1) Even the simplest types of infectious agents must have DNA, RNA, and protein.
a) Not all infectious agents have DNA, RNA, and protein. Some viruses have RNA and protein
only, viroids have only protein, and prions are composed only of RNA.
b) Not all infectious agents have DNA, RNA, and protein. Some viruses have DNA and protein
only, viroids have only protein, and prions are composed only of DNA.
c) Not all infectious agents have DNA, RNA, and protein. Some viruses have RNA and protein
only, viroids have only DNA, and prions are composed only of protein.
d) Not all infectious agents have DNA, RNA, and protein. Some viruses have DNA and protein
only, viroids have only RNA, and prions are composed only of protein.
2) A large amount of the DNA in eukaryotic cells has no function and is called \"junk DNA\".
a) Although some DNA in cells has been called \"junk DNA\" and have no function, their
concentraition maintains constant to allow other DNA act normally.
b) Although some DNA in cells has been called \"junk DNA\", it may simply be that we do not
yet understand its function.
c) Although some DNA in cells has been called \"junk DNA\", they have well known function to
store energy in the cells.
d) Although some DNA in cells has been called \"junk DNA\", but this is due to their mutation
and they could mutate again to restore their functions.
Solution
1. Even the simplest types of infectious agents must have DNA, RNA, and protein.
ANSWER: The above statement is false. This must be corrected as:
Even the simplest types of infectious agents must have either DNA and protein coat , RNA only
or Protein Coat only.
So the answer is:
D. Not all infectious agents have DNA, RNA, and protein. Some viruses have DNA and protein
only, viroids have only RNA, and prions are composed only of protein.
2.
Junk DNA/Non Coding DNA
The DNA sequences that does not code for a protein, that usually occurs in repetitive sequences
of nucleotides, and does not seem to serve any useful purpose are called Junk DNA.
So the answer is:
B. Although some DNA in cells has been called \"junk DNA\", it may simply be that we do not
yet understand its function..
DNA organization or Genetic makeup in Prokaryotic and Eukaryotic SystemsBir Bahadur Thapa
DNA organization or Genetic makeup in Prokaryotic and Eukaryotic Systems!! It is prepared under the syllabus of Tribhuwan University, Nepal, MSc. 3rd Semester as a lecture class!!
DNA organization or Genetic makeup in Prokaryotic and Eukaryotic SystemsBir Bahadur Thapa
DNA organization or Genetic makeup in Prokaryotic and Eukaryotic Systems!! It is prepared under the syllabus of Tribhuwan University, Nepal, MSc. 3rd Semester as a lecture class!!
In-class introduction to basic Punnett square set-up and problem s.docxbradburgess22840
In-class introduction to basic Punnett square set-up and problem solving, Part 1
Problem-solving tips:
· A Punnett square allows you to predict the possible genetic outcome of children based on the genetic make-up of the parents.
· First, read the problem and figure out whether the trait of interest or genetic disorder is found on the dominant allele or the recessive allele because that will have an impact on how you interpret the results of the Punnett square.
· Select a letter to represent the trait or disorder and define the dominant and recessive alleles. For example: For eye color, B (dominant) = brown eyes and b (recessive) = blue eyes. For achondroplasia (dwarfism), A (dominant) = achondroplasia and a (recessive) = normal allele.
· If it is a sex-linked question, remember to include the sexual genotypes of the parents (XX for mom and XY for dad).
· Write down all possible genotypes & phenotypes and use this information to help you set up the Punnett square.
1. Practice question on a human trait. In reality, eye color is controlled by multiple genes and is a complex trait. For simplicity, we’ll assume that brown eyes are dominant to blue eyes. Answer the questions below.
a) Select a letter for this trait and define the dominant and recessive alleles.
B (dominant) =
b (recessive) =
b) Write down all possible genotypes and phenotypes for individuals in the population
Possible genotypes
(the 2 alleles an individual has)
Possible phenotypes (the physical appearance of a trait)
Homozygous dominant individuals
Homozygous recessive individuals
Heterozygous individuals
c) Set up the Punnett square and solve this problem. Kristy is heterozygous and Mark has blue eyes. What percentage of their offspring will have blue eyes?
Kristy's genotype
Mark's genotype
a) Select a letter for this genetic condition and define the dominant and recessive alleles.
F (dominant) =
f (recessive) =
b) Write down all possible genotypes and phenotypes for individuals in the population
Possible genotypes
(the 2 alleles an individual has)
Possible phenotypes (the physical appearance of a trait)
Homozygous dominant individuals
Homozygous recessive individuals
Heterozygous individuals
c) Set up the Punnett square and solve this problem. Kristy and Mark are carriers for cystic fibrosis. The term carrier is only used when a condition is on the recessive allele. Carriers are heterozygous individuals who are normal and show no symptoms of the disorder, but they have the ability to pass on the mutated recessive allele to their offspring. What percentage of their children will be normal? What percentage of their children will be carriers?
Kristy's genotype
Mark's genotype
2. Practice question on a genetic condition. Cystic fibrosis (CF) is an autosomal, recessive condition that results in mucus buildup in the lungs and digestive system organs. As a result, CF .
