two single strands. DNA has four bases called adenine (A), thymine (T), cytosine (C) and guanine (G) that form pairs between the two strands. Adenine only pairs with thymine and cytosine only binds with guanine. In order to unwind DNA, these interactions between base pairs must be broken. This is performed by an enzyme known as DNA helicase. DNA helicase disrupts the hydrogen bonding between base pairs to separate the strands into a Y shape known as the replication fork. This area will be the template for replication to begin.
Step 2: Primer Binding
The leading strand is the simplest to replicate. Once the DNA strands have been separated, a short piece of RNA called a primer binds to the 3' end of the strand. The primer always binds as the starting point for replication. Primers are generated by the enzyme DNA primase.
Step 3: Elongation
Enzymes known as DNA polymerases are responsible creating the new strand by a process called elongation. There are five different known types of DNA polymerases in bacteria and human cells. In bacteria such as E. coli, polymerase III is the main replication enzyme, while polymerase I, II, IV and V are responsible for error checking and repair. DNA polymerase III binds to the strand at the site of the primer and begins adding new base pairs complementary to the strand during replication. In eukaryotic cells, polymerases alpha, delta, and epsilon are the primary polymerases involved in DNA replication. Because replication proceeds in the 5' to 3' direction on the leading strand, the newly formed strand is continuous.
Step 4: Termination
Once both the continuous and discontinuous strands are formed, an enzyme called exonuclease removes all RNA primers from the original strands. These primers are then replaced with appropriate bases. Another exonuclease “proofreads” the newly formed DNA to check, remove and replace any errors. Another enzyme called DNA ligase joins Okazaki fragments together forming a single unified strand. The ends of the linear DNA present a problem as DNA polymerase can only add nucleotides in the 5′ to 3′ direction. The ends of the parent strands consist of repeated DNA sequences called telomeres. Telomeres act as protective caps at the end of chromosomes to prevent nearby chromosomes from fusing.
So if that here are the functions of enzymes used:
DNA helicase - unwinds and separates double stranded DNA as it moves along the DNA. It forms the replication fork by breaking hydrogen bonds between nucleotide pairs in DNA.
DNA primase - a type of RNA polymerase that generates RNA primers. Primers are short RNA molecules that act as templates for the starting point of DNA replication.
DNA polymerases - synthesize new DNA molecules by adding nucleotides to leading and lagging DNA strands.
Topoisomerase or DNA Gyrase - unwinds and rewinds DNA strands to prevent the DNA from becoming tangled or supercoiled.
Exonucleases - group of enzymes that remove nucleotide bases from the end of a DNA chain.
DNA ligase - joins DNA fragments together by forming phosphodiester bonds between nucleotides.
Have a nice day
Answer: D- Parasitism
Explanation: Parasitism is an intimate and long-term impact or interaction that a couple of organisms living in the same habitat have on themselves between two distinct living beings, where one organism lives on another living being which gives it food, eater and shelter at the expense of the other organism.
The organism that lives in the host organism causes it some damage and gives no benefits at all. Also, the parasites are usually insignificant in size when compared to the organisms they live and feed on so they are able to survive in a large category of environmental situations. In this case, the dwarf mistletoes are the parasites while the trees are the hosts.
option C is correct........
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