Meselson and Stahl in 1957 gave experimental evidence that each DNA strand served as a
template for new DNA synthesis, a process called semi-conservative replication. At that time,
there were three proposed models for DNA replication put forward by the scientific community
after DNA structure had been discovered- Semi-conservative, conservative and dispersive
replication.
Semi-conservative replication: In this model, the two strands of DNA unwind from each other,
and each acts as a template for synthesis of a new, complementary strand. This results in two
DNA molecules with one original strand and one new strand.
Conservative replication. In this model, DNA replication results in one molecule that consists of
both original DNA strands and another molecule that consists of two new strands with exactly
the same sequences as the original molecule.
Meselson and Stahl used the density gradient sedimentation experiment (done with the E.Coli
bacteria) to establish that DNA replicates using the semi-conservative model of replication as
follows:
Results of analysis:
Solution
Meselson and Stahl in 1957 gave experimental evidence that each DNA strand served as a
template for new DNA synthesis, a process called semi-conservative replication. At that time,
there were three proposed models for DNA replication put forward by the scientific community
after DNA structure had been discovered- Semi-conservative, conservative and dispersive
replication.
Semi-conservative replication: In this model, the two strands of DNA unwind from each other,
and each acts as a template for synthesis of a new, complementary strand. This results in two
DNA molecules with one original strand and one new strand.
Conservative replication. In this model, DNA replication results in one molecule that consists of
both original DNA strands and another molecule that consists of two new strands with exactly
the same sequences as the original molecule.
Meselson and Stahl used the density gradient sedimentation experiment (done with the E.Coli
bacteria) to establish that DNA replicates using the semi-conservative model of replication as
follows:
Results of analysis:.
Prokaryotic DNA replication : These slides contains basics of the prokaryotic DNA replication for S.Y.B.Sc and T.Y.B.Sc students of Microbiology and biotechnology
It covers topics like Enzymes used in replication, Semiconservative replication, Meselson and Stahl experiment, Termination of replication, modes of replication: theta and rolling circle, basic rules of replication
Prokaryotic DNA replication : These slides contains basics of the prokaryotic DNA replication for S.Y.B.Sc and T.Y.B.Sc students of Microbiology and biotechnology
It covers topics like Enzymes used in replication, Semiconservative replication, Meselson and Stahl experiment, Termination of replication, modes of replication: theta and rolling circle, basic rules of replication
In Meselson and Stahls experiment, cells were grown in media contai.pdfmalavshah9013
In Meselson and Stahl\'s experiment, cells were grown in media containing^15 N in place of^14
N. and then the cells\' DNA was separated by equilibrium centrifugation. This experiment
demonstrated that DNA is replicated by DNA polymerase replication is conservative.
replication is semiconservative. forms double helices by means of hydrogen bonding between
base pairs. replication occurs in opposite directions on complementary strands. Deduction of the
structure of DNA was based on a series of key X-ray crystallographic studies done by Avery,
MacLeod, and McCarty, Beadle and Tatum, Meselson and Stahl, Wilkins and Franklin, Watson
and Crick.
Solution
2) The correct option is option C - Peas.
Gregor Mendel was an Austrian monk who established the classical tenets of genetics based on
his experiments on the garden pea (Pisum sativum). He studied various characters such as height
of plant, pea pod shape and colour, flower colour etc. Based on his observations, he came up
with the 3 basic laws of genetics - The law of dominance, The law of segregation and The law of
independent assortment.
3) The correct option is option C- replication is semi-conservative
This is another classical example in molecular genetics performed by Messelson and Stahl. The
aim of the experiment was to establish the semi-conservative replication status of DNA strands.
Basically, it means that during replication of DNA, the double helix unwinds and separates into
two strands. Each strand then acts as a guiding strand ( template strand) over which the new
strand is synthesized. At the end of the replication cycle, one DNA molecule would have
replicated into two DNA molecules. Each molecule of DNA would have two strands- one
original parent strand and the other being the newly synthesized daughter strand.
4) The correct option is option D- Wilkins and Franklin
Maurice Wilkins and Rosalind Franklin performed the key X-ray diffraction experiments on
DNA. Their data was ultimately helpful for Watson and Crick to deduce the structure of DNA..
