Mark leaves the house for school walking at the rate of 4 kph. Two h.pdf
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Mark leaves the house for school walking at the rate of 4 kph. Two hours later Joyce follows Mark at 4.5 kph. How long will it take Joyce to overtake Mark? Solution Mark\'s rate is 4 kph Mark\'s time time is t+2 Joyce\'s rate is 4.5kph Joyce\'s time is t when Joyce overtakes Mark, they will be at the same distance, so we can use the formula rate * time = distance and set their distance = to one another: 4(t+2)=4.5t 4t+8=4.5t 8=.5t 16=t this makes sense because in 16 hours time, Joyce will has walked 4.5*16 = 72km and in 18 hours time (he started two hours before her) Mark has walked 4*18=72km.
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how do mutations in the DNA sequence affect this type of restric.pdf
how do mutations in the DNA sequence affect this type of restriction enzyme analysis
how do mutations in the DNA sequence affect this type of restriction enzyme analysis
how do mutations in the DNA sequence affect this type of restriction enzyme analysis
Solution
A restriction enzyme or restriction endonuclease is an enzyme that cuts DNA at or near specific
recognition nucleotides sequences known as restriction sites. These enzymes are routinely used
for DNA modification in Laboratories and or a vital tool in molecular cloning. Restriction
enzymes are used to digest genomic DNA gene analysis by southern blot this technique allows
researchers to identify how many copies of a gene are present in the genome of one individual or
how many gene mutations have occured within a population. Restriction enzyme digestion has
played and continuous to play a major role in analysing the genetic changes in Cancer the
availability of 200 different restriction enzymes is recognising different sequences in DNA has
been invaluable in studying cancer genetics.Due to mutations in the DNA sequence the chances
of these studies are minimised definitely..
Fill out the following table, using examples were appropriate. S.pdf
Fill out the following table, using examples were appropriate.
Solution
CharacteristicLiverwortsMossesHorsetailsFernsBody formFlattened thallusUprightUpright
Above ground, flat/uprightPresence of
Roots, Stems
and leavesYes, stems and leaves not clearly differentiated Yes, allYes, Leaves reduced, usually
non-photosyntheticYes, allMethod of asexual reproductionGemmae and tips of forked thalli give
rise to progenyGemmaevegetative, via rhizomessporophyte, diploidMethod of sexual
reproductionMale organs: Antheridia
Female organs: Archegonia,
Production of male and female gametes; male sperms swim towards female eggs for
fertilizationMale organs: Antheridia
Female organs: Archegonia,
Production of male and female gametes; male sperms swim towards female eggs for
fertilization, formation of sporophyte after fertilizationspores germinate to form male and
female prothallifree-living gametophyte, haploid. Union of gametes to produce offspring.
Exercise 4-10Larkspur, Inc., opened an incorporated dental practic.pdf
Exercise 4-10
Larkspur, Inc., opened an incorporated dental practice on January 1, 2017. During the first month
of operations, the following transactions occurred.
Prepare the adjusting entries on January 31. Account titles are Accumulated
Depreciation—Equipment, Depreciation Expense, Service Revenue, Accounts Receivable,
Insurance Expense, Interest Expense, Interest Payable, Prepaid Insurance, Supplies, Supplies
Expense, Utilities Expense, and Accounts Payable. (If no entry is required, select \"No Entry\"
for the account titles and enter 0 for the amounts. Credit account titles are automatically indented
when the amount is entered. Do not indent manually.)
No.
Date
Account Titles and Explanation
Debit
Credit1.Performed services for patients who had dental plan insurance. At January 31, $770 of
such services was completed but not yet billed to the insurance companies.2.Utility expenses
incurred but not paid prior to January 31 totaled $780.3.Purchased dental equipment on January
1 for $86,500, paying $22,200 in cash and signing a $64,300, 3-year note payable (interest is
paid each December 31). The equipment depreciates $530 per month. Interest is $690 per
month.4.Purchased a 1-year malpractice insurance policy on January 1 for $24,000.5.Purchased
$2,650 of dental supplies (recorded as increase to Supplies). On January 31, determined that
$500 of supplies were on hand.
Solution
1 31-Jan Accounts receivable 770 Service revenue 770 2 31-Jan Utilities
expense 780 Accounts payable 780 3 31-Jan Depreciation expense 530
Accumulated depreciation-equipment 530 Interest expense 690 Interest payable
690 4 31-Jan Insurance expense 2000 =24000/12 Prepaid insurance 2000 5
31-Jan Supplies expense 2150 =2650-500 Supplies 2150.
Does human DNA possess operonsSolutionAnswerOperons (clust.pdf
Does human DNA possess operons?
Solution
Answer:
Operons (clusters of co-regulated genes with related functions) are a well-known feature of
prokaryotic genomes. Polycistronic transcription is a characteristic of bacteria and archaea,
where many of the genes are clustered in operons composed of two to more genes.
The genes of eukaryotes (including humans) are generally considered to be monocistronic, each
with its own promoter at the 5\' end and a transcription terminator at the 3\' end. However, it has
recently become clear that not all eukaryotic genes are transcribed monocistronically. Functional
gene clustering also occurs in eukaryotes, from yeasts to filamentous fungi, mammals,
nematodes, and plants.
These gene clusters therefore represent functional gene organizations with operon-like features,
although the genes are not usually transcribed as a single mRNA as is the case in prokaryotes.
Like bacterial operons, eukaryotic operons often result in co-expression of functionally related
proteins.
Thus, human DNA possess operon like properties but not the ideal operon system present in
prokaryotes..
Explain the relationship between population growth and social differ.pdf
Explain the relationship between population growth and social differentiation. How does
population growth results in social differentiation? In turn, how does social differentiation allow
for more population growth?
Solution
As the population grows, it is possible that there will be more social differentiation made.
Different ethic groups may come together in populations that are increasing. Social
differentiation is the distinction between system and environment to determine that society is a
complex system which replicates the system/environment distinction to form internal
subsystems. Population gowth has resulted both in intragroup and intergroup differentiation in
societies. Intragroup and intergroup differentiation are two basic types of social differentiation.
Intragroup differentiation represents a division of the group into subgroups that perform different
functions in the group without being superior or inferior to each other. Intergroup differentiation,
on the other hand, is represented by the millions of various organized, semiorganized, and “as if
organized” groups, or social systems, into which the total human population is divided and
structured. On the hand, as societies differentiate more and more, there is an urge to increase the
population strength of each individual society. So, over time, each society grows and this
increases the total population..
differentiate between Bergeyso Manual of systematic bacteriology a.pdf
differentiate between Bergey\'so Manual of systematic bacteriology and Bergey\'s Manual of
Determinative Bacteriology.
Solution
ANS:
David Berger published a book (1923) on bacterial classification based on bacterial species. The
book is named as “THE BERGEY’S MANUAL OF DETERMINATIVE BACTERIOLOGY”.
This book is the \"BIBLE\" of bacterial taxonomy. In his first edition he classified bacteria on
basis of phenetics, but the second is based on phylogenetic studies.
Bergey\'s Manual of systematic bacteriology:
In this bacteria’s are classified on the basis of genetic relationship and its evolution.
It explains eu- and archaebacteria phylogenetic information based on r RNA sequencing.
Bergey\'s Manual of Determinative Bacteriology:
In this bacteria’s are classified on the based on cell wall composition, morphology, biochemical
tests and differential staining etc...
