List and define the significance of the following proteins/structures involved in DNA
replication: DNA polymerase III, DNA polymerase I, Primase, Single stranded binding proteins
(SSBPs), Helicase, Ligase, Primers, Topoisomerase, Anti-parallel ds DNA, Leading strand of
DNA, Lagging Strand of DNA, replication bubble, and deoxyribonucleotide tri-phosphates
(dNTPs). Describe the process of DNA replication using all of the terms that were listed (and
defined) above. Make sure you include the difference in how the leading strand and lagging
strand are synthesized. Also, include any differences (covered in class) between bacterial DNA
replication and eukaryotic DNA replication. Lastly, include a hand drawn diagram with each of
these structures and enzymes in different colors to illustrate how this process occurs.
Solution
DNA Replication: it is the biological process in which two identical replicas of DNA are
produced. It involves many steps and components.
Protein/ structures involved in DNA replication are as follows:
DNA polymerase III: it was discovered by Thomas Kornberg and Malcolm Gefter. It is an
enzyme which has its role to play in prokaryotic DNA replication. It has high processivity which
is the number of nucleotides added per binding event. It can also perform proof reading which is
the correction of wrongly added nucleotides is
DNA polymerase I: it was first discovered polymerase enzyme and Arthur Korenberg discovered
it. It does not have a role in replication but poses exonuclease activity that is removal of
nucleotides during DNA repair.
Primase: it is a type of RNA polymerase. This is very important in initiation of DNA replication
because no polymerase can synthesise a DNA without small single stranded RNA. Thus a
primase help in synthesising such small single stranded RNA complementary to the template
DNA. After initiation, this RNA piece is removed.
Single-stranded binding proteins (SSBPs): these are the class of proteins that prevent premature
binding of nucleotides to template DNA. They protect single stranded DNA from being digested
by nucleases and also prevents looping of single stranded DNA, so that other enzymes can
perform their function.
Helicases: these are the group of enzymes which help in separating two strands of DNA for the
process of replication.
Ligases: another class of enzymes which help in binding of two strands of DNA by catalysing
formation of phosphodiester bond between the two strands.
Primer: a short single stranded DNA/RNA that serves as a starting point for the DNA synthesis.
Primase helps in the formation of primer.
Topoisomerases: during replication DNA over wounds ahead the replication point making
further replication difficult, in such instance topoisomerase help in unwinding supercoiled DNA.
Anti-parallel double strand: DNA is made of two strands running in opposite directions and
hence anti parallel one runs from 5’ to 3’ and the other from 3’ to 5’.
Leading strand: DNA replication on leading stra.
DNA replication is fundamental process occurring in all living organism to copy their DNA. The process is called replication in sense that each strand of dsDNA serve as template for reproduction of complementary strand.
DNA replication is fundamental process occurring in all living organism to copy their DNA. The process is called replication in sense that each strand of dsDNA serve as template for reproduction of complementary strand.
Eukaryotic DNA replication is a conserved mechanism that restricts DNA replication to once per cell cycle. Eukaryotic DNA replication of chromosomal DNA is central for the duplication of a cell and is necessary for the maintenance of the eukaryotic genome.
DNA replication in eukaryotes occurs in three stages: initiation, elongation, and termination, which are aided by several enzymes. Because eukaryotic genomes are quite
complex, DNA replication is a very complicated process that involves several enzymes and other proteins. It occurs in three main stages: initiation, elongation, and termination.
Problem I. As shown in the pedigree below, a pair of first cousins (.pdfsivakumar19831
Problem I. As shown in the pedigree below, a pair of first cousins (I and K) marry an unrelated
pair of first cousins (J and L). Their children M and N produce child O.
Q1. Calculate the inbreeding coefficient (f) of O (express answers to Q1 and Q3 to three decimal
places). Assume that f= 0 for all ancestors. (Hint: To help you out, I have shaded individuals
who cannot contribute to autozygosity.) To show your work, use the letter names to indicate each
path/loop that contributes to autozygosity. For example: path 1 = M-I-E…
Q2. Interpret your result. What does this value of f indicate?