If the mechanism of DNA replication (semi-conservative, conservative.pdfarpitcomputronics
If the mechanism of DNA replication (semi-conservative, conservative, and dispersive) was not
yet known:
a. What experiment(s) could distinguish between the three possible mechanisms. Explain how
this method can distinguish between the 3 models.
b. Compare your experiment with the original mechanism. What are the advantages,
disadvantages to each
Solution
If the DNA replication mechanism is not yet known Messelson-Stahl experiment set forward to
distinguish the three mechanisms; conservative, semi- conservative and dispersive mode of
replication.
Messelson-Stahl experiment model used non radioactive heavy isotope of Nitrogen as they found
nitrogen is abundant element in DNA. They labelled the DNA of E.coli bacteria with heavy
isotope 15N and grew in a medium. Soon after many generations 15N labelled DNA were
synthesised. Then they removed the 15N labelled DNA from the medium and placed in normal
14N medium. The cells were then extracted from this medium and were placed in CsCl solution
and are centrifused..
this file includes the packaging of DNA in Prokaryotic and Eukaryotic cell, the difference between DNA and RNA, Semi-conservative mode of DNA replication, and the mechanism of DNA replication.
1) Phosphodiester bonds in DNA are advantageous because they are m.pdfanujmkt
1)
Phosphodiester bonds in DNA are advantageous because they are more stable chemically when
compared with ester bonds. This stability is rendered because of the negative charge present on a
phosphodiester bond. This charge repels similarly charged ions like OH- ions which have the
ability to carry out a nucleophilic hydrolysis reaction and hence break the bond. Hence such
hydrolysis cannot occur in a phosphodiester bond and it is more stable.
2)
The semi-conservative nature of DNA replication means that during DNA replication, each
daughter DNA has one strand inherited from the parent as it is and the other strand of the double
helix is newly synthesised. This was initially proved by Meselson and Stahl in 1958.
The experiment can be described in the following steps:
These results confirmed the hypothesis made by the 2 scientists as the initial band of heavy DNA
indicated all the DNA of same kind. The intermediate band in second generation indicated that
half the DNA was original (with heavy isotope) and the other half was with normal N. And then
the increase in band-width of normal N containing cells indicated that all new cells now
manufactured their DNA from medium having normal N source.
3)
Topoisomers are DNA molecules with same chemical structure but different topology i.e. they
have different geometrical properties and spatial relations. There are two major Topoisomers of
DNA:
Solution
1)
Phosphodiester bonds in DNA are advantageous because they are more stable chemically when
compared with ester bonds. This stability is rendered because of the negative charge present on a
phosphodiester bond. This charge repels similarly charged ions like OH- ions which have the
ability to carry out a nucleophilic hydrolysis reaction and hence break the bond. Hence such
hydrolysis cannot occur in a phosphodiester bond and it is more stable.
2)
The semi-conservative nature of DNA replication means that during DNA replication, each
daughter DNA has one strand inherited from the parent as it is and the other strand of the double
helix is newly synthesised. This was initially proved by Meselson and Stahl in 1958.
The experiment can be described in the following steps:
These results confirmed the hypothesis made by the 2 scientists as the initial band of heavy DNA
indicated all the DNA of same kind. The intermediate band in second generation indicated that
half the DNA was original (with heavy isotope) and the other half was with normal N. And then
the increase in band-width of normal N containing cells indicated that all new cells now
manufactured their DNA from medium having normal N source.
3)
Topoisomers are DNA molecules with same chemical structure but different topology i.e. they
have different geometrical properties and spatial relations. There are two major Topoisomers of
DNA:.
phosphate group and deoxyribose The groups are .pdfanwarfoot
phosphate group and deoxyribose The groups are 1. Phosphate 2. Deoxyribose
sugar 3. Nitrogen base The phosphates and deoxyribose sugars make up the sides of the
\"ladder\" (alternating one after the other) and nitrogen bases are the \"rungs\" of the ladder.
Solution
phosphate group and deoxyribose The groups are 1. Phosphate 2. Deoxyribose
sugar 3. Nitrogen base The phosphates and deoxyribose sugars make up the sides of the
\"ladder\" (alternating one after the other) and nitrogen bases are the \"rungs\" of the ladder..