DNA is a double helical structure that transfers the genetic information from one generation to another. it consists of two strands with the four nucleotide basis .The four nucleotides contains adenine, cytosine, guanine, thymine .These four nuclic basis are such arranged and coiled with the help of hydrogen bonds and forms the helical structure of DNA. In RNA the thymine is replaced with uracil. Here you will learn the replication ,transcription and translation process in DNA.
2.)a.) Are viruses that infect Archaea different from other bacter.pdfpigsleypeyno82989
2.)
a.) Are viruses that infect Archaea different from other bacterial viruses?
Yes, they have only DNA genomes.
b.) Of the following examples, which is not a common part of many viral life cycles?
Once an adenovirus virion is inside a cell, host RNA polymerases transcribe early genes
c.) Some viruses are associated with cancer. There are a variety of ways in which viruses may
affect a cell in a way that may lead to cancer. Based on your knowledge of the life cycles of
viruses and of molecular genetics, what is one way that this might happen?
d.) Which statement describes the difference between the normal neuron protein PrPC and the
pathogenic version PrPSc?Yes, they do not have protein coats because of the extreme conditions
where they live.No, they consist of nucleic acid and protein just like other bacterial viruses.Yes,
the structure of archaeal viruses is always completely different from other bacterial viruses.
Yes, they have only DNA genomes.
b.) Of the following examples, which is not a common part of many viral life cycles?The pox
virus genome is replicated by DNA polymerase.Adenoviruses attach to the cell and the
nucleocapsid is taken up by the cell.Pox viruses replicate in the cytoplasm of the cell.
Once an adenovirus virion is inside a cell, host RNA polymerases transcribe early genes
c.) Some viruses are associated with cancer. There are a variety of ways in which viruses may
affect a cell in a way that may lead to cancer. Based on your knowledge of the life cycles of
viruses and of molecular genetics, what is one way that this might happen?Viral genetic
material inserts into a host genome close to the termination sequence, far downstream of the
promoter, and causes a silent mutation.Viral genetic material inserts into a host genome in a
repetitive noncoding region.Viral genetic material inserts into a host genome, placing a
promoter in a new location and turning on a gene that had previously been turned off.Viral
genetic material inserts into a host genome and causes a mutation in a promoter, turning off a
gene that had previously been turned on.
d.) Which statement describes the difference between the normal neuron protein PrPC and the
pathogenic version PrPSc?PrPC is just a protein and the PrPSc is a protein with nucleic acid
inside.The PrPC protein contains more beta sheets while the PrPSc protein contains more alpha
helices.The three-dimensional conformation is identical, but the amino acid sequence is
different.The amino acid sequence is identical, but the three-dimensional conformation is
different.
Solution
Archaea can be infected by double stranded DNA virus that are unrelated to any other form of
virus and have variety unusual shapes. These are single stranded and double stranded DNA virus.
Species of the Ligameneviral and the families Ampullaviridae, Biccaudaviridae and
Clavaviridea, infects hyperthermophiic archaea species. These viruses have lipid bilayer coat
inside the icosahedral protein coat, and some of the viru.
In 9. 3-1 BALANCE SHEET The assets of Dallas & Associates consist e.pdfeyewatchsystems
In 9. 3-1 BALANCE SHEET The assets of Dallas & Associates consist entirely of current assets
and net plant and equipment. The firm has total assets of $2.5 million and net plant and
equipment equals $2 million. It has notes payable of $150,000, long-term debt of $750,000, and
total common equity of S1.5 million. The firm does have accounts payable and accruals on its
balance sheet. The finm only finances with debt and common equity, so it has no preferred stock
on its balance sheet. a. What is the company\'s total debt? b. What is the amount of total
liabilities and equity that appears on the firm\'s balance sheet? c. What is the balance of current
assets on the firm\'s balance sheet? d. What is the balance of current liabilities on the firm\'s
balance sheet? e. What is the amount of accounts payable and accruals on its balance sheet?
(Hint: Consider this as a single line item on the firm\'s balance sheet.) f. What is the firm\'s net
working capital? g. What is the firm\'s net operating working capital? h. What is the explanation
for the difference in your answers to parts f and g?
Solution
(a) Total Assets = $ 2.5 million, The firm\'s total assets are entirely composed of current assets
and net plant and equipment.
Net Plant and Equipment = $ 2 million and Current Assets = Total Assets - Net Plant and
Equipment = 2.5 - 2 = $ 0.5 million
Notes Payable = $ 150000 and Long-Term Debt = $ 750000
Total Common Equity = $ 1.5 million
Owner\'s Equity + Liabilities = Notes Payable + Long-Term Debt + Accruals + Accounts
Payable + Total Common Equity = Total Assets = $ 2.5 million
0.15 + 0.75 + Accounts Payable + Accruals + 1.5 = 2.5
Accounts Payable + Accruals = $ 0.1 million
Total Debt includes long-term capital financing debt and short-term interest bearing debt only.
Accounts Payable and Accruals do not form a part of a firm\'s total debt.
Hence, total debt = Notes Payable + Long-Term Debt = 150000 + 750000 = $ 900000
(b) Total Liabilities = Accounts Payable + Accruals + Long-Term Debt + Notes Payable = 0.1 +
0.75 + 0.15 = $ 1 million
Owner\'s Equity = Total Assets - Total Liabilities = 2.5 - 1 = $ 1.5 million
(c) Current Assets = Total Assets - Net Plant and Equipment = 2.5 - 2 = $ 0.5 million
(d) Current Laibility = Accruals + Accounts Payable = $ 0.1 million
NOTE: The current portion of notes payable(amount due within one year or accounting period)
is treated as current liability and the remaining portion is treated as non-current liability. As the
notes payable portion does not mention the amount due within a year, the entire value of notes
payable is treated as non-current laibility.
NOTE: Please raise separate queries for solutions to the remaining sub-parts..
I need help understanding the Pan Tompkins algorythm, and Id also .pdfeyewatchsystems
I need help understanding the Pan Tompkins algorythm, and I\'d also like to know how to detect
QRS peaks of the ECG (with circles, for example). I\'m using the script in this website as a
guidance, but I have trouble understanding what it\'s doing. Perhaps there\'s a simpler way to do
it? http://cnx.org/contents/YR1BUs9_@1/QRS-Detection-Using-Pan-Tompki
Our teacher has otherwise explained to us FIR filters, and wants us to use them instead of the
low pass, high pass that are mentioned in the website and though I do understand how they work,
I don\'t really know which window is the best, or how many coefficients the filter needs. I\'ve
applying the cut-off frequencies 5-15 because I\'ve seen that\'s where the QRS peaks are
concentrated. Is there any way to improve this as well?
TB=1;
TBn=TB/Fs;
N2=ceil(3.3/TBn);
Fc1 =5;
Fc2 =15 ;
wc1 = 2*pi*Fc1;
wc2 = 2*pi*Fc2;
wcn1 = 2*pi*Fc1/Fs;
wcn2 = 2*pi*Fc2/Fs;
Bz=fir1(N2,[wcn1 wcn2]);
y3=filter(Bz,1,x1);
figure; plot(x1); figure; plot(y3);
Solution
Save the function and use it as given below ::
pan_tompkin(ECG1,200,1) 1st is ECG raw data , 2nd is sampling frequency, 3rd is
for plots -always set it to 1
Matlab Code ------------------------------------ function .
function [qrs_amp_raw,qrs_i_raw,delay]=pan_tompkin(ecg,fs,gr)
%% function [qrs_amp_raw,qrs_i_raw,delay]=pan_tompkin(ecg,fs)
% Complete implementation of Pan-Tompkins algorithm
%% Inputs
% ecg : raw ecg vector signal 1d signal
% fs : sampling frequency e.g. 200Hz, 400Hz and etc
% gr : flag to plot or not plot (set it 1 to have a plot or set it zero not
% to see any plots
%% Outputs
% qrs_amp_raw : amplitude of R waves amplitudes
% qrs_i_raw : index of R waves
% delay : number of samples which the signal is delayed due to the
% filtering
if ~isvector(ecg)
error(\'ecg must be a row or column vector\');
end
if nargin < 3
gr = 1; % on default the function always plots
end
ecg = ecg(:); % vectorize
%% Initialize
qrs_c =[]; %amplitude of R
qrs_i =[]; %index
SIG_LEV = 0;
nois_c =[];
nois_i =[];
delay = 0;
skip = 0; % becomes one when a T wave is detected
not_nois = 0; % it is not noise when not_nois = 1
selected_RR =[]; % Selected RR intervals
m_selected_RR = 0;
mean_RR = 0;
qrs_i_raw =[];
qrs_amp_raw=[];
ser_back = 0;
test_m = 0;
SIGL_buf = [];
NOISL_buf = [];
THRS_buf = [];
SIGL_buf1 = [];
NOISL_buf1 = [];
THRS_buf1 = [];
%% Plot differently based on filtering settings
if gr
if fs == 200
figure, ax(1)=subplot(321);plot(ecg);axis tight;title(\'Raw ECG Signal\');
else
figure, ax(1)=subplot(3,2,[1 2]);plot(ecg);axis tight;title(\'Raw ECG Signal\');
end
end
%% Noise cancelation(Filtering) % Filters (Filter in between 5-15 Hz)
if fs == 200
%% Low Pass Filter H(z) = ((1 - z^(-6))^2)/(1 - z^(-1))^2
b = [1 0 0 0 0 0 -2 0 0 0 0 0 1];
a = [1 -2 1];
h_l = filter(b,a,[1 zeros(1,12)]);
ecg_l = conv (ecg ,h_l);
ecg_l = ecg_l/ max( abs(ecg_l));
delay = 6; %based on the paper
if gr
ax(2)=subplot(322);plot(ecg_l);axis tight;title(\'Low pass filtered\');
end
%% High Pas.
More Related Content
Similar to 1) Even the simplest types of infectious agents must have DNA, RNA, .pdf
In-class introduction to basic Punnett square set-up and problem s.docxbradburgess22840
In-class introduction to basic Punnett square set-up and problem solving, Part 1
Problem-solving tips:
· A Punnett square allows you to predict the possible genetic outcome of children based on the genetic make-up of the parents.
· First, read the problem and figure out whether the trait of interest or genetic disorder is found on the dominant allele or the recessive allele because that will have an impact on how you interpret the results of the Punnett square.
· Select a letter to represent the trait or disorder and define the dominant and recessive alleles. For example: For eye color, B (dominant) = brown eyes and b (recessive) = blue eyes. For achondroplasia (dwarfism), A (dominant) = achondroplasia and a (recessive) = normal allele.
· If it is a sex-linked question, remember to include the sexual genotypes of the parents (XX for mom and XY for dad).
· Write down all possible genotypes & phenotypes and use this information to help you set up the Punnett square.
1. Practice question on a human trait. In reality, eye color is controlled by multiple genes and is a complex trait. For simplicity, we’ll assume that brown eyes are dominant to blue eyes. Answer the questions below.
a) Select a letter for this trait and define the dominant and recessive alleles.
B (dominant) =
b (recessive) =
b) Write down all possible genotypes and phenotypes for individuals in the population
Possible genotypes
(the 2 alleles an individual has)
Possible phenotypes (the physical appearance of a trait)
Homozygous dominant individuals
Homozygous recessive individuals
Heterozygous individuals
c) Set up the Punnett square and solve this problem. Kristy is heterozygous and Mark has blue eyes. What percentage of their offspring will have blue eyes?
Kristy's genotype
Mark's genotype
a) Select a letter for this genetic condition and define the dominant and recessive alleles.
F (dominant) =
f (recessive) =
b) Write down all possible genotypes and phenotypes for individuals in the population
Possible genotypes
(the 2 alleles an individual has)
Possible phenotypes (the physical appearance of a trait)
Homozygous dominant individuals
Homozygous recessive individuals
Heterozygous individuals
c) Set up the Punnett square and solve this problem. Kristy and Mark are carriers for cystic fibrosis. The term carrier is only used when a condition is on the recessive allele. Carriers are heterozygous individuals who are normal and show no symptoms of the disorder, but they have the ability to pass on the mutated recessive allele to their offspring. What percentage of their children will be normal? What percentage of their children will be carriers?
Kristy's genotype
Mark's genotype
2. Practice question on a genetic condition. Cystic fibrosis (CF) is an autosomal, recessive condition that results in mucus buildup in the lungs and digestive system organs. As a result, CF .
DNA is a double helical structure that transfers the genetic information from one generation to another. it consists of two strands with the four nucleotide basis .The four nucleotides contains adenine, cytosine, guanine, thymine .These four nuclic basis are such arranged and coiled with the help of hydrogen bonds and forms the helical structure of DNA. In RNA the thymine is replaced with uracil. Here you will learn the replication ,transcription and translation process in DNA.
2.)a.) Are viruses that infect Archaea different from other bacter.pdfpigsleypeyno82989
2.)
a.) Are viruses that infect Archaea different from other bacterial viruses?
Yes, they have only DNA genomes.
b.) Of the following examples, which is not a common part of many viral life cycles?
Once an adenovirus virion is inside a cell, host RNA polymerases transcribe early genes
c.) Some viruses are associated with cancer. There are a variety of ways in which viruses may
affect a cell in a way that may lead to cancer. Based on your knowledge of the life cycles of
viruses and of molecular genetics, what is one way that this might happen?
d.) Which statement describes the difference between the normal neuron protein PrPC and the
pathogenic version PrPSc?Yes, they do not have protein coats because of the extreme conditions
where they live.No, they consist of nucleic acid and protein just like other bacterial viruses.Yes,
the structure of archaeal viruses is always completely different from other bacterial viruses.
Yes, they have only DNA genomes.
b.) Of the following examples, which is not a common part of many viral life cycles?The pox
virus genome is replicated by DNA polymerase.Adenoviruses attach to the cell and the
nucleocapsid is taken up by the cell.Pox viruses replicate in the cytoplasm of the cell.
Once an adenovirus virion is inside a cell, host RNA polymerases transcribe early genes
c.) Some viruses are associated with cancer. There are a variety of ways in which viruses may
affect a cell in a way that may lead to cancer. Based on your knowledge of the life cycles of
viruses and of molecular genetics, what is one way that this might happen?Viral genetic
material inserts into a host genome close to the termination sequence, far downstream of the
promoter, and causes a silent mutation.Viral genetic material inserts into a host genome in a
repetitive noncoding region.Viral genetic material inserts into a host genome, placing a
promoter in a new location and turning on a gene that had previously been turned off.Viral
genetic material inserts into a host genome and causes a mutation in a promoter, turning off a
gene that had previously been turned on.
d.) Which statement describes the difference between the normal neuron protein PrPC and the
pathogenic version PrPSc?PrPC is just a protein and the PrPSc is a protein with nucleic acid
inside.The PrPC protein contains more beta sheets while the PrPSc protein contains more alpha
helices.The three-dimensional conformation is identical, but the amino acid sequence is
different.The amino acid sequence is identical, but the three-dimensional conformation is
different.
Solution
Archaea can be infected by double stranded DNA virus that are unrelated to any other form of
virus and have variety unusual shapes. These are single stranded and double stranded DNA virus.
Species of the Ligameneviral and the families Ampullaviridae, Biccaudaviridae and
Clavaviridea, infects hyperthermophiic archaea species. These viruses have lipid bilayer coat
inside the icosahedral protein coat, and some of the viru.
In 9. 3-1 BALANCE SHEET The assets of Dallas & Associates consist e.pdfeyewatchsystems
In 9. 3-1 BALANCE SHEET The assets of Dallas & Associates consist entirely of current assets
and net plant and equipment. The firm has total assets of $2.5 million and net plant and
equipment equals $2 million. It has notes payable of $150,000, long-term debt of $750,000, and
total common equity of S1.5 million. The firm does have accounts payable and accruals on its
balance sheet. The finm only finances with debt and common equity, so it has no preferred stock
on its balance sheet. a. What is the company\'s total debt? b. What is the amount of total
liabilities and equity that appears on the firm\'s balance sheet? c. What is the balance of current
assets on the firm\'s balance sheet? d. What is the balance of current liabilities on the firm\'s
balance sheet? e. What is the amount of accounts payable and accruals on its balance sheet?
(Hint: Consider this as a single line item on the firm\'s balance sheet.) f. What is the firm\'s net
working capital? g. What is the firm\'s net operating working capital? h. What is the explanation
for the difference in your answers to parts f and g?
Solution
(a) Total Assets = $ 2.5 million, The firm\'s total assets are entirely composed of current assets
and net plant and equipment.
Net Plant and Equipment = $ 2 million and Current Assets = Total Assets - Net Plant and
Equipment = 2.5 - 2 = $ 0.5 million
Notes Payable = $ 150000 and Long-Term Debt = $ 750000
Total Common Equity = $ 1.5 million
Owner\'s Equity + Liabilities = Notes Payable + Long-Term Debt + Accruals + Accounts
Payable + Total Common Equity = Total Assets = $ 2.5 million
0.15 + 0.75 + Accounts Payable + Accruals + 1.5 = 2.5
Accounts Payable + Accruals = $ 0.1 million
Total Debt includes long-term capital financing debt and short-term interest bearing debt only.
Accounts Payable and Accruals do not form a part of a firm\'s total debt.
Hence, total debt = Notes Payable + Long-Term Debt = 150000 + 750000 = $ 900000
(b) Total Liabilities = Accounts Payable + Accruals + Long-Term Debt + Notes Payable = 0.1 +
0.75 + 0.15 = $ 1 million
Owner\'s Equity = Total Assets - Total Liabilities = 2.5 - 1 = $ 1.5 million
(c) Current Assets = Total Assets - Net Plant and Equipment = 2.5 - 2 = $ 0.5 million
(d) Current Laibility = Accruals + Accounts Payable = $ 0.1 million
NOTE: The current portion of notes payable(amount due within one year or accounting period)
is treated as current liability and the remaining portion is treated as non-current liability. As the
notes payable portion does not mention the amount due within a year, the entire value of notes
payable is treated as non-current laibility.
NOTE: Please raise separate queries for solutions to the remaining sub-parts..
I need help understanding the Pan Tompkins algorythm, and Id also .pdfeyewatchsystems
I need help understanding the Pan Tompkins algorythm, and I\'d also like to know how to detect
QRS peaks of the ECG (with circles, for example). I\'m using the script in this website as a
guidance, but I have trouble understanding what it\'s doing. Perhaps there\'s a simpler way to do
it? http://cnx.org/contents/YR1BUs9_@1/QRS-Detection-Using-Pan-Tompki
Our teacher has otherwise explained to us FIR filters, and wants us to use them instead of the
low pass, high pass that are mentioned in the website and though I do understand how they work,
I don\'t really know which window is the best, or how many coefficients the filter needs. I\'ve
applying the cut-off frequencies 5-15 because I\'ve seen that\'s where the QRS peaks are
concentrated. Is there any way to improve this as well?
TB=1;
TBn=TB/Fs;
N2=ceil(3.3/TBn);
Fc1 =5;
Fc2 =15 ;
wc1 = 2*pi*Fc1;
wc2 = 2*pi*Fc2;
wcn1 = 2*pi*Fc1/Fs;
wcn2 = 2*pi*Fc2/Fs;
Bz=fir1(N2,[wcn1 wcn2]);
y3=filter(Bz,1,x1);
figure; plot(x1); figure; plot(y3);
Solution
Save the function and use it as given below ::
pan_tompkin(ECG1,200,1) 1st is ECG raw data , 2nd is sampling frequency, 3rd is
for plots -always set it to 1
Matlab Code ------------------------------------ function .
function [qrs_amp_raw,qrs_i_raw,delay]=pan_tompkin(ecg,fs,gr)
%% function [qrs_amp_raw,qrs_i_raw,delay]=pan_tompkin(ecg,fs)
% Complete implementation of Pan-Tompkins algorithm
%% Inputs
% ecg : raw ecg vector signal 1d signal
% fs : sampling frequency e.g. 200Hz, 400Hz and etc
% gr : flag to plot or not plot (set it 1 to have a plot or set it zero not
% to see any plots
%% Outputs
% qrs_amp_raw : amplitude of R waves amplitudes
% qrs_i_raw : index of R waves
% delay : number of samples which the signal is delayed due to the
% filtering
if ~isvector(ecg)
error(\'ecg must be a row or column vector\');
end
if nargin < 3
gr = 1; % on default the function always plots
end
ecg = ecg(:); % vectorize
%% Initialize
qrs_c =[]; %amplitude of R
qrs_i =[]; %index
SIG_LEV = 0;
nois_c =[];
nois_i =[];
delay = 0;
skip = 0; % becomes one when a T wave is detected
not_nois = 0; % it is not noise when not_nois = 1
selected_RR =[]; % Selected RR intervals
m_selected_RR = 0;
mean_RR = 0;
qrs_i_raw =[];
qrs_amp_raw=[];
ser_back = 0;
test_m = 0;
SIGL_buf = [];
NOISL_buf = [];
THRS_buf = [];
SIGL_buf1 = [];
NOISL_buf1 = [];
THRS_buf1 = [];
%% Plot differently based on filtering settings
if gr
if fs == 200
figure, ax(1)=subplot(321);plot(ecg);axis tight;title(\'Raw ECG Signal\');
else
figure, ax(1)=subplot(3,2,[1 2]);plot(ecg);axis tight;title(\'Raw ECG Signal\');
end
end
%% Noise cancelation(Filtering) % Filters (Filter in between 5-15 Hz)
if fs == 200
%% Low Pass Filter H(z) = ((1 - z^(-6))^2)/(1 - z^(-1))^2
b = [1 0 0 0 0 0 -2 0 0 0 0 0 1];
a = [1 -2 1];
h_l = filter(b,a,[1 zeros(1,12)]);
ecg_l = conv (ecg ,h_l);
ecg_l = ecg_l/ max( abs(ecg_l));
delay = 6; %based on the paper
if gr
ax(2)=subplot(322);plot(ecg_l);axis tight;title(\'Low pass filtered\');
end
%% High Pas.
I believe my chosen answer is correct, but Im not entirely positiv.pdfeyewatchsystems
I believe my chosen answer is correct, but I\'m not entirely positive. Any help is appreciated. The
enzyme glucose-6-phosphate dehydrogenase deficiency (G6PD) is inherited as a recessive gene
on the X chromosome in humans. Below is a pedigree of a family affected by G6PD. X-
represents a mutant X chromosome. What is I-1\'s genotype? XX. XX- X-X- could be A or B
Solution
G6PD is recessive traits disease and so in I-2 the Male would be having diseased X chromosome
recessive traits now for being disease in II a male must have a single disease X trait for that if
female is homozygous for non disease trait of heterzygous for disease trait in both condition in I
is possible.
So condition A and B both is correct.
how does genies transcript illustrate the statement we learn to .pdfeyewatchsystems
how does genie\'s transcript illustrate the statement \"we learn to be human.\" aren\'t we human?
Solution
The transcript of genie illustrates that a wild child living in captivity, when given proper care and
attention can learn to be human. initially, she does not respond to the methods that were applied
for the betterment of her and to let her learn to talk. But slowly she was left to learn at her own
pace and understanding. The study on her suggested that we the humans have the innate ability
to learn words and use them correctly. So when genie was provided the situation she steadily
learned the ability to talk..
How many different alleles can a gene have Explain.SolutionEa.pdfeyewatchsystems
How many different alleles can a gene have? Explain.
Solution
Each gene can have many alleles. But a diploid organism typically contains two alleles in each
gene. This because in sexual reproduction one allele is received from each parent. If a gene
contains more than two alleles, it causes chromosomal abnormalities and thus diseases..
Comment on the behavior of the different types of flexures (crab-leg.pdfeyewatchsystems
Comment on the behavior of the different types of flexures (crab-leg, folded flexure, serpentine,
cartwheel, straight, notched). Which would have the highest cross-axis sensitivity? The lowest?
Which would you use for a device that experiences large displacements?
Solution
Serpentine have highest cross axis sensitivity.straight have lowest sensitivity. And crab leg is
gives promising displacements in three directions... So its best suited for device..
human color-blindness mutations result from A.reciprocal translocat.pdfeyewatchsystems
human color-blindness mutations result from: A.reciprocal translocation B. dosge compensation
C. Unequal crossing over D. Chromosome nondisjunction E. Fragile X syndrome I think answer
is B. dosage compensation???
Solution
Human Color blindness results from unequal crossing over . Gene\'s for red and green Color
vision lie adjacent on x chromosome . Evolving from a common ancestor they are similar enough
to mispair occasionally which results in the production of gametes lacking some or all of the one
of the two genes resulting in red green color blindness..
Generation 1 2 3 4 6 5 6 7 SolutionThe chances of their first c.pdfeyewatchsystems
Generation 1 2 3 4 6 5 6 7
Solution
The chances of their first child showing the trait is 50%
As this trait is being carried by the females in generation II, and in generation III due to the
congenial marriage among the family members the homozygosity of trait increases.
Due to which this trait again shows up again generation IV. If the individual IV-4 and IV-7 get
married, which is again congenial marriage among the family member. The risk factor for the
children having this trait increases..
Explain how cell division occurs in the germ (meiosis) cells of anim.pdfeyewatchsystems
Explain how cell division occurs in the germ (meiosis) cells of animals. Be sure to state and
describe each stage and phase of the cell cycle.
Solution
Germ cell division occurs by Meiosis. The division is explained with oogenesis as example.
Diploid primordial germ cells that form the germinal epithelium of fetal ovary divide repeatedly
by mitosis to produce diploid oogonia.
Diploid oogonia produced in groups just under the surface of the ovary. In contrast to
spermatogenesis, only one oogonium from each group normally grows into a diploid primary
oocyte. Each primary oocyte then enters prophase of Meiosis I and remains in this arrested state
until puberty.
As puberty begins, a primary oocyte completes Meiosis I each month to produce a secondary
oocyte and a polar body.
The secondary oocyte undergoes Meiosis II at fertilization to produce an ovum and a second
polar body..
Ecologists are sampling mice from different parks in New York City. M.pdfeyewatchsystems
Ecologists are sampling mice from different parks in New York City. Most of the mice these
researchers collect demonstrate straight\" tails; however, few mice exhibit \"curly\" tails. In this
scenario, a \"straight tail would be considered the henotype, and a \"curly\" tail would be
considered a phenotype. O heterogametic; homogametic O polymorphism; wild-type O mutant;
wild-type O wild-type; mutant None of the above.
Solution
The answer to your first question is wrong. Most of the researchers find mice with straight tails,
while a few find mice with curly tails. However, this finding does not prove that the curly-tailed
mice are mutants. A mutant phenotype is much rarer than the wild type. The curly phenotype
could simply be a recessive phenotype with a phenotypic frequency lower than that of the
straight phenotype, which might be the dominant trait. Thus, the answer should be \'None of the
above\'.
The answer to the second question is correct. Since the parents\' and the child\'s DNA sequences
differ in only a single nucleotide, it is a point mutation.
The answer to the third question is wrong. Since the flower colour is determined by a gene with
2 alleles, 1 of which is dominant, to get a red colour, the orchid just needs a single dominant
allele. Hence the answer is 1.
Your answer to the fourth question is wrong. The parent and daughter cells may not be
completely identical, and we must take into account the de-novo mutations. Therefore, the
answer should be \'The DNA sequences of the parent and daughter cells were largely similar,
with the exception of any de-novo mutations\'..
Describe the electronic configuration of a given element. Why are we.pdfeyewatchsystems
Describe the electronic configuration of a given element. Why are we interested in electrons (two
primary answers). (With regard to Human Anatomy and Physiology, what is the importance of
electrons?)
Solution
Electronic configuration is the arrangement of electron around the nucleus of an atom. Number
of electrons orbiting the nucleus determines the energy level of an atom. This electron makes the
bonds like, ionic bond, covalent bond and htdrogen bond. These bonds help in electrostatic
interactions between the molecules and helps molecules like haemoglobin, myoglobin to be
folded and form a three dimensional structure and function properly. these electrons helps in
phosphorylation, hydrolysisetc,..
Assets are listed on the balance sheet in order ofincreasing size .pdfeyewatchsystems
Assets are listed on the balance sheet in order of
increasing size and relative life.
decreasing size.
relative life.
decreasing liquidity.a.
increasing size and relative life.b.
decreasing size.c.
relative life.d.
decreasing liquidity.
Solution
Liquidity of the the assets means how quick an asset can be converted to cash. More quicker it
can be higher the liquidity.
Standard format of balance states assets to be reported on the basis of their liquidity. Assets with
highest liquidity should be reported on Top
Therefore in Assets we report first current assets and then PPE. Even in Current assets we report
Cash and Marketable securities (most liquid aset) on top. There fore
Assets are listed on the balance sheet in order of
decreasing liquidity.d.
decreasing liquidity..
1-A) What types of data document patterns and processes of evoluti.pdfeyewatchsystems
1-
A) What types of data document patterns and processes of evolution?
B) What does homology mean and how is it important with respect to evolution? How types of
homology are for providing evidence of evolution?
C) does the fossil record provide support for the of evolution? Why do sugar gliders and flying
squirrels look so much alike?
D) What do we call the similarity between flying squirrels and sugar gliders?
E) Provide a complete explanation for the question, what is evolution Vocabulary
Solution
Answer A) There are mainly four types of data which document patterns and process of
evolution; Homology, fossil record, direct observations and biogeography.
Answer B) Homology means sharing a similar structure due to relatedness. this is important with
respect to evolution because homology tells us that two different animals which shares the
homology structures had a common ancestor during evolution. there are two types of homology
in evolution; paralogs, which tells us that two copy of one gene located at two different positions
on genome inside a single organism has evolved as a duplication event and such genes are called
paralogous. while orthologs tells us that if two genes are similar in two different organism\'s
genomes then they were originated in their ancestor.
Answer C) Yes, fossil record is one of the data documentation for the evolutionary studies and
they help to find the missing links between various points of evolution. sugar gliders and flying
squirrels look alike because they share similarity in structures such as white belly, big eyes, same
size, and a thin layer of skin under the legs for gliding in the air.
Answer D) We call this kind of similarity in same functional structures as analogy..
Which of the following does not take place in a signaltransduction pa.pdfeyewatchsystems
Which of the following does not take place in a signaltransduction pathway? O An external
signaling molecule binds to a specific receptor. O An activated membrane-bound receptor
undergoes a conformational change. O Proteins in the pathway add or remove phosphate groups
from other proteins. O Activated proteins move into the nucleus to activate transcription factors.
O Proteins are activated when histones are deacetylated.
Solution
1) Protiens are activated when histones are deacetylated.
Histones are not involved in signal transduction.
2) The mutation could allow telomerase to be expressed when it should not.
Cancerous cells continously express telomerase as a result of which these cells have their
telomeres in tact and thus they constantly divide.
3)Retinoblsstoma results when an idividual inherits mutant gene from both the parents both of
which are necessary for cancer to develop.
4). DNA repair genes..
Which term indicates the evolution of many closely related species fr.pdfeyewatchsystems
Which term indicates the evolution of many closely related species from one or a few ancestral
species in a relatively short period of time? microevolution preadaptations adaptive radiation
allometric growth allopatric speciation A significant reduction in the amount of which gas
would allow more ultraviolet radiation to reach Earth\'s surface? oxygen methane ozone
cyanide carbon dioxide The most common outcome of the process in the accompanying figure
is no viable gametes. This happens because of hybrid inviability defective meiosis. gametic
isolation species incompatibility.
Solution
1. microevolution
2. defects in ozone gas layer allows more UV rays to come to the earth
3. because of defects in meiosis no viable gametes are formed.
you inoculate an LB tube with an aerobic bacterium and incubate it f.pdfeyewatchsystems
you inoculate an LB tube with an aerobic bacterium and incubate it for 24 hours. what kind of
growth do you expect to see and why?
Solution
If a liquid media in a tube is inoculated with aerobic bacteria and incubated for 24hrs, the growth
can be seen as a surface pellicle. Bacteria grows on the surface of the liquid media. This is
because it is the area where oxygen is available for the bacteria to utilize. Whereas anaerobic
bacteria grow at the bottom of the tube where air is absent..
What is meant by the relationship or the two strands of DNA The rig.pdfeyewatchsystems
What is meant by the relationship or the two strands of DNA? The right handed twisted
configuration of the strands. The arrangement branching pattern of the strands of DNA. The
arrangement where A is matched with T and G is matched with C. The staircase arrangement of
paired purine and pyrimidine bases linked by ionic bonds. The alignment or strands so one
strand starts with a 3. carbon, the other with a 5 carbon. integrative you have discovered a new
species bacteria. To begin your investigation of this content of ONA from the bacterial cells 10%
what is the adenine content? 10% 20% 40% 60% it is impossible to calculate this number for
eukaryotes. What happens during the process of transcription? Proteins are synthesized from
RNA molecules RNA is synthesized from protein DNA is synthesized from protein RNA is
synthesized from DNA Proteins are synthesized from DNA molecules What is a TATA box? a
nucleotide sequence for a codon on mRNA a sequence in the promoter of DNA a nucleotide
sequence for an anticodon on IRNA molecule a nucleotide sequence an indicating a stop codon
for translation a special protein in the enhancer region that is required for the initiation of
transcription What is an exon? A protein that is edited out of RNA after translation. A part of
RNA that is removed during the processing of an RNA molecule in the nucleus. A transfer RNA
that binds to the codon and carries an amino acid. A series of amino acids at the 3\' end of the
mRNA molecule that protect transcript. A part of an intact. Mature mRNA that leaves the
nucleus. What account for the uniform diameter of the DNA molecule? The two sides of the
DNA molecule are held together by hydrogen bonds. A purine always bonds with a pyrimidine.
One side of the DNA molecule has an unconnected 5\' phosphate group and the opposite end has
an unconnected 3\' hydroxyl group. The 3\' carbon of one deoxyribose and the 5\' carbon of
another deoxyribose bond together. The alternating sugar and phosphate backbone colls around
the outside of the hellx. If you isolate a single nucleotide from a nucleic acid chain and
determine that the nitrogenous ring structure is guanine, what can you say with certainty about
the type of molecule that the nucleotide could have come from? The molecule could be DNA
but not RNA. The molecule could be RNA but not DNA. The molecule is neither DNA nor
RNA. The molecule could be either DNA or RNA. The molecule could be double-stranded
DNA but not single-stranded DNA.
Solution
34. The correct answer is C.
35. The correct answer is B. ATP, RNA, and DNA
36. The correct answer is B. Purine always bonds with Pyrimidine
37. The correct answer is D. The molecule could either be DNA or RNA..
Which of the following categories of control can include the logical.pdfeyewatchsystems
Which of