According to this eubacteria and archaebacteria is identified by different identification schemes..
Describe examples of variant histones and the chromosomal structures.pdf
Describe examples of variant histones and the chromosomal structures and/or processes with
which each is associated?
Solution
Answer : Histones are the basic proteins that contain high amounts of basic amino acids such as
arginine and lysine. They play a role in packaging eukaryotic DNA into structural units called
nucleosomes. There are five types of histone proteins in eukaryotic chromosomes namely, H1,
H2A, H2B, H3 and H4. H1 is called the linker histone and highly conserved and the last four
histones in copies of two forms the histone octamer found at the centre of nucleosome. Apart
from its involvement in DNA packaging they also have a vital role in eukaryotic gene regulation
which is brought about by acetylation and deacetylation of histone proteins.
Create a system to simulate vehicles at an intersection. Assume th.pdf
\"Create a system to simulate vehicles at an intersection. Assume that there is one lane going in
each of four directions, with stoplights facing each direction. Vary the arrival average of vehicles
in each direction and the frequency of the light changes to view the \"behavior\" of the
intersection.\"
You canNOT use the Java built-in Queue or Deque class. Use an Array or Linked List
implementation of a queue as explained in the text. This means a jsjf package containing
QueueADT.java file with CircularArrayQueue.java or LinkedQueue.java and LinearNode.java.
Include user-defined exceptions and javadoc comments.
Be sure to test thoroughly, and document your results.
Be sure to label your outputs so that when I run your code I can readily understand what you\'re
doing.
Determine what is the optimal light-change time to maximize car throughput.
WANT KNOW HOW CAN I MAKE IT RUN!
import java.lang.Math;
import java.io.*;
import jsjf.*;
import jsjf.exceptions.*;
import java.util.*;
import java.net.*;
import java.io.*;
/**
*This class is where all action resides. It contains all the queues that represent the lanes at the
intersection.
* It also contains all methods that pertain to populating the intersection and moving the vehicles
in their
* retrospective directions.
*/
public class Simulator {
/** This queue will hold only vehicle objects that enter in the left lane on Main Street headed
East. */
private LinkedQueue EMainL = new LinkedQueue();
/** This queue will hold only vehicle objects that enter in the right lane on Main Street headed
East. */
private LinkedQueue EMainR = new LinkedQueue();
/** This queue will hold only vehicle objects that enter in the left lane on Main Street headed
West. */
private LinkedQueue WMainL = new LinkedQueue();
/** This queue will hold only vehicle objects that enter in the right lane on Main Street headed
West. */
private LinkedQueue WMainR = new LinkedQueue();
/** This queue will hold only vehicle objects that enter in the left lane on Church Street
headed North. */
private LinkedQueue NChurchL = new LinkedQueue();
/** This queue will hold only vehicle objects that enter in the right lane on Church Street
headed North. */
private LinkedQueue NChurchR = new LinkedQueue();
/** This queue will hold only vehicle objects that enter in the left lane on Church Street
headed South. */
private LinkedQueue SChurchL = new LinkedQueue();
/** This queue will hold only vehicle objects that enter in the right lane on Church Street
headed South. */
private LinkedQueue SChurchR = new LinkedQueue();
/** This integer value will keep track of the current time during the simulation.*/
int time = 0;
/** This integer value will be used to make sure only two vehicles move during each light
change for each lane. */
int i = 0;
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intersection. */
int vehicleNum = 1;
/** This creates a filewriter object that allows us for the ability to write streams o.
Recommended
how do mutations in the DNA sequence affect this type of restric.pdf
how do mutations in the DNA sequence affect this type of restriction enzyme analysis
how do mutations in the DNA sequence affect this type of restriction enzyme analysis
how do mutations in the DNA sequence affect this type of restriction enzyme analysis
Solution
A restriction enzyme or restriction endonuclease is an enzyme that cuts DNA at or near specific
recognition nucleotides sequences known as restriction sites. These enzymes are routinely used
for DNA modification in Laboratories and or a vital tool in molecular cloning. Restriction
enzymes are used to digest genomic DNA gene analysis by southern blot this technique allows
researchers to identify how many copies of a gene are present in the genome of one individual or
how many gene mutations have occured within a population. Restriction enzyme digestion has
played and continuous to play a major role in analysing the genetic changes in Cancer the
availability of 200 different restriction enzymes is recognising different sequences in DNA has
been invaluable in studying cancer genetics.Due to mutations in the DNA sequence the chances
of these studies are minimised definitely..
Fill out the following table, using examples were appropriate. S.pdf
Fill out the following table, using examples were appropriate.
Solution
CharacteristicLiverwortsMossesHorsetailsFernsBody formFlattened thallusUprightUpright
Above ground, flat/uprightPresence of
Roots, Stems
and leavesYes, stems and leaves not clearly differentiated Yes, allYes, Leaves reduced, usually
non-photosyntheticYes, allMethod of asexual reproductionGemmae and tips of forked thalli give
rise to progenyGemmaevegetative, via rhizomessporophyte, diploidMethod of sexual
reproductionMale organs: Antheridia
Female organs: Archegonia,
Production of male and female gametes; male sperms swim towards female eggs for
fertilizationMale organs: Antheridia
Female organs: Archegonia,
Production of male and female gametes; male sperms swim towards female eggs for
fertilization, formation of sporophyte after fertilizationspores germinate to form male and
female prothallifree-living gametophyte, haploid. Union of gametes to produce offspring.
Exercise 4-10Larkspur, Inc., opened an incorporated dental practic.pdf
Exercise 4-10
Larkspur, Inc., opened an incorporated dental practice on January 1, 2017. During the first month
of operations, the following transactions occurred.
Prepare the adjusting entries on January 31. Account titles are Accumulated
Depreciation—Equipment, Depreciation Expense, Service Revenue, Accounts Receivable,
Insurance Expense, Interest Expense, Interest Payable, Prepaid Insurance, Supplies, Supplies
Expense, Utilities Expense, and Accounts Payable. (If no entry is required, select \"No Entry\"
for the account titles and enter 0 for the amounts. Credit account titles are automatically indented
when the amount is entered. Do not indent manually.)
No.
Date
Account Titles and Explanation
Debit
Credit1.Performed services for patients who had dental plan insurance. At January 31, $770 of
such services was completed but not yet billed to the insurance companies.2.Utility expenses
incurred but not paid prior to January 31 totaled $780.3.Purchased dental equipment on January
1 for $86,500, paying $22,200 in cash and signing a $64,300, 3-year note payable (interest is
paid each December 31). The equipment depreciates $530 per month. Interest is $690 per
month.4.Purchased a 1-year malpractice insurance policy on January 1 for $24,000.5.Purchased
$2,650 of dental supplies (recorded as increase to Supplies). On January 31, determined that
$500 of supplies were on hand.
Solution
1 31-Jan Accounts receivable 770 Service revenue 770 2 31-Jan Utilities
expense 780 Accounts payable 780 3 31-Jan Depreciation expense 530
Accumulated depreciation-equipment 530 Interest expense 690 Interest payable
690 4 31-Jan Insurance expense 2000 =24000/12 Prepaid insurance 2000 5
31-Jan Supplies expense 2150 =2650-500 Supplies 2150.
Does human DNA possess operonsSolutionAnswerOperons (clust.pdf
Does human DNA possess operons?