More Related Content
Similar to Meselson and Stahl in 1957 gave experimental evidence that each DNA .pdf
In Meselson and Stahls experiment, cells were grown in media contai.pdfmalavshah9013
In Meselson and Stahl\'s experiment, cells were grown in media containing^15 N in place of^14
N. and then the cells\' DNA was separated by equilibrium centrifugation. This experiment
demonstrated that DNA is replicated by DNA polymerase replication is conservative.
replication is semiconservative. forms double helices by means of hydrogen bonding between
base pairs. replication occurs in opposite directions on complementary strands. Deduction of the
structure of DNA was based on a series of key X-ray crystallographic studies done by Avery,
MacLeod, and McCarty, Beadle and Tatum, Meselson and Stahl, Wilkins and Franklin, Watson
and Crick.
Solution
2) The correct option is option C - Peas.
Gregor Mendel was an Austrian monk who established the classical tenets of genetics based on
his experiments on the garden pea (Pisum sativum). He studied various characters such as height
of plant, pea pod shape and colour, flower colour etc. Based on his observations, he came up
with the 3 basic laws of genetics - The law of dominance, The law of segregation and The law of
independent assortment.
3) The correct option is option C- replication is semi-conservative
This is another classical example in molecular genetics performed by Messelson and Stahl. The
aim of the experiment was to establish the semi-conservative replication status of DNA strands.
Basically, it means that during replication of DNA, the double helix unwinds and separates into
two strands. Each strand then acts as a guiding strand ( template strand) over which the new
strand is synthesized. At the end of the replication cycle, one DNA molecule would have
replicated into two DNA molecules. Each molecule of DNA would have two strands- one
original parent strand and the other being the newly synthesized daughter strand.
4) The correct option is option D- Wilkins and Franklin
Maurice Wilkins and Rosalind Franklin performed the key X-ray diffraction experiments on
DNA. Their data was ultimately helpful for Watson and Crick to deduce the structure of DNA..
If the mechanism of DNA replication (semi-conservative, conservative.pdfarpitcomputronics
If the mechanism of DNA replication (semi-conservative, conservative, and dispersive) was not
yet known:
a. What experiment(s) could distinguish between the three possible mechanisms. Explain how
this method can distinguish between the 3 models.
b. Compare your experiment with the original mechanism. What are the advantages,
disadvantages to each
Solution
If the DNA replication mechanism is not yet known Messelson-Stahl experiment set forward to
distinguish the three mechanisms; conservative, semi- conservative and dispersive mode of
replication.
Messelson-Stahl experiment model used non radioactive heavy isotope of Nitrogen as they found
nitrogen is abundant element in DNA. They labelled the DNA of E.coli bacteria with heavy
isotope 15N and grew in a medium. Soon after many generations 15N labelled DNA were
synthesised. Then they removed the 15N labelled DNA from the medium and placed in normal
14N medium. The cells were then extracted from this medium and were placed in CsCl solution
and are centrifused..
this file includes the packaging of DNA in Prokaryotic and Eukaryotic cell, the difference between DNA and RNA, Semi-conservative mode of DNA replication, and the mechanism of DNA replication.
1) Phosphodiester bonds in DNA are advantageous because they are m.pdfanujmkt
1)
Phosphodiester bonds in DNA are advantageous because they are more stable chemically when
compared with ester bonds. This stability is rendered because of the negative charge present on a
phosphodiester bond. This charge repels similarly charged ions like OH- ions which have the
ability to carry out a nucleophilic hydrolysis reaction and hence break the bond. Hence such
hydrolysis cannot occur in a phosphodiester bond and it is more stable.
2)
The semi-conservative nature of DNA replication means that during DNA replication, each
daughter DNA has one strand inherited from the parent as it is and the other strand of the double
helix is newly synthesised. This was initially proved by Meselson and Stahl in 1958.
The experiment can be described in the following steps:
These results confirmed the hypothesis made by the 2 scientists as the initial band of heavy DNA
indicated all the DNA of same kind. The intermediate band in second generation indicated that
half the DNA was original (with heavy isotope) and the other half was with normal N. And then
the increase in band-width of normal N containing cells indicated that all new cells now
manufactured their DNA from medium having normal N source.
3)
Topoisomers are DNA molecules with same chemical structure but different topology i.e. they
have different geometrical properties and spatial relations. There are two major Topoisomers of
DNA:
Solution
1)
Phosphodiester bonds in DNA are advantageous because they are more stable chemically when
compared with ester bonds. This stability is rendered because of the negative charge present on a
phosphodiester bond. This charge repels similarly charged ions like OH- ions which have the
ability to carry out a nucleophilic hydrolysis reaction and hence break the bond. Hence such
hydrolysis cannot occur in a phosphodiester bond and it is more stable.