Q3. Recalculate f for child O when f=1/2 for great-great grandmother D (f=0 for all other
ancestors). Hint: After you have calculated all the loops above, you’ll adjust just the loop that
includes ancester D to account for her ancestral inbreeding using the formula presented in class
(and text). B 0271 M H K E
Solution
Path 1:
· O-M-I-E-A-F-K-N-O
· O-M-I-E-B-F-K-N-O
So, the probability that a copy of gene in A makes the way down to O via E and a copy of gene
in B makes the way down to O via F =(½)n = (½)7=0.00781. Here n= node=7
Path 2:
· O-N-L-H-D-G-J-M-O
· O-N-L-H-C-G-J-M-O
Similarly, like the first one the total probability in path 2 will be same that is 0.00781.
Thus inbreeding co-efficient = 0.00781+0.00781=0.01562.
2. Generally, the higher the level of inbreeding closer the coefficient of relationship approaches a
value of 1 and approaches a value of 0 for individuals with remote common ancestors.
As here the value of inbreeding co-efficient is approximately 0 so we can conclude that
1. the common ancestors are not inbred.
2. O has remote common ancestors means they are distantly related.
3. When a common ancestor is itself inbred the inbreeding coefficient is = (1/2)n (1+ inbreeding
co-efficient of common ancestor) ,
The probability that a copy of gene in D makes the way to O = (1/2)7 = 0.00781
Therefore, inbreeding co-efficient is = 0.00781 * (1 + 1/2) =0.01172.
Points are assigned both on the correctness of the answer and the com.pdfsivakumar19831
Points are assigned both on the correctness of the answer and the completeness of the problem -
show all givens, state all assumptions, show all calculations. 1. Transportation affects society in
many ways and is frequently in the news. Find a transportation news article for a local, regional,
or national source (newspaper, credible internet site, etc.). Provide a link to the article and
summarize the article and explain the importance.
Solution
Hello,
As stated I am providing you with a link to the *.Pdf file regarding Transportation and its role in
society.
https://nptel.ac.in/courses/105101087/downloads/Lec-3.pdf
This is a very credible source.
Transportation is a non separable part of any society. It exhibits a very close relation to the style
of life, the range and location of activities and the goods and services which will be available for
consumption. Advances in transportation has made possible changes in the way of living and the
way in which societies are organized and therefore have a great inuence in the development of
civilizations. This chapter conveys an understanding of the importance of transportation in the
modern society by presenting selected characteristics of existing transportation systems, their use
and relationships to other human activities. Transportation is responsible for the development of
civilizations from very old times by meeting travel requirement of people and transport
requirement of goods. Such movement has changed the way people live and travel. In developed
and developing nations, a large fraction of people travel daily for work,shopping and social
reasons. But transport also consumes a lot of resources like time,fuel, materials and land..
More Related Content
Similar to List and define the significance of the following proteinsstructures.pdf
Eukaryotic DNA replication is a conserved mechanism that restricts DNA replication to once per cell cycle. Eukaryotic DNA replication of chromosomal DNA is central for the duplication of a cell and is necessary for the maintenance of the eukaryotic genome.
DNA replication in eukaryotes occurs in three stages: initiation, elongation, and termination, which are aided by several enzymes. Because eukaryotic genomes are quite
complex, DNA replication is a very complicated process that involves several enzymes and other proteins. It occurs in three main stages: initiation, elongation, and termination.
Problem I. As shown in the pedigree below, a pair of first cousins (.pdfsivakumar19831
Problem I. As shown in the pedigree below, a pair of first cousins (I and K) marry an unrelated
pair of first cousins (J and L). Their children M and N produce child O.