Solution
Answer:
Operons (clusters of co-regulated genes with related functions) are a well-known feature of
prokaryotic genomes. Polycistronic transcription is a characteristic of bacteria and archaea,
where many of the genes are clustered in operons composed of two to more genes.
The genes of eukaryotes (including humans) are generally considered to be monocistronic, each
with its own promoter at the 5\' end and a transcription terminator at the 3\' end. However, it has
recently become clear that not all eukaryotic genes are transcribed monocistronically. Functional
gene clustering also occurs in eukaryotes, from yeasts to filamentous fungi, mammals,
nematodes, and plants.
These gene clusters therefore represent functional gene organizations with operon-like features,
although the genes are not usually transcribed as a single mRNA as is the case in prokaryotes.
Like bacterial operons, eukaryotic operons often result in co-expression of functionally related
proteins.
Thus, human DNA possess operon like properties but not the ideal operon system present in
prokaryotes..
Explain the relationship between population growth and social differ.pdf
Explain the relationship between population growth and social differentiation. How does
population growth results in social differentiation? In turn, how does social differentiation allow
for more population growth?
Solution
As the population grows, it is possible that there will be more social differentiation made.
Different ethic groups may come together in populations that are increasing. Social
differentiation is the distinction between system and environment to determine that society is a
complex system which replicates the system/environment distinction to form internal
subsystems. Population gowth has resulted both in intragroup and intergroup differentiation in
societies. Intragroup and intergroup differentiation are two basic types of social differentiation.
Intragroup differentiation represents a division of the group into subgroups that perform different
functions in the group without being superior or inferior to each other. Intergroup differentiation,
on the other hand, is represented by the millions of various organized, semiorganized, and “as if
organized” groups, or social systems, into which the total human population is divided and
structured. On the hand, as societies differentiate more and more, there is an urge to increase the
population strength of each individual society. So, over time, each society grows and this
increases the total population..
differentiate between Bergeyso Manual of systematic bacteriology a.pdf
differentiate between Bergey\'so Manual of systematic bacteriology and Bergey\'s Manual of
Determinative Bacteriology.
Solution
ANS:
David Berger published a book (1923) on bacterial classification based on bacterial species. The
book is named as “THE BERGEY’S MANUAL OF DETERMINATIVE BACTERIOLOGY”.
This book is the \"BIBLE\" of bacterial taxonomy. In his first edition he classified bacteria on
basis of phenetics, but the second is based on phylogenetic studies.
Bergey\'s Manual of systematic bacteriology:
In this bacteria’s are classified on the basis of genetic relationship and its evolution.
It explains eu- and archaebacteria phylogenetic information based on r RNA sequencing.
Bergey\'s Manual of Determinative Bacteriology:
In this bacteria’s are classified on the based on cell wall composition, morphology, biochemical
tests and differential staining etc...
According to this eubacteria and archaebacteria is identified by different identification schemes..
Describe examples of variant histones and the chromosomal structures.pdf
Describe examples of variant histones and the chromosomal structures and/or processes with
which each is associated?
Solution
Answer : Histones are the basic proteins that contain high amounts of basic amino acids such as
arginine and lysine. They play a role in packaging eukaryotic DNA into structural units called
nucleosomes. There are five types of histone proteins in eukaryotic chromosomes namely, H1,
H2A, H2B, H3 and H4. H1 is called the linker histone and highly conserved and the last four
histones in copies of two forms the histone octamer found at the centre of nucleosome. Apart
from its involvement in DNA packaging they also have a vital role in eukaryotic gene regulation
which is brought about by acetylation and deacetylation of histone proteins.
Create a system to simulate vehicles at an intersection. Assume th.pdf
\"Create a system to simulate vehicles at an intersection. Assume that there is one lane going in
each of four directions, with stoplights facing each direction. Vary the arrival average of vehicles
in each direction and the frequency of the light changes to view the \"behavior\" of the
intersection.\"
You canNOT use the Java built-in Queue or Deque class. Use an Array or Linked List
implementation of a queue as explained in the text. This means a jsjf package containing
QueueADT.java file with CircularArrayQueue.java or LinkedQueue.java and LinearNode.java.
Include user-defined exceptions and javadoc comments.
Be sure to test thoroughly, and document your results.
Be sure to label your outputs so that when I run your code I can readily understand what you\'re
doing.
Determine what is the optimal light-change time to maximize car throughput.
WANT KNOW HOW CAN I MAKE IT RUN!
import java.lang.Math;
import java.io.*;
import jsjf.*;
import jsjf.exceptions.*;
import java.util.*;
import java.net.*;
import java.io.*;
/**
*This class is where all action resides. It contains all the queues that represent the lanes at the
intersection.
* It also contains all methods that pertain to populating the intersection and moving the vehicles
in their
* retrospective directions.
*/
public class Simulator {
/** This queue will hold only vehicle objects that enter in the left lane on Main Street headed
East. */
private LinkedQueue EMainL = new LinkedQueue();
/** This queue will hold only vehicle objects that enter in the right lane on Main Street headed
East. */
private LinkedQueue EMainR = new LinkedQueue();
/** This queue will hold only vehicle objects that enter in the left lane on Main Street headed
West. */
private LinkedQueue WMainL = new LinkedQueue();
/** This queue will hold only vehicle objects that enter in the right lane on Main Street headed
West. */
private LinkedQueue WMainR = new LinkedQueue();
/** This queue will hold only vehicle objects that enter in the left lane on Church Street
headed North. */
private LinkedQueue NChurchL = new LinkedQueue();
/** This queue will hold only vehicle objects that enter in the right lane on Church Street
headed North. */
private LinkedQueue NChurchR = new LinkedQueue();
/** This queue will hold only vehicle objects that enter in the left lane on Church Street
headed South. */
private LinkedQueue SChurchL = new LinkedQueue();
/** This queue will hold only vehicle objects that enter in the right lane on Church Street
headed South. */
private LinkedQueue SChurchR = new LinkedQueue();
/** This integer value will keep track of the current time during the simulation.*/
int time = 0;
/** This integer value will be used to make sure only two vehicles move during each light
change for each lane. */
int i = 0;
/** This integer value will keep track of the vehicle numbers of the cars that arrive at the
intersection. */
int vehicleNum = 1;
/** This creates a filewriter object that allows us for the ability to write streams o.
Darwin believed that women were less evolved than men, and because o.pdf
Darwin believed that women were less evolved than men, and because of their smaller brains
were eternally primitive. He argued that male intelligence was consistently sharpened through
the struggle for possession of females through sexual selection and through hunting and other
activities such as defending females and young which is natural selection. Men had ultimately
become superior to women.
Address each of the following: What do you think were the implications of these statements from
Darwin for women in science and mathematics following their publication in 1871? Do you
think it helped or made it worse for women trying to break into this male dominated
environment? What types of implications might these statements have caused? Do you think
Darwin\'s statements have lingering influences in today\'s science profession? Describe.
Solution
As indicated by Darwinian hypothesis, ladies were less developed than men, and due to their
littler brains, they were \"forever primitive,\" virtuous, less otherworldly, more materialistic, and
\"a genuine risk to contemporary civilization.\" The assumed insight crevice that many driving
Darwinists accepted existed between human guys and females was large to the point that some
driving Darwinists characterized them as two unmistakable species—guys as Homo frontalis and
females as Homo parietalis. The distinctions were great to the point that Darwin was
flabbergasted \"such extraordinary creatures have a place with the same species.\"
Purposes behind male predominance incorporated the conclusion that war and chasing pruned
the weaker men, permitting just the most fit to return home and reproduce. Ladies, conversely,
were not subject to these determination weights but rather were ensured by men, permitting the
frail to survive. Darwin\'s compositions and those of his devotees uncover that the conviction of
ladies inadequacy was vital to early transformative hypothesis. In The Descent of Man Darwin
contended that grown-up females of most species looked like the youthful of both genders and
that \"guys are more developmentally progressed than females.
This corrupting perspective of ladies quickly spread to Darwin\'s logical and scholarly peers
amid the late Victorian period and intensely affected numerous scholars who had a noteworthy
part in forming the accompanying era—from Sigmund Freud to Havelock Ellis. Evelleen
Richards infers that Darwin\'s perspectives about ladies coherently took after from his
hypothesis, \"in this manner feeding a few eras of logical sexism.\"Importantly, subsequently of
Darwin\'s thoughts, researchers were roused to utilize science to manufacture bolster for the
position that ladies were \"plainly mediocre and irreversibly subordinate\" to men, hence
affecting all of society. Darwinian intuition upheld the possibility that men were better than
ladies, which was at that point inside many societies. Darwin essentially offered belief to this
unbiblical instructing.
a) become more similar to one another. b) become more different th.pdf
a) become more similar to one another.
b) become more different than the mean.
c) cluster around the mean.
d) are more accurately represented by the mean.As the standard deviation gets larger, the scores
in the distribution
a) become more similar to one another.
b) become more different than the mean.
c) cluster around the mean.
d) are more accurately represented by the mean.
Solution
b) become more different than the mean.
Why are the shape and slope of a dose-response curve importantS.pdf
Why are the shape and slope of a dose-response curve important?
Solution
Answer:
It is important for to predict the toxicity of a substance at specific dose levels. Major differences
among drugs may exist not only in the point at which the threshold is reached but also in the
percent of population responding per unit change in dose (i.e., the slope). The slope of the dose
response curve represents rate of buildup of toxic effects..
When streaking a plate, the idea is to dilute the bacteria sufficie.pdf
When streaking a plate, the idea is to dilute the bacteria sufficiently so that by the 3^rd phase
of the plate you have isolated colonies; a plate contains sufficient surface area to allow isolated
colonies to appear, whereas a slant does not. _ Enrichment media contains substances that
various bacteria require for growth; blood agar is considered enrichment media as it detects
hemolysins.
Solution
the answers are true, streaking provides more surface area for the microbes and we can isolate
individual colonies but a slant cannot. In a slant we can store the bacterias for long term.
Enrichment media is enriched with specific nutrients to support the growth of specific
microorganism and inhibits other microbes.Blood agar is a enriched midia.
Why is it bad for a group to be polyphyleticWhat type of traits c.pdf
Why is it bad for a group to be polyphyletic?
What type of traits can be used in a cladistic analysis?
Solution
Polyphyletic group doesnot involve grouping of all the desendents of a common ancestor. This
kind of grouping arises when analogous characters of the organisms were misunderstood. The
organisms placed under polyphyletic group have similar characteristics, however they donot
share these characters from a common ancestor.
Cladistic analysis takes into consideration of shared characters i.e; synapomorphies, evolutionary
relationship, common ancestor.
When recruiting a sample of individuals to participate in a study on.pdf
When recruiting a sample of individuals to participate in a study on reaction time, researchers
used the most recent census to identify local older adults, who were
between the ages of 60 and 65, between 65 and 70, between 70 and 75, between 75 and
80, and between 80 and 85. Once they had identified these subgroups, the researchers randomly
selected 20 people from within each age group to participate in the study. What kind of sampling
method did the researchers use?
a. convenience sampling
b. random sampling
c. stratified sampling
d. cluster sampling
Solution
d. cluster sampling.
Where does the Calvin cycle take place stroma of the chloroplast t.pdf
Where does the Calvin cycle take place? stroma of the chloroplast thylakoid membrane
thylakoid space outer membrane of the chloroplast cytoplasm In a plant cell, where are ATP
synthase complexes located? A. thylakoid membrane B. Inner mitochondrial membrane C.
plasma membrane D. A & B E. A, B, and C
Solution
9. Stroma of the chloroplasts Calvin cycle takes place inside the stroma of the chloroplasts. 10.
The answer is D. Both Aand B. In plants ATP synthase complexes are present in the thylakoids
membrane and in the inner mitochondrial membrane..
What are the pros and cons of using the mode as your measure of cent.pdf
What are the pros and cons of using the mode as your measure of central tendency?
Solution
The mode is that observation that occurs most often.
It is usually not unique, and is therefore not that often used.
Advantages :
1) It has the advantage that it applies to numerical as well as categorical variables.
2) As with the median, the mode is easy to find if the data is small and sorted.
Disadvantages :
1) The mode only refers to the element that ocurs the maximum number of times. So, it doe snot
apply when the data set has many outliers or if the data set has a very large number of elements.
We presented a list of capabilities that should be provided by the DB.pdf
We presented a list of capabilities that should be provided by the DBMS software to the DBA,
database designers, and end users to help them design, administer, and use a database. Then we
gave a brief historical perspective on the evolution of database applications. We pointed out the
recent rapid growth of the amounts and types of data that must be stored in databases, and we
discussed the emergence of new systems for handling \"big data\" applications. Finally, we
discussed the overhead costs of using a DBMS and discussed some situations in which it may not
be advantageous to use one. Define the following terms: data, database, DBMS, database
system, database catalog, program-data independence, user view, DBA, end user, canned
transaction, deductive database system, persistent object, meta-data, and transaction-processing
application. What four main types of actions involve databases? Briefly discuss each. Discuss
the main characteristics of the database approach and how it differs from traditional file systems.
What are the responsibilities of the DBA and the database designers? What are the different
types of database end users? Discuss the main activities of each. Discuss the capabilities that
should be provided by a DBMS. Discuss the differences between database systems and
information retrieval systems. Identify some informal queries and update operations that you
would expect to apply to the database shown in Figure 1.2. What is the difference between
controlled and uncontrolled redundancy? Illustrate with examples. Specify all the relationships
among the records of the database shown in Figure 1.2.
Solution
I.
Data: data is simply called as collection of objects.
Data base : data base is collection of information which is organised in so the information can be
accessed very easily.
Data base Management System: it is a computer software. it creates and manages database.it also
capture and analyses the data.
Data base System : it is a highlevel definition which relates the structure and relationship
between stored data.
Database catalog : database catalog consists of tables, views, synonyms, value ranges, indexes,
users, and user groups are stored.
program data independence: Data independence is the type of data transparency that matters for a
DBMS.
User View: It is some set screens which appears for a user to select a certain actiomn or option.
DBA: DBA refers to data base adminstrator he controls all the action with respect to data base.
End User: end user is a person he will use the product after it is developed fully.
Canned Transaction : It is a standard type of query and updates that have been carefully
programmed and tested.
deductive database: A deductive database is a database system that can make deductions (i.e.,
conclude additional facts) based on rules and facts.
Persistent objects: These are the fundamental logical units of data storage in a database.
Meta data: meta data is a data which provides the information of other .
Two loudspeakers on elevated platforms are at opposite ends of a fie.pdf
Two loudspeakers on elevated platforms are at opposite ends of a field. Each broadcasts equally
in all directions. The sound intensity level at a point halfway between the loudspeakers is 71.6
dB .
Q:What is the sound intensity level at a point one-quarter of the way from one speaker to the
other along the line joining them?
Solution
We know that the net intensity level due to multiple source is given as: 71.6 = 10 log ( Inet / I0);
71.6 = 10 log(2I/Io)
Now, what we have is that the Inet = 2I where I is the intensity due to one of the speakers.
[Since the two loudspeakers are the same, we take Intensity at the centre of the field due be
same]
Further, we know that Intensity due to a sound source is inversely proportional to the distance
from the source.
That is the net intensity at a point one quarter away from one of the sources = 4I + 4I/9 = 40 I/9
[Since the distance from one of the speakers will be halved, the intensity becomes 4 times, while
for other the distance will become 3/2 times, hence the intensity will become 4/9 times]
Hence, the new sound intensity level = 10 log(40 I / 9 Io) = 10log [(20/9)*(2I/Io)] = 10log(20/9)
+ 71.6
Therefore, the net sound intensity level = 75.0679 dB.
Tube and Contents Molality 1 10 g dl NaCl 2 36 gal Na B 0.8 gdl. Na .pdf
Tube and Contents Molality 1 10 g dl NaCl 2 36 gal Na B 0.8 g/dl. Na 0.145 m 0.45 g dL NaCl
Approximately the concentration of seawater. Note: dL 100 mL. Milli osmolality 290 mosm
Visual Appearanc of RBCs
Solution
S. No. Tube and contents molarity osmolarity 1 10 g/dL 1.711 3.422 2 3.5 g/dL 0.599 1.198
3 0.85 g/dL 0.145 0.29 4 0.45 g/dL 0.077 0.154 5 0.20 g/dL 0.034 0.068.
At 50°C, PCR primers are able to access target DNA before template s.pdf
At 50°C, PCR primers are able to access target DNA before template strands can rejoin due
to...A. ) Template being forced apart by single stranded binding protein
Solution
Correct answer is option A and B. Template being forced apart by single stranded binding
proteins. Primers being more abundant than template.
SsBP do not allow the dna strands to anneal until the primer attaches to the template strand. They
help in more accurate PCR. Large number of primers means more chances of primers binding to
target DNA..
The questions are..2. Which cell cycle phase requires the longest .pdf
The questions are..
2. Which cell cycle phase requires the longest time for completion?
3. Which mitotic phase requires the longest time for completion?
4. What important changes are occurring in the nucleus during the longest phase of mitosis?
Does this justify the amount of time spent in the phase?
5. Which mitotic phase requires the next longest time for completion?
6. Which mitotic phase requires the shortest time for completion? 7. Are there any
assumptions made in using this method to estimate the timing of phases within the cell cycle?
What are the most important assumptions?
In the scientific method (hypothetico-deductive method), complementary hypotheses are
evaluated to determine the best explanation for a natural phenomenon. The scientific researcher
develops both a testable hypothesis (sometimes called an alternative hypothesis) and a
complementary null hypothesis. Then the researcher collects data to evaluate both hypotheses,
and if the data is sufficient, determines that one of the hypotheses is the better explanation.
8. Cancer is a disease that involves irregular accumulation of cells within the body tissues.
Develop a pair of hypotheses about cancer that address the length of the phases within the cell
cycle and the length of the cell cycle itself. Please list your hypotheses along with a brief (only
one or two sentences) explanation explaining why you have chosen those hypotheses. 2.
Solution
ans 2: Interphase requires the longest time for completion,about 78 % of the time of Cell cycle
the cell remain in this non dividing stage.IN this phase DNA replicates cells, become larger
accumulate substances that are essential for cell division, growth and development.
ans 3: Prophase is the longest mitotic phase, which requires about 14 % of the time of cell
cycle.
ans 4: Imortant changes occuring during prophase are:
During early prophase:
The chromosome begins to condense.
The mitotic spindle begins to form. As we know spindle fibers plays an imortant role of
organizing the chromosomes and move around during the mitotic phase.
The nucleolus disappears,this shows that the protein synthesis stops and the nucleus is ready to
break down.
During late prophase:
Chromosomes become compact
Nuclear envelop breakes down releasing the chromosomes.
Mitotic spindleelongates and gets attached to chromosome.
This phase takes longer than the others because the nuclear envelope fragments and the
microtubules have to attach to the chromosomes.
Yes, time spend in this phase is justified as without spindle fibers attachment to the
chromosomes and nuclear envelope breakdown the cell division is not possible.
Ans 5: The Metaphase of the mitotic phase requires the next longest time for completion about
4% of the cellcycle time is consumed during this phase. In this phase lining up of chromosome at
metaplate occurs, and the check point for two kinetochores of each chromosome are correctly
attached to microtubules from opposite spindle poles or not.
Ans 6.
The Fibonacci numbers are an interesting sequence of positive integer.pdf
The Fibonacci numbers are an interesting sequence of positive integers where each value is the
sum of the previous two values. We begin with F_1=F_2=1 (4.3) Each successive value is then
obtained from the formula F_i=F_i+1+F_i-2, i= 3, 4, 5, ... (4.4) Create an Excel worksheet
containing the first 20 Fibonacci numbers. Test each Fibonacci number to determine if it is odd
or even. Place the Fibonacci numbers in column A, and the corresponding odd/even labels in
column B.
Solution
0 FALSE even 1 TRUE odd 1 TRUE odd 2 FALSE even 3 TRUE odd 5 TRUE odd 8
FALSE even 13 TRUE odd 21 TRUE odd 34 FALSE even 55 TRUE odd 89 TRUE odd
144 FALSE even 233 TRUE odd 377 TRUE odd 610 FALSE even 987 TRUE odd 1597
TRUE odd 2584 FALSE even 4181 TRUE odd.
The graph of a function f is shown in Compute the Riemann sum when .pdf
The graph of a function f is shown in Compute the Riemann sum when [0, 10] is subdivided
into ten equal subdivided into ten equal subintervals [xi-1, xi] and Riemann sum = 51 Riemann
sum = 47 Riemann sum = 49 Riemann sum = 48
Solution
Riemann sum = sum of area of all rectangles
= 6(1)+7(1)+5+4+5+4+3+4+6+7 =51
Hence option A is right..
Suppose Venus rotated as rapidly as Earth. Briefly explain how and w.pdf
Suppose Venus rotated as rapidly as Earth. Briefly explain how and why you would expect it to
be different in terms of each of the following: geological processes, atmospheric circulation,
magnetic field, and climate history.
Solution
The same levels of volcanism and tectonics, and a higher level of erosion..
Suppose that f X rightarrow Y and g Y rightarrow Z are bijections.pdf
Suppose that f: X rightarrow Y and g: Y rightarrow Z are bijections of sets. Prove that the
composite g f: X rightarrow Z is also a bijection and that (g f)^-1 = f^-1 g^-1: z rightarrow Y
rightarrow X.
Solution
As they are bijections, they have inverses f1, g1 (i.e. the respective inverses are defined) from B
to A and from C to B respectively.
(gf)(f1g1)=(g1C)g1=1C(gf)(f1g1)=(g1C)g1=1C
(f1g1)(gf)=(f11A)f=1A(f1g1)(gf)=(f11A)f=1A
so gf has an inverse and thus is bijective.
Hence proved.
In the answer written above, A and C are subscripts.(please note).
Q1 Three deeply influential social organizations in culture areA.pdf
Q1: Three deeply influential social organizations in culture are:
A/ Family, State, & Education. B/ Family, Religion, & Education. C/ Religion, Education, &
State. D/ Family, State, & Religion.
Solution
I strongly believe it is B. Family, Religion & Education.
Family is the root for any individual, where he is influenced since young & Religion also counts
in as this also comes pretty early in life and influences to great extent.
Not the least education, influences in the developing stages but it has an cementing effect in
shaping the culture of an individua..
processes or cel germ line leading to gametes. A B 5. Natural selec.pdf
processes or cel germ line\' leading to gametes. A B 5. Natural selection operates on individuals
either favorable or not, but it is the population that evolves over time. A B 6. Natural selection
can occur in a population with or without variation in heritable traits. A B 7. Nuclear DNA is
found in the nucleus of cells and in mitochondria. A (B 8. A nucleotide is a DNA unit made of
one sugar molecule, one phosphate group and one nitrogen base. A B 9. Humans can synthesize
all naturally occurring amino acids necessary for their body to function. A B 10. The genetic
code is universal in all vertebrates in which the same DNA triplets specify the same amino acids.
A B 11. Homeobox genes are an evolutionarily ancient, critically important type of regulatory
genes. A B 12. Gregor Mendel discovered that traits are inherited as discrete units\' later termed
genes\' A B 13. The relationship between dominant and recessive alleles in simple Mendelian
traits at the phenotype level is always about 4:1, i e. dominant to recessive expression. A B 14. A
genetic bottleneck describes a situation when a large population is reduced to few breeding
members resulting in reduced genetic diversity.
Solution
5. True.
It is the population that evolves and not its individual members.
6. False.
Variation is necessary for evolution to occur.
7. False.
Nuclear DNA is found in nucleus only. Mitochondria has organellar DNA.
8. True.
Nucleotide is formed of a pentose sugar, a nitrogen base and a phosphate group.
9. False.
Their are amino acids called as essential amino acids which cannot be formed by body. They
must be taken through diet.
10. True.
Genetic code is universal meaning same triplet codon codes for same amino acid in all
organisms.
11.
12. True.
Gregor Mendel discovered heridity units and named them as factors which later became known
as genes.
13. False.
The ratio is 3:1.
13.
Question 3 of 5 Barry recently took a hit of LSD. Immediately afterw.pdf
Question 3 of 5 Barry recently took a hit of LSD. Immediately afterward he experiences
sympathomimetic effects such as an elevation of body temperature and dilation of his pupils. He
is most likely in the __________ phase of the LSD experience.
A. psychic
B. perceptual
C. somatic
D. sensory
Solution
The correct answer is option C. The elevation of body temperature and dilation of pupils occurs
in somatic phase of LSD experience.
Stages of LSD experience:
Stage I (Within 1 hour) : Primarily Somatic-
Sympathetic nervous system activity: Increase in temperature, Pupil dilation, increase in blood
pressure and heart rate, increase in glucose level, decrease appetite, dizziness, drowsiness,
headache, tremor.
Stage II (Within 2-3 hours of ingestion) :
Visual Illusions: wave-like, vortex of color, grids, checkerboard, flashing, increased brightness.
Stage II (3 and above hours after ingestion):
Psychic stage, maximum effect, mood changes, swings, panic, unexpected thought or memory,
visible hallucinations, etc..
How to Make a Field invisible in Odoo 17
It is possible to hide or invisible some fields in odoo. Commonly using “invisible” attribute in the field definition to invisible the fields. This slide will show how to make a field invisible in odoo 17.
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Darwin believed that women were less evolved than men, and because o.pdf
Darwin believed that women were less evolved than men, and because of their smaller brains
were eternally primitive. He argued that male intelligence was consistently sharpened through
the struggle for possession of females through sexual selection and through hunting and other
activities such as defending females and young which is natural selection. Men had ultimately
become superior to women.
Address each of the following: What do you think were the implications of these statements from
Darwin for women in science and mathematics following their publication in 1871? Do you
think it helped or made it worse for women trying to break into this male dominated
environment? What types of implications might these statements have caused? Do you think
Darwin\'s statements have lingering influences in today\'s science profession? Describe.
Solution
As indicated by Darwinian hypothesis, ladies were less developed than men, and due to their
littler brains, they were \"forever primitive,\" virtuous, less otherworldly, more materialistic, and
\"a genuine risk to contemporary civilization.\" The assumed insight crevice that many driving
Darwinists accepted existed between human guys and females was large to the point that some
driving Darwinists characterized them as two unmistakable species—guys as Homo frontalis and
females as Homo parietalis. The distinctions were great to the point that Darwin was
flabbergasted \"such extraordinary creatures have a place with the same species.\"
Purposes behind male predominance incorporated the conclusion that war and chasing pruned
the weaker men, permitting just the most fit to return home and reproduce. Ladies, conversely,
were not subject to these determination weights but rather were ensured by men, permitting the
frail to survive. Darwin\'s compositions and those of his devotees uncover that the conviction of
ladies inadequacy was vital to early transformative hypothesis. In The Descent of Man Darwin
contended that grown-up females of most species looked like the youthful of both genders and
that \"guys are more developmentally progressed than females.
This corrupting perspective of ladies quickly spread to Darwin\'s logical and scholarly peers
amid the late Victorian period and intensely affected numerous scholars who had a noteworthy
part in forming the accompanying era—from Sigmund Freud to Havelock Ellis. Evelleen
Richards infers that Darwin\'s perspectives about ladies coherently took after from his
hypothesis, \"in this manner feeding a few eras of logical sexism.\"Importantly, subsequently of
Darwin\'s thoughts, researchers were roused to utilize science to manufacture bolster for the
position that ladies were \"plainly mediocre and irreversibly subordinate\" to men, hence
affecting all of society. Darwinian intuition upheld the possibility that men were better than
ladies, which was at that point inside many societies. Darwin essentially offered belief to this
unbiblical instructing.
a) become more similar to one another. b) become more different th.pdf
a) become more similar to one another.
b) become more different than the mean.
c) cluster around the mean.
d) are more accurately represented by the mean.As the standard deviation gets larger, the scores
in the distribution
a) become more similar to one another.
b) become more different than the mean.
c) cluster around the mean.
d) are more accurately represented by the mean.
Solution
b) become more different than the mean.
Why are the shape and slope of a dose-response curve importantS.pdf
Why are the shape and slope of a dose-response curve important?
Solution
Answer:
It is important for to predict the toxicity of a substance at specific dose levels. Major differences
among drugs may exist not only in the point at which the threshold is reached but also in the
percent of population responding per unit change in dose (i.e., the slope). The slope of the dose
response curve represents rate of buildup of toxic effects..
When streaking a plate, the idea is to dilute the bacteria sufficie.pdf
When streaking a plate, the idea is to dilute the bacteria sufficiently so that by the 3^rd phase
of the plate you have isolated colonies; a plate contains sufficient surface area to allow isolated
colonies to appear, whereas a slant does not. _ Enrichment media contains substances that
various bacteria require for growth; blood agar is considered enrichment media as it detects
hemolysins.
Solution
the answers are true, streaking provides more surface area for the microbes and we can isolate
individual colonies but a slant cannot. In a slant we can store the bacterias for long term.
Enrichment media is enriched with specific nutrients to support the growth of specific
microorganism and inhibits other microbes.Blood agar is a enriched midia.
Why is it bad for a group to be polyphyleticWhat type of traits c.pdf
Why is it bad for a group to be polyphyletic?
What type of traits can be used in a cladistic analysis?
Solution
Polyphyletic group doesnot involve grouping of all the desendents of a common ancestor. This
kind of grouping arises when analogous characters of the organisms were misunderstood. The
organisms placed under polyphyletic group have similar characteristics, however they donot
share these characters from a common ancestor.
Cladistic analysis takes into consideration of shared characters i.e; synapomorphies, evolutionary
relationship, common ancestor.
When recruiting a sample of individuals to participate in a study on.pdf
When recruiting a sample of individuals to participate in a study on reaction time, researchers
used the most recent census to identify local older adults, who were
between the ages of 60 and 65, between 65 and 70, between 70 and 75, between 75 and
80, and between 80 and 85. Once they had identified these subgroups, the researchers randomly
selected 20 people from within each age group to participate in the study. What kind of sampling
method did the researchers use?
a. convenience sampling
b. random sampling
c. stratified sampling
d. cluster sampling
Solution
d. cluster sampling.
Where does the Calvin cycle take place stroma of the chloroplast t.pdf
Where does the Calvin cycle take place? stroma of the chloroplast thylakoid membrane
thylakoid space outer membrane of the chloroplast cytoplasm In a plant cell, where are ATP
synthase complexes located? A. thylakoid membrane B. Inner mitochondrial membrane C.
plasma membrane D. A & B E. A, B, and C
Solution
9. Stroma of the chloroplasts Calvin cycle takes place inside the stroma of the chloroplasts. 10.
The answer is D. Both Aand B. In plants ATP synthase complexes are present in the thylakoids
membrane and in the inner mitochondrial membrane..
What are the pros and cons of using the mode as your measure of cent.pdf
What are the pros and cons of using the mode as your measure of central tendency?
Solution
The mode is that observation that occurs most often.
It is usually not unique, and is therefore not that often used.
Advantages :
1) It has the advantage that it applies to numerical as well as categorical variables.
2) As with the median, the mode is easy to find if the data is small and sorted.
Disadvantages :
1) The mode only refers to the element that ocurs the maximum number of times. So, it doe snot
apply when the data set has many outliers or if the data set has a very large number of elements.
We presented a list of capabilities that should be provided by the DB.pdf
We presented a list of capabilities that should be provided by the DBMS software to the DBA,
database designers, and end users to help them design, administer, and use a database. Then we
gave a brief historical perspective on the evolution of database applications. We pointed out the
recent rapid growth of the amounts and types of data that must be stored in databases, and we
discussed the emergence of new systems for handling \"big data\" applications. Finally, we
discussed the overhead costs of using a DBMS and discussed some situations in which it may not
be advantageous to use one. Define the following terms: data, database, DBMS, database
system, database catalog, program-data independence, user view, DBA, end user, canned
transaction, deductive database system, persistent object, meta-data, and transaction-processing
application. What four main types of actions involve databases? Briefly discuss each. Discuss
the main characteristics of the database approach and how it differs from traditional file systems.
What are the responsibilities of the DBA and the database designers? What are the different
types of database end users? Discuss the main activities of each. Discuss the capabilities that
should be provided by a DBMS. Discuss the differences between database systems and
information retrieval systems. Identify some informal queries and update operations that you
would expect to apply to the database shown in Figure 1.2. What is the difference between
controlled and uncontrolled redundancy? Illustrate with examples. Specify all the relationships
among the records of the database shown in Figure 1.2.
Solution
I.
Data: data is simply called as collection of objects.
Data base : data base is collection of information which is organised in so the information can be
accessed very easily.
Data base Management System: it is a computer software. it creates and manages database.it also
capture and analyses the data.
Data base System : it is a highlevel definition which relates the structure and relationship
between stored data.
Database catalog : database catalog consists of tables, views, synonyms, value ranges, indexes,
users, and user groups are stored.
program data independence: Data independence is the type of data transparency that matters for a
DBMS.
User View: It is some set screens which appears for a user to select a certain actiomn or option.
DBA: DBA refers to data base adminstrator he controls all the action with respect to data base.
End User: end user is a person he will use the product after it is developed fully.
Canned Transaction : It is a standard type of query and updates that have been carefully
programmed and tested.
deductive database: A deductive database is a database system that can make deductions (i.e.,
conclude additional facts) based on rules and facts.
Persistent objects: These are the fundamental logical units of data storage in a database.
Meta data: meta data is a data which provides the information of other .
Two loudspeakers on elevated platforms are at opposite ends of a fie.pdf
Two loudspeakers on elevated platforms are at opposite ends of a field. Each broadcasts equally
in all directions. The sound intensity level at a point halfway between the loudspeakers is 71.6
dB .
Q:What is the sound intensity level at a point one-quarter of the way from one speaker to the
other along the line joining them?
Solution
We know that the net intensity level due to multiple source is given as: 71.6 = 10 log ( Inet / I0);
71.6 = 10 log(2I/Io)
Now, what we have is that the Inet = 2I where I is the intensity due to one of the speakers.
[Since the two loudspeakers are the same, we take Intensity at the centre of the field due be
same]
Further, we know that Intensity due to a sound source is inversely proportional to the distance
from the source.
That is the net intensity at a point one quarter away from one of the sources = 4I + 4I/9 = 40 I/9
[Since the distance from one of the speakers will be halved, the intensity becomes 4 times, while
for other the distance will become 3/2 times, hence the intensity will become 4/9 times]
Hence, the new sound intensity level = 10 log(40 I / 9 Io) = 10log [(20/9)*(2I/Io)] = 10log(20/9)
+ 71.6
Therefore, the net sound intensity level = 75.0679 dB.
Tube and Contents Molality 1 10 g dl NaCl 2 36 gal Na B 0.8 gdl. Na .pdf
Tube and Contents Molality 1 10 g dl NaCl 2 36 gal Na B 0.8 g/dl. Na 0.145 m 0.45 g dL NaCl
Approximately the concentration of seawater. Note: dL 100 mL. Milli osmolality 290 mosm
Visual Appearanc of RBCs
Solution
S. No. Tube and contents molarity osmolarity 1 10 g/dL 1.711 3.422 2 3.5 g/dL 0.599 1.198
3 0.85 g/dL 0.145 0.29 4 0.45 g/dL 0.077 0.154 5 0.20 g/dL 0.034 0.068.
At 50°C, PCR primers are able to access target DNA before template s.pdf
At 50°C, PCR primers are able to access target DNA before template strands can rejoin due
to...A. ) Template being forced apart by single stranded binding protein
Solution
Correct answer is option A and B. Template being forced apart by single stranded binding
proteins. Primers being more abundant than template.
SsBP do not allow the dna strands to anneal until the primer attaches to the template strand. They
help in more accurate PCR. Large number of primers means more chances of primers binding to
target DNA..
The questions are..2. Which cell cycle phase requires the longest .pdf
The questions are..
2. Which cell cycle phase requires the longest time for completion?
3. Which mitotic phase requires the longest time for completion?
4. What important changes are occurring in the nucleus during the longest phase of mitosis?
Does this justify the amount of time spent in the phase?
5. Which mitotic phase requires the next longest time for completion?
6. Which mitotic phase requires the shortest time for completion? 7. Are there any
assumptions made in using this method to estimate the timing of phases within the cell cycle?
What are the most important assumptions?
In the scientific method (hypothetico-deductive method), complementary hypotheses are
evaluated to determine the best explanation for a natural phenomenon. The scientific researcher
develops both a testable hypothesis (sometimes called an alternative hypothesis) and a
complementary null hypothesis. Then the researcher collects data to evaluate both hypotheses,
and if the data is sufficient, determines that one of the hypotheses is the better explanation.
8. Cancer is a disease that involves irregular accumulation of cells within the body tissues.
Develop a pair of hypotheses about cancer that address the length of the phases within the cell
cycle and the length of the cell cycle itself. Please list your hypotheses along with a brief (only
one or two sentences) explanation explaining why you have chosen those hypotheses. 2.
Solution
ans 2: Interphase requires the longest time for completion,about 78 % of the time of Cell cycle
the cell remain in this non dividing stage.IN this phase DNA replicates cells, become larger
accumulate substances that are essential for cell division, growth and development.
ans 3: Prophase is the longest mitotic phase, which requires about 14 % of the time of cell
cycle.
ans 4: Imortant changes occuring during prophase are:
During early prophase:
The chromosome begins to condense.
The mitotic spindle begins to form. As we know spindle fibers plays an imortant role of
organizing the chromosomes and move around during the mitotic phase.
The nucleolus disappears,this shows that the protein synthesis stops and the nucleus is ready to
break down.
During late prophase:
Chromosomes become compact
Nuclear envelop breakes down releasing the chromosomes.
Mitotic spindleelongates and gets attached to chromosome.
This phase takes longer than the others because the nuclear envelope fragments and the
microtubules have to attach to the chromosomes.
Yes, time spend in this phase is justified as without spindle fibers attachment to the
chromosomes and nuclear envelope breakdown the cell division is not possible.
Ans 5: The Metaphase of the mitotic phase requires the next longest time for completion about
4% of the cellcycle time is consumed during this phase. In this phase lining up of chromosome at
metaplate occurs, and the check point for two kinetochores of each chromosome are correctly
attached to microtubules from opposite spindle poles or not.
Ans 6.
The Fibonacci numbers are an interesting sequence of positive integer.pdf
The Fibonacci numbers are an interesting sequence of positive integers where each value is the
sum of the previous two values. We begin with F_1=F_2=1 (4.3) Each successive value is then
obtained from the formula F_i=F_i+1+F_i-2, i= 3, 4, 5, ... (4.4) Create an Excel worksheet
containing the first 20 Fibonacci numbers. Test each Fibonacci number to determine if it is odd
or even. Place the Fibonacci numbers in column A, and the corresponding odd/even labels in
column B.
Solution
0 FALSE even 1 TRUE odd 1 TRUE odd 2 FALSE even 3 TRUE odd 5 TRUE odd 8
FALSE even 13 TRUE odd 21 TRUE odd 34 FALSE even 55 TRUE odd 89 TRUE odd
144 FALSE even 233 TRUE odd 377 TRUE odd 610 FALSE even 987 TRUE odd 1597
TRUE odd 2584 FALSE even 4181 TRUE odd.
The graph of a function f is shown in Compute the Riemann sum when .pdf
The graph of a function f is shown in Compute the Riemann sum when [0, 10] is subdivided
into ten equal subdivided into ten equal subintervals [xi-1, xi] and Riemann sum = 51 Riemann
sum = 47 Riemann sum = 49 Riemann sum = 48
Solution
Riemann sum = sum of area of all rectangles
= 6(1)+7(1)+5+4+5+4+3+4+6+7 =51
Hence option A is right..
Suppose Venus rotated as rapidly as Earth. Briefly explain how and w.pdf
Suppose Venus rotated as rapidly as Earth. Briefly explain how and why you would expect it to
be different in terms of each of the following: geological processes, atmospheric circulation,
magnetic field, and climate history.
Solution
The same levels of volcanism and tectonics, and a higher level of erosion..
Suppose that f X rightarrow Y and g Y rightarrow Z are bijections.pdf
Suppose that f: X rightarrow Y and g: Y rightarrow Z are bijections of sets. Prove that the
composite g f: X rightarrow Z is also a bijection and that (g f)^-1 = f^-1 g^-1: z rightarrow Y
rightarrow X.
Solution
As they are bijections, they have inverses f1, g1 (i.e. the respective inverses are defined) from B
to A and from C to B respectively.
(gf)(f1g1)=(g1C)g1=1C(gf)(f1g1)=(g1C)g1=1C
(f1g1)(gf)=(f11A)f=1A(f1g1)(gf)=(f11A)f=1A
so gf has an inverse and thus is bijective.
Hence proved.
In the answer written above, A and C are subscripts.(please note).
Q1 Three deeply influential social organizations in culture areA.pdf
Q1: Three deeply influential social organizations in culture are:
A/ Family, State, & Education. B/ Family, Religion, & Education. C/ Religion, Education, &
State. D/ Family, State, & Religion.
Solution
I strongly believe it is B. Family, Religion & Education.
Family is the root for any individual, where he is influenced since young & Religion also counts
in as this also comes pretty early in life and influences to great extent.
Not the least education, influences in the developing stages but it has an cementing effect in
shaping the culture of an individua..
processes or cel germ line leading to gametes. A B 5. Natural selec.pdf
processes or cel germ line\' leading to gametes. A B 5. Natural selection operates on individuals
either favorable or not, but it is the population that evolves over time. A B 6. Natural selection
can occur in a population with or without variation in heritable traits. A B 7. Nuclear DNA is
found in the nucleus of cells and in mitochondria. A (B 8. A nucleotide is a DNA unit made of
one sugar molecule, one phosphate group and one nitrogen base. A B 9. Humans can synthesize
all naturally occurring amino acids necessary for their body to function. A B 10. The genetic
code is universal in all vertebrates in which the same DNA triplets specify the same amino acids.
A B 11. Homeobox genes are an evolutionarily ancient, critically important type of regulatory
genes. A B 12. Gregor Mendel discovered that traits are inherited as discrete units\' later termed
genes\' A B 13. The relationship between dominant and recessive alleles in simple Mendelian
traits at the phenotype level is always about 4:1, i e. dominant to recessive expression. A B 14. A
genetic bottleneck describes a situation when a large population is reduced to few breeding
members resulting in reduced genetic diversity.
Solution
5. True.
It is the population that evolves and not its individual members.
6. False.
Variation is necessary for evolution to occur.
7. False.
Nuclear DNA is found in nucleus only. Mitochondria has organellar DNA.
8. True.
Nucleotide is formed of a pentose sugar, a nitrogen base and a phosphate group.
9. False.
Their are amino acids called as essential amino acids which cannot be formed by body. They
must be taken through diet.
10. True.
Genetic code is universal meaning same triplet codon codes for same amino acid in all
organisms.
11.
12. True.
Gregor Mendel discovered heridity units and named them as factors which later became known
as genes.
13. False.
The ratio is 3:1.
13.
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