2)
The semi-conservative nature of DNA replication means that during DNA replication, each
daughter DNA has one strand inherited from the parent as it is and the other strand of the double
helix is newly synthesised. This was initially proved by Meselson and Stahl in 1958.
The experiment can be described in the following steps:
These results confirmed the hypothesis made by the 2 scientists as the initial band of heavy DNA
indicated all the DNA of same kind. The intermediate band in second generation indicated that
half the DNA was original (with heavy isotope) and the other half was with normal N. And then
the increase in band-width of normal N containing cells indicated that all new cells now
manufactured their DNA from medium having normal N source.
3)
Topoisomers are DNA molecules with same chemical structure but different topology i.e. they
have different geometrical properties and spatial relations. There are two major Topoisomers of
DNA:.
phosphate group and deoxyribose The groups are .pdfanwarfoot
phosphate group and deoxyribose The groups are 1. Phosphate 2. Deoxyribose
sugar 3. Nitrogen base The phosphates and deoxyribose sugars make up the sides of the
\"ladder\" (alternating one after the other) and nitrogen bases are the \"rungs\" of the ladder.
Solution
phosphate group and deoxyribose The groups are 1. Phosphate 2. Deoxyribose
sugar 3. Nitrogen base The phosphates and deoxyribose sugars make up the sides of the
\"ladder\" (alternating one after the other) and nitrogen bases are the \"rungs\" of the ladder..
in case of solid oxygen the atoms of oxygen are s.pdfanwarfoot
in case of solid oxygen the atoms of oxygen are sharing elctrons and have a
covalent bond between them so there is a dipole dipole bond
Solution
in case of solid oxygen the atoms of oxygen are sharing elctrons and have a
covalent bond between them so there is a dipole dipole bond.
Yeast is a microorganism and doesnt itself rise.pdfanwarfoot
Yeast is a microorganism and doesn\'t itself rise. Yeast eats carbohydrate and makes
Carbon Dioxide (CO2) - the gas in soda water - as a waste product and it is this gas that makes
the bread or cake \"rise\".
Solution
Yeast is a microorganism and doesn\'t itself rise. Yeast eats carbohydrate and makes
Carbon Dioxide (CO2) - the gas in soda water - as a waste product and it is this gas that makes
the bread or cake \"rise\"..
First, lets analyze what each component of the .pdfanwarfoot
First, let\'s analyze what each component of the solution is (strong base/acid, weak
base/acid, salt?) HF: This is a weak acid (something you should probably know off the top of
your head, but if you look at Ka, it\'s a very small value which indicates that it is weak). NaOH:
A strong base (once again, something you should know off the top of your head. Virtually any
ionic compound with Na is strong). Since it\'s strong, we can assume that however much NaOH
we have will give us an equivalent amount of OH- ions. NaF: This is the salt for HF (as
previously mentioned, this is completely soluble in water and adds F- ions, which will affect the
equilibrium of HF). Like before, we can assume that however much NaF we have gives us the
same amount of F-, since all of it dissolves. Now, let\'s think about the reactions that are
happening. First, whenever you have an acid and a base, you have a neutralization reaction (I\'m
leaving Na out of all the following reactions since it is a spectator ion): HF(aq) + OH-(aq) ==>
F-(aq) + H2O(l) This means that the OH- we\'ve added will neutralize our initial amount of HF
into F- and water. Since we have more HF than OH-, all of the OH- will go towards neutralizing
the HF. Thus, 2.00M HF and 1.00M NaOH is essentially equivalent to 1.00M HF and 0 NaOH.
Now, all we have is HF and F-. Recall the Henderson-Hasselbalch equation, which tells you how
to calculate pH based on the ratio of an acid and its salt. In the case of HF, it will be pH = pKa +
log([F-]/[HF]) You have [F-] in the form of [NaF], and you have [HF] which we just
determined. You can calculate pKa from the given Ka value. Thus, you should be able to
calculate pH.