Q1. Calculate the inbreeding coefficient (f) of O (express answers to Q1 and Q3 to three decimal
places). Assume that f= 0 for all ancestors. (Hint: To help you out, I have shaded individuals
who cannot contribute to autozygosity.) To show your work, use the letter names to indicate each
path/loop that contributes to autozygosity. For example: path 1 = M-I-E…
Q2. Interpret your result. What does this value of f indicate?
Q3. Recalculate f for child O when f=1/2 for great-great grandmother D (f=0 for all other
ancestors). Hint: After you have calculated all the loops above, you’ll adjust just the loop that
includes ancester D to account for her ancestral inbreeding using the formula presented in class
(and text). B 0271 M H K E
Solution
Path 1:
· O-M-I-E-A-F-K-N-O
· O-M-I-E-B-F-K-N-O
So, the probability that a copy of gene in A makes the way down to O via E and a copy of gene
in B makes the way down to O via F =(½)n = (½)7=0.00781. Here n= node=7
Path 2:
· O-N-L-H-D-G-J-M-O
· O-N-L-H-C-G-J-M-O
Similarly, like the first one the total probability in path 2 will be same that is 0.00781.
Thus inbreeding co-efficient = 0.00781+0.00781=0.01562.
2. Generally, the higher the level of inbreeding closer the coefficient of relationship approaches a
value of 1 and approaches a value of 0 for individuals with remote common ancestors.
As here the value of inbreeding co-efficient is approximately 0 so we can conclude that
1. the common ancestors are not inbred.
2. O has remote common ancestors means they are distantly related.
3. When a common ancestor is itself inbred the inbreeding coefficient is = (1/2)n (1+ inbreeding
co-efficient of common ancestor) ,
The probability that a copy of gene in D makes the way to O = (1/2)7 = 0.00781
Therefore, inbreeding co-efficient is = 0.00781 * (1 + 1/2) =0.01172.
Points are assigned both on the correctness of the answer and the com.pdfsivakumar19831
Points are assigned both on the correctness of the answer and the completeness of the problem -
show all givens, state all assumptions, show all calculations. 1. Transportation affects society in
many ways and is frequently in the news. Find a transportation news article for a local, regional,
or national source (newspaper, credible internet site, etc.). Provide a link to the article and
summarize the article and explain the importance.
Solution
Hello,
As stated I am providing you with a link to the *.Pdf file regarding Transportation and its role in
society.
https://nptel.ac.in/courses/105101087/downloads/Lec-3.pdf
This is a very credible source.
Transportation is a non separable part of any society. It exhibits a very close relation to the style
of life, the range and location of activities and the goods and services which will be available for
consumption. Advances in transportation has made possible changes in the way of living and the
way in which societies are organized and therefore have a great inuence in the development of
civilizations. This chapter conveys an understanding of the importance of transportation in the
modern society by presenting selected characteristics of existing transportation systems, their use
and relationships to other human activities. Transportation is responsible for the development of
civilizations from very old times by meeting travel requirement of people and transport
requirement of goods. Such movement has changed the way people live and travel. In developed
and developing nations, a large fraction of people travel daily for work,shopping and social
reasons. But transport also consumes a lot of resources like time,fuel, materials and land..
Predefined Characters in C Type. Please make sure the program compil.pdfsivakumar19831
Predefined Characters in C Type. Please make sure the program compiles thanks :) // This
program reads one character from the keyboard and will // - convert it to uppercase, if it is
lowercase // - convert it to lowercase, if it is uppercase // - display a message, if it is not a letter
of the alphabet #include #include using namespace std; int main() char c coutc if (isalpha(c))
//check to see if it is a letter of alphabet if (isupper(c))//check to see if it is uppercase cout
Solution
#include
#include
using namespace std;
int main()
{
char c;
cin.get(c); //cin.get() input the character which can be of any type
if(isalpha(c))
{
if(isupper(c))
{
cout<<\"Your character \"<.
Objective Binary Search Tree traversal (2 points)Use traversal.pp.pdfsivakumar19831
Objective: Binary Search Tree traversal (2 points)
Use traversal.pptx as guidance to write a program to build a binary search tree Dictionary. Input
records from inventory.txt. Both key and Element of BST have the same data from each input
record.
public interface BinNode {
/** Get and set the element value */
public E element();
public void setElement(E v);
/** @return The left child */
public BinNode left();
/** @return The right child */
public BinNode right();
/** @return True if a leaf node, false otherwise */
public boolean isLeaf();
}
import java.lang.Comparable;
/** Binary Search Tree implementation for Dictionary ADT */
class BST, E>
implements Dictionary {
private BSTNode root; // Root of the BST
int nodecount; // Number of nodes in the BST
/** Constructor */
BST() { root = null; nodecount = 0; }
/** Reinitialize tree */
public void clear() { root = null; nodecount = 0; }
/** Insert a record into the tree.
@param k Key value of the record.
@param e The record to insert. */
public void insert(Key k, E e) {
root = inserthelp(root, k, e);
nodecount++;
}
// Return root
public BSTNode getRoot()
{
return root;
}
/** Remove a record from the tree.
@param k Key value of record to remove.
@return The record removed, null if there is none. */
public E remove(Key k) {
E temp = findhelp(root, k); // First find it
if (temp != null) {
root = removehelp(root, k); // Now remove it
nodecount--;
}
return temp;
}
/** Remove and return the root node from the dictionary.
@return The record removed, null if tree is empty. */
public E removeAny() {
if (root == null) return null;
E temp = root.element();
root = removehelp(root, root.key());
nodecount--;
return temp;
}
/** @return Record with key value k, null if none exist.
@param k The key value to find. */
public E find(Key k) { return findhelp(root, k); }
/** @return The number of records in the dictionary. */
public int size() { return nodecount; }
private E findhelp(BSTNode rt, Key k) {
if (rt == null) return null;
if (rt.key().compareTo(k) > 0)
return findhelp(rt.left(), k);
else if (rt.key().compareTo(k) == 0) return rt.element();
else return findhelp(rt.right(), k);
}
/** @return The current subtree, modified to contain
the new item */
private BSTNode inserthelp(BSTNode rt,
Key k, E e) {
if (rt == null) return new BSTNode(k, e);
if (rt.key().compareTo(k) > 0)
rt.setLeft(inserthelp(rt.left(), k, e));
else
rt.setRight(inserthelp(rt.right(), k, e));
return rt;
}
/** Remove a node with key value k
@return The tree with the node removed */
private BSTNode removehelp(BSTNode rt,Key k) {
if (rt == null) return null;
if (rt.key().compareTo(k) > 0)
rt.setLeft(removehelp(rt.left(), k));
else if (rt.key().compareTo(k) < 0)
rt.setRight(removehelp(rt.right(), k));
else { // Found it
if (rt.left() == null) return rt.right();
else if (rt.right() == null) return rt.left();
else { // Two children
BSTNode temp = getmin(rt.right());
rt.setElement(temp.element());
rt.setKey(temp.key());
rt.setRight(deletemin(rt.right(.
On January 1, 2017, Pinnacle Corporation exchanged $3,527,500 cash f.pdfsivakumar19831
On January 1, 2017, Pinnacle Corporation exchanged $3,527,500 cash for 100 percent of the
outstanding voting stock of Strata Corporation. On the acquisition date, Strata had the following
balance sheet:
Cash
$
284,000
Accounts payable
$
383,000
Accounts receivable
311,000
Long-term debt
3,060,000
Inventory
443,000
Common stock
1,500,000
Buildings (net)
1,920,000
Retained earnings
1,205,000
Licensing agreements
3,190,000
$
6,148,000
$
6,148,000
Pinnacle prepared the following fair-value allocation:
Fair value of Strata (consideration transferred)
$
3,527,500
Carrying amount acquired
2,705,000
Excess fair value
$
822,500
to buildings (undervalued)
$
480,000
to licensing agreements (overvalued)
(110,000
)
370,000
to goodwill (indefinite life)
$
452,500
At the acquisition date, Strata’s buildings had a 10-year remaining life and its licensing
agreements were due to expire in 5 years. At December 31, 2018, Strata’s accounts payable
included an $86,600 current liability owed to Pinnacle. Strata Corporation continues its separate
legal existence as a wholly owned subsidiary of Pinnacle with independent accounting records.
Pinnacle employs the initial value method in its internal accounting for its investment in Strata.
The separate financial statements for the two companies for the year ending December 31, 2018,
follow. Credit balances are indicated by parentheses.
Pinnacle
Strata
Sales
$
(7,683,000
)
$
(3,328,000
)
Cost of goods sold
4,940,000
1,835,000
Interest expense
338,000
207,000
Depreciation expense
630,000
448,000
Amortization expense
638,000
Dividend income
(35,000
)
Net income
$
(1,810,000
)
$
(200,000
)
Retained earnings 1/1/18
$
(5,160,000
)
$
(1,560,000
)
Net income
(1,810,000
)
(200,000
)
Dividends declared
600,000
35,000
Retained Earnings 12/31/18
$
(6,370,000
)
$
(1,725,000
)
Cash
$
282,000
$
547,000
Accounts receivable
1,255,000
305,000
Inventory
1,295,000
1,565,000
Investment in Strata
3,527,500
Buildings (net)
6,060,000
2,084,000
Licensing agreements
1,914,000
Goodwill
408,000
Total assets
$
12,827,500
$
6,415,000
Accounts payable
$
(337,500
)
$
(950,000
)
Long-term debt
(3,120,000
)
(2,240,000
)
Common stock
(3,000,000
)
(1,500,000
)
Retained earnings 12/31/18
(6,370,000
)
(1,725,000
)
Total Liabilities and OE
$
(12,827,500
)
$
(6,415,000
)
a. Prepare a worksheet to consolidate the financial information for these two companies.
PINNACLE COMPANY AND SUBSIDIARY STRATA
Consolidation Worksheet
For Year December 31, 2018
Consolidation Entries
Accounts
Pinnacle
Strata
Debit
Credit
Consolidated Totals
Sales
$(7,683,000)
$(3,328,000)
Cost of goods sold
4,940,000
1,835,000
Interest expense
338,000
207,000
Depreciation expense
630,000
448,000
Amortization expense
638,000
Dividend income
(35,000)
Net income
$(1,810,000)
$(200,000)
$0
Retained earnings 1/1/18
(5,160,000)
(1,560,000)
Net income
(1,810,000)
(200,000)
0
Dividends declared
600,000
35,000
Retained earnings 12/31/18
$(6,370,000)
$(1,725,000)
$0
Cash
$282,000
$547,000
Accounts rec.
Most of the functions of a cell membrane are performed by proteins ch.pdfsivakumar19831
Most of the functions of a cell membrane are performed by proteins cholesterol None of this
glycolysis Select all of the groups of membrane proteins All of the statements are correct
Solution
13) It is a transmembrane protein
14) Transmembrane protein pass through both sides of the lipid bilayer and protude from both
extracellular and cytoplasmic side of the membrane. It acts as a gateway to permit transport of
proteins across membrane.
17) Transmembrane proteins are amphipathic as they are present in the bilayer and also protrude
out of the membrane
12) Membrane proteins can be peripheral, intergral and lipid anchored..
Objectives 1. Understand the design, implementation, usage and limita.pdfsivakumar19831
Objectives 1. Understand the design, implementation, usage and limitation of a Linked List
ADT 2. Gain experience implementing abstract data t types using already developed data
structures Recognize and practice possible linked list applications. 3. Overview In this project,
you should write a simple line editor. Keep the entire text on a linked list, one line in a separate
node. Start the program with entering EDIT file, after which a prompt appears along with the line
number. If the letter I is entered with a number n following it, then insert the text to be followed
before line n. If I is not followed by a number, then insert the text before the current line. If D is
entered with two numbers n and m, one n, or no number following it, then delete lines n through
m, line n, or the current line. Do the same with the command L, which stands for listing lines. If
A is entered, then append the text to the existing lines. Entry E signifies exit and saving the text
in a file. Here is an Example: EDIT testfile 1> The first line 2> 3> And another line 4>I 3 3 The
second line 4> One more line 5> L 1> The first line 3 The second line 4> One more line 5 And
another line 5> D 2 4> L // This is now line 5, not 3; // line 5, since L was issued from line 5
//line 4, since one line was deleted; 1 The first line 2> The second line 3> One more line 4> And
another line 4> E // this and the following lines // now have new numbers
Solution.
New Nirvana Ltd is a company controlled by the members of the hard r.pdfsivakumar19831
New Nirvana Ltd is a company controlled by the members of the hard rock band, N/N. A
number of wholly owned subsidiaries of New Nirvana Ltd are involved in setting up and running
the band’s concerts. One of the subsidiary companies, Nuclear Blast Sounds Pty Ltd, is
responsible for setting up the sound equipment at N/N concerts in Australia. At a recent N/N
concert in Sydney, Nuclear Blast Sounds Pty Ltd negligently set the sound levels too 2 high with
the result that five audience members suffered permanent hearing loss. Unfortunately for those
audience members, Nuclear Blast Sounds Pty Ltd had no negligence insurance and cannot pay
the likely damages claims. Advise the injured audience members whether they can make New
Nirvana Ltd liable for Nuclear Blast Sounds Pty Ltd’s negligence.
Solution
Facts of the Case: Nuclear Blast Sounds Pty Ltd which is a wholly owned subsidiary of New
Nirvana Ltd negligently set the sound levels too high with the result that five audience members
suffered permanent hearing loss.
Case Law Reference:The Case of Chandler vs Cape plc has established that a parent enterprise
may be held liable for breaches of health and safety laws by a subsidiary Enterprise.
Explanation: In the given case, Nuclear Blast Sounds Pty Ltd is a wholly owned subsidiary of
New Nirvana Ltd.It naturally acts on the directions of holding company. Since the permanent
hearing loss caused to the audience is due to ngligence of Nuclear Blast Sounds Pty Ltd the
audience has full rights to sue New Nirvana Ltd if in case the subsidiary is unable to honour their
claims.
New Nirvana Ltd has no right to use Corporate Veil in this case as the subsidiary is a wholly
owned subsidiary and it acts on direction of New Nirvana Ltd.
Conclusion: Injured audience members can make New Nirvana Ltd liable for Nuclear Blast
Sounds Pty Ltd’s negligence..
Malaria being spread by mosquitoes is an example of which type of dis.pdfsivakumar19831
Malaria being spread by mosquitoes is an example of which type of disase
transmission? Vehicle-borne Zeonosis Fontice-borne Vector-borne The disease-
evoking power of a pathogen is called which of the following?
Invasiveness Variability Virulence Communicability
The likelihood of a pathogen or agent being trans-from one infected person to another susceptive
person is referred to as which of the following?
Invasiveness Variability Virulence Communicability
Metabolic diseases are best defined by which of the
following? A collective name that refers to a group
of many diseases with one common characteristic-
uncontrolled growth of mutated cells A lower level of mental, physical,
or moral state than is normal Diseases existing at birth and often before birth, or
that develop during the first month of life Caused by
the body reacting to an invasion of or injury by a
foreign object or substance Any of the diseases or
disorders that disrupt the process of converting food
to energy on a cellular level; affects the ability of the
cell to perform critical biochemical reactions that
involve the processing or transport of proteins (amino acids),
carbohydrates (sugars and starches), or lipids (fatty acids) As a general rule,
a disease is included on a state\'s list of notifiable diseases under what
conditions? Causes serious morbidity or death Has
the potential to spread Can be controlled with
appropriate intervention All of the above conditions
warrant reporting the disease. The natural course of
communicable disease involves a susceptible host; a
point of exposure; a cube clinical disease phase; the
clinical disease phase; and a phase of recovery, disability, or death. Which phase is related to the
latency period? Susceptibility phase Subclinical phase Clinical disease phase Recovery,
disability, or death phase Which term refers to quarantine-type activities
that are conducted on an inpatient basis in a hospital
or nursing home? Quarantine Isolation Good
hygiene Immunization Chemoprophylaxis What herd
immunity threshold did Jonas Salk suggest would
make it infeasible for there to be a polio epidemic? 55% 65% 75% 85%
Solution
13. b)- zoonosis
14. c)-virulence
15. d)- communicability
16. e)
17. d) - all of the above conditions
18. b) - subclinical phase
19. b) - isolation
20. d) - 85%.
In oxidative phosphorylation and phosphorylation, the enzyme ATP synt.pdfsivakumar19831
In oxidative phosphorylation and phosphorylation, the enzyme ATP syntheses couples the
synthesis of ATP to which one of the following processes? A. The facilitated diffusion of
protons B. The splitting of water with the release of O_2 C. The excilation of the Chlorophyll
reaction center D. The reduction of NADP\' through NADP* reductase E. carboxylation of
RuBP using the enzyme rubisco During which process(es) does the splitting of carbon dioxide
to form oxygen gas and carbon A. photosynthesis B. both photosynthesis and respiration C.
respiration D. neither photosynthesis nor respiration E. photorespiration. Which of the
following describes a property of water necessary for life? A. Water changes temperature
rapidly. B. Water is non-cohesive. C. Water is most dense as a solid. D. Water has constant
high pH. E. Water can dissolve many chemicals. What is the lowest level of biological
organization that can perform all of the activities required for life? A. organelle B. molecule C.
cell D. tissue E. organism
Solution
30. The proton gradient generated during electron transport drives the synthesis of ATP through
ATP synthase. This process is similar in both oxidative phosphorylation and
photophosphorylation.
Excitation of the chlorophyll reaction center, the splitting of water with the release of O2 and, the
reduction of NADP+ through NADP+ reductase occur only during the process of
photophosphorylation.
Carboxylation of RuBP using the enzyme rubisco takes place during Calvin cycle.
Therefore, option A is correct.
The enzyme ATP synthase couples the synthesis of ATP to the facilitated diffusion of protons
during oxidative phosphorylation, and photophosphorylation.
31. Photosynthesis is the process in which light energy is used to reduce carbon dioxide to yield
carbohydrates.
During respiration, oxygen is consumed and carbon dioxide is released.
Therefore, option A is correct.
The splitting of carbon dioxide to form oxygen gas and yield carbohydrates occurs during the
process of photosynthesis.
32. Water helps to regulate the temperature of the body.
Water does not have high ph. It has a neutral ph.
Water can dissolve many chemicals. It helps to transport essential molecules and chemicals
throughout the body.
Therefore, option E is correct.
33. Cell is the lowest level of biological organization that can perform all of the activities
required for life.
Cells that perform the same function are grouped together to form tissues,
An aggregation of different tissues for a specialized function is called as an organ.
Therefore, option C is correct.
Cell is the lowest level of biological organization that can perform all of the activities required
for life..
In this TEM image, what does the TEM allow you to view that would no.pdfsivakumar19831
In this TEM image, what does the TEM allow you to view that would not be visible using the
Compound Microscope?
The human cell
SELECT ALL ANSWERS THAT APPLYThe structure of internal cell organellesThe Human
Immunodeficiency Virus
The human cell
SELECT ALL ANSWERS THAT APPLY
Solution
Answers are
The structure of internal cell organelles
And
The human immuno deficiency virus.
Transmission electron microscope TEM can be used to see extremely small things like cell
organelles and viruses. With compound Microscope we cannot see those. Human cell can be
seen with compound Microscope..
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2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
Synthetic Fiber Construction in lab .pptxPavel ( NSTU)
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
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List and define the significance of the following proteinsstructures.pdf
1. List and define the significance of the following proteins/structures involved in DNA
replication: DNA polymerase III, DNA polymerase I, Primase, Single stranded binding proteins
(SSBPs), Helicase, Ligase, Primers, Topoisomerase, Anti-parallel ds DNA, Leading strand of
DNA, Lagging Strand of DNA, replication bubble, and deoxyribonucleotide tri-phosphates
(dNTPs). Describe the process of DNA replication using all of the terms that were listed (and
defined) above. Make sure you include the difference in how the leading strand and lagging
strand are synthesized. Also, include any differences (covered in class) between bacterial DNA
replication and eukaryotic DNA replication. Lastly, include a hand drawn diagram with each of
these structures and enzymes in different colors to illustrate how this process occurs.
Solution
DNA Replication: it is the biological process in which two identical replicas of DNA are
produced. It involves many steps and components.
Protein/ structures involved in DNA replication are as follows:
DNA polymerase III: it was discovered by Thomas Kornberg and Malcolm Gefter. It is an
enzyme which has its role to play in prokaryotic DNA replication. It has high processivity which
is the number of nucleotides added per binding event. It can also perform proof reading which is
the correction of wrongly added nucleotides is
DNA polymerase I: it was first discovered polymerase enzyme and Arthur Korenberg discovered
it. It does not have a role in replication but poses exonuclease activity that is removal of
nucleotides during DNA repair.
Primase: it is a type of RNA polymerase. This is very important in initiation of DNA replication
because no polymerase can synthesise a DNA without small single stranded RNA. Thus a
primase help in synthesising such small single stranded RNA complementary to the template
DNA. After initiation, this RNA piece is removed.
Single-stranded binding proteins (SSBPs): these are the class of proteins that prevent premature
binding of nucleotides to template DNA. They protect single stranded DNA from being digested
by nucleases and also prevents looping of single stranded DNA, so that other enzymes can
perform their function.
Helicases: these are the group of enzymes which help in separating two strands of DNA for the
process of replication.
Ligases: another class of enzymes which help in binding of two strands of DNA by catalysing
formation of phosphodiester bond between the two strands.
Primer: a short single stranded DNA/RNA that serves as a starting point for the DNA synthesis.
Primase helps in the formation of primer.
2. Topoisomerases: during replication DNA over wounds ahead the replication point making
further replication difficult, in such instance topoisomerase help in unwinding supercoiled DNA.
Anti-parallel double strand: DNA is made of two strands running in opposite directions and
hence anti parallel one runs from 5’ to 3’ and the other from 3’ to 5’.
Leading strand: DNA replication on leading strand is continuous. The parent strand runs 3’ to 5’
direction and polymerases add nucleotides in 5’ to 3’ direction continuously.
Lagging strand: as polymerase cannot add nucleotides in 3’ to 5’ direction, one of the two parent
strands of DNA which runs from 5’ to 3’ direction, opens at multiple sites and nucleotides are
added. Thus DNA synthesise on lagging strand is not continuous and are produced like
fragments called Okazaki fragments.
Replication Bubble: A DNA has to open up for the process of replication. Small unzipped
portion of DNA is called replication bubble.
Deoxy Nucleoside triphosphate (dNTP): it is a monomer or single unit of DNA. Each dNTP
contains a nitrogenous base, a deoxyribose sugar, and one phosphate group.