Solution
First, let\'s analyze what each component of the solution is (strong base/acid, weak
base/acid, salt?) HF: This is a weak acid (something you should probably know off the top of
your head, but if you look at Ka, it\'s a very small value which indicates that it is weak). NaOH:
A strong base (once again, something you should know off the top of your head. Virtually any
ionic compound with Na is strong). Since it\'s strong, we can assume that however much NaOH
we have will give us an equivalent amount of OH- ions. NaF: This is the salt for HF (as
previously mentioned, this is completely soluble in water and adds F- ions, which will affect the
equilibrium of HF). Like before, we can assume that however much NaF we have gives us the
same amount of F-, since all of it dissolves. Now, let\'s think about the reactions that are
happening. First, whenever you have an acid and a base, you have a neutralization reaction (I\'m
leaving Na out of all the following reactions since it is a spectator ion): HF(aq) + OH-(aq) ==>
F-(aq) + H2O(l) This means that the OH- we\'ve added will neutralize our initial amount of HF
into F- and water. Since we have more HF than OH-, all of the OH- will go towards neutralizing
the HF. Thus, 2.00M HF and 1.00M NaOH is essentially equivalent to 1.00M HF and 0 NaOH.
Now,.
D) Not B) because Cl-Benzene bond develops a doub.pdfanwarfoot
D) Not B) because Cl-Benzene bond develops a double bond character due to
resonance
Solution
D) Not B) because Cl-Benzene bond develops a double bond character due to
resonance.
YesBecause (1,1) is missing, it is not reflexive though (3,3) (2,2.pdfanwarfoot
Yes
Because (1,1) is missing, it is not reflexive though (3,3) (2,2) are there
Solution
Yes
Because (1,1) is missing, it is not reflexive though (3,3) (2,2) are there.
We know that from Nernst Equation ,E o cell = E o – ( 0.059 n ) .pdfanwarfoot
We know that from Nernst Equation ,
E o cell = E o – ( 0.059 / n ) log ( [ products ] / [reactants ])
Given Eo cell = 0.17 V
Eo = 0.24 V
[products ] = [Cd 2+ ] = ?
[reactants ] = [ Ni 2+ ] = 1.0 M
N = no . of electrons transferred = 2
Solution
We know that from Nernst Equation ,
E o cell = E o – ( 0.059 / n ) log ( [ products ] / [reactants ])
Given Eo cell = 0.17 V
Eo = 0.24 V
[products ] = [Cd 2+ ] = ?
[reactants ] = [ Ni 2+ ] = 1.0 M
N = no . of electrons transferred = 2.
lower.. think about it.. take a straw draw a line.pdfanwarfoot
lower.. think about it.. take a straw draw a line on it and fill it with water, then look
at it from different angles
Solution
lower.. think about it.. take a straw draw a line on it and fill it with water, then look
at it from different angles.
From left to right iodocyclopropane; 1-bromo-3-m.pdfanwarfoot
From left to right: iodocyclopropane; 1-bromo-3-methylcyclopentane; 1-iodo-2,2-
dimethylpropane;
Solution
From left to right: iodocyclopropane; 1-bromo-3-methylcyclopentane; 1-iodo-2,2-
dimethylpropane;.
For inorganic compounds Chemical properties remai.pdfanwarfoot
For inorganic compounds Chemical properties remain same but colour or physical
appearance changes. CuSO4 .5H2O is blue and crystalline but CuSO4 anhydrous is
colourless.Colour change can be explained by theories of Co ordination compounds. In organic
compounds ; chemical properties also change as dehydration changes the chemical nature;
C2H5OH---> C2H4+H2O
Solution
For inorganic compounds Chemical properties remain same but colour or physical
appearance changes. CuSO4 .5H2O is blue and crystalline but CuSO4 anhydrous is
colourless.Colour change can be explained by theories of Co ordination compounds. In organic
compounds ; chemical properties also change as dehydration changes the chemical nature;
C2H5OH---> C2H4+H2O.
Over forty interstellar molecules have been found, ranging from simp.pdfanwarfoot
Over forty interstellar molecules have been found, ranging from simple OH radicals to molecules
with over a dozen different atoms, including a few polycyclic. Their significance lies in that
planets are formed from such nebulas( area of the cloud), so this tells us what may be found.
Solution
Over forty interstellar molecules have been found, ranging from simple OH radicals to molecules
with over a dozen different atoms, including a few polycyclic. Their significance lies in that
planets are formed from such nebulas( area of the cloud), so this tells us what may be found..
One of the four prominent symptoms of typhoid is coated tongue. Even.pdfanwarfoot
One of the four prominent symptoms of typhoid is coated tongue. Even before the onset of fever,
the tongue appears furred, with centrally coated and a red margin around. There are other illness
also which shows coated tongue, but with different appearance. In malaria, the tongue is
uniformly coated whereas in candidia thrush, a thick white layer of which patch is formed. In
typhoid, the patch is centrally located on the tongue surrounded by a red margin.
During the first week of typhoid, the dorsum of tongue is covered with a thin dirty white fur,
which occurs due to the toxins released by the pathogen. The coat gradually clears from the tip
and edges and completely disappears by the thid week of infection, leaving a glazed, dry or
smooth red tongue.
By observing the coat pattern, the Doctor would have diagnosed the fever well in advance.
Solution
One of the four prominent symptoms of typhoid is coated tongue. Even before the onset of fever,
the tongue appears furred, with centrally coated and a red margin around. There are other illness
also which shows coated tongue, but with different appearance. In malaria, the tongue is
uniformly coated whereas in candidia thrush, a thick white layer of which patch is formed. In
typhoid, the patch is centrally located on the tongue surrounded by a red margin.
During the first week of typhoid, the dorsum of tongue is covered with a thin dirty white fur,
which occurs due to the toxins released by the pathogen. The coat gradually clears from the tip
and edges and completely disappears by the thid week of infection, leaving a glazed, dry or
smooth red tongue.
By observing the coat pattern, the Doctor would have diagnosed the fever well in advance..
Introduction One of the key goals for the Windows Subsystem for Li.pdfanwarfoot
Introduction
One of the key goals for the Windows Subsystem for Linux is to allow users to work with their
files as they would on Linux, while giving full interoperability with files the user already has on
their Windows machine. Unlike a virtual machine, where you have to use network shares or
other solutions to share files between the host and guest OS, WSL has direct access to all your
Windows drives to allow for easy interop.
Windows file systems differ substantially from Linux file systems, and this post looks into how
WSL bridges those two worlds.
File systems on Linux
Linux abstracts file systems operations through the Virtual File System (VFS), which provides
both an interface for user mode programs to interact with the file system (through system calls
such as open, read, chmod, stat, etc.) and an interface that file systems have to implement. This
allows multiple file systems to coexist, providing the same operations and semantics, with VFS
giving a single namespace view of all these file systems to the user.
File systems are mounted on different directories in this namespace. For example, on a typical
Linux system your hard drive may be mounted at the root, /, with directories such as /dev, /proc,
/sys, and /mnt/cdrom all mounting different file systems which may be on different devices.
Examples of file systems used on Linux include ext4, rfs, FAT, and others.
VFS implements the various system calls for file system operations by using a number of data
structures such as inodes, directory entries and files, and related callbacks that file systems must
implement.
Inodes
The inode is the central data structure used in VFS. It represents a file system object such as a
regular file, directory, symbolic link, etc. An inode contains information about the file type, size,
permissions, last modified time, and other attributes. For many common Linux disk file systems
such as ext4, the on-disk data structures used to represent file metadata directly correspond to the
inode structure used by the Linux kernel.
While an inode represents a file, it does not represent a file name. A single file may have
multiple names, or hard links, but only one inode.
File systems provide a lookup callback to VFS which is used to retrieve an inode for a particular
file, based on the parent inode and the child name. File systems must implement a number of
other inode operations such as chmod, stat, open, etc.
Directory entries
VFS uses a directory entry cache to represent your file system namespace. Directory entries only
exist in memory, and contain a pointer to the inode for the file. For example, if you have a path
like /home/user/foo, there is a directory entry for home, user, and foo, each with a pointer to an
inode. Directory entries are cached for fast lookup, but if an entry is not yet in the cache, the
inode lookup operation is used to retrieve the inode from the file system so a new directory entry
can be created.
File objects
When an inode is opened, .
Honest Reviews of Tim Han LMA Course Program.pptxtimhan337
Personal development courses are widely available today, with each one promising life-changing outcomes. Tim Han’s Life Mastery Achievers (LMA) Course has drawn a lot of interest. In addition to offering my frank assessment of Success Insider’s LMA Course, this piece examines the course’s effects via a variety of Tim Han LMA course reviews and Success Insider comments.
Introduction to AI for Nonprofits with Tapp NetworkTechSoup
Dive into the world of AI! Experts Jon Hill and Tareq Monaur will guide you through AI's role in enhancing nonprofit websites and basic marketing strategies, making it easy to understand and apply.
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...Levi Shapiro
Letter from the Congress of the United States regarding Anti-Semitism sent June 3rd to MIT President Sally Kornbluth, MIT Corp Chair, Mark Gorenberg
Dear Dr. Kornbluth and Mr. Gorenberg,
The US House of Representatives is deeply concerned by ongoing and pervasive acts of antisemitic
harassment and intimidation at the Massachusetts Institute of Technology (MIT). Failing to act decisively to ensure a safe learning environment for all students would be a grave dereliction of your responsibilities as President of MIT and Chair of the MIT Corporation.
This Congress will not stand idly by and allow an environment hostile to Jewish students to persist. The House believes that your institution is in violation of Title VI of the Civil Rights Act, and the inability or
unwillingness to rectify this violation through action requires accountability.
Postsecondary education is a unique opportunity for students to learn and have their ideas and beliefs challenged. However, universities receiving hundreds of millions of federal funds annually have denied
students that opportunity and have been hijacked to become venues for the promotion of terrorism, antisemitic harassment and intimidation, unlawful encampments, and in some cases, assaults and riots.
The House of Representatives will not countenance the use of federal funds to indoctrinate students into hateful, antisemitic, anti-American supporters of terrorism. Investigations into campus antisemitism by the Committee on Education and the Workforce and the Committee on Ways and Means have been expanded into a Congress-wide probe across all relevant jurisdictions to address this national crisis. The undersigned Committees will conduct oversight into the use of federal funds at MIT and its learning environment under authorities granted to each Committee.
• The Committee on Education and the Workforce has been investigating your institution since December 7, 2023. The Committee has broad jurisdiction over postsecondary education, including its compliance with Title VI of the Civil Rights Act, campus safety concerns over disruptions to the learning environment, and the awarding of federal student aid under the Higher Education Act.
• The Committee on Oversight and Accountability is investigating the sources of funding and other support flowing to groups espousing pro-Hamas propaganda and engaged in antisemitic harassment and intimidation of students. The Committee on Oversight and Accountability is the principal oversight committee of the US House of Representatives and has broad authority to investigate “any matter” at “any time” under House Rule X.
• The Committee on Ways and Means has been investigating several universities since November 15, 2023, when the Committee held a hearing entitled From Ivory Towers to Dark Corners: Investigating the Nexus Between Antisemitism, Tax-Exempt Universities, and Terror Financing. The Committee followed the hearing with letters to those institutions on January 10, 202
Embracing GenAI - A Strategic ImperativePeter Windle
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
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Meselson and Stahl in 1957 gave experimental evidence that each DNA .pdf
1. Meselson and Stahl in 1957 gave experimental evidence that each DNA strand served as a
template for new DNA synthesis, a process called semi-conservative replication. At that time,
there were three proposed models for DNA replication put forward by the scientific community
after DNA structure had been discovered- Semi-conservative, conservative and dispersive
replication.
Semi-conservative replication: In this model, the two strands of DNA unwind from each other,
and each acts as a template for synthesis of a new, complementary strand. This results in two
DNA molecules with one original strand and one new strand.
Conservative replication. In this model, DNA replication results in one molecule that consists of
both original DNA strands and another molecule that consists of two new strands with exactly
the same sequences as the original molecule.
Meselson and Stahl used the density gradient sedimentation experiment (done with the E.Coli
bacteria) to establish that DNA replicates using the semi-conservative model of replication as
follows:
Results of analysis:
Solution
Meselson and Stahl in 1957 gave experimental evidence that each DNA strand served as a
template for new DNA synthesis, a process called semi-conservative replication. At that time,
there were three proposed models for DNA replication put forward by the scientific community
after DNA structure had been discovered- Semi-conservative, conservative and dispersive
replication.
Semi-conservative replication: In this model, the two strands of DNA unwind from each other,
and each acts as a template for synthesis of a new, complementary strand. This results in two
DNA molecules with one original strand and one new strand.
Conservative replication. In this model, DNA replication results in one molecule that consists of
both original DNA strands and another molecule that consists of two new strands with exactly
the same sequences as the original molecule.
Meselson and Stahl used the density gradient sedimentation experiment (done with the E.Coli
bacteria) to establish that DNA replicates using the semi-conservative model of replication as
follows:
Results of analysis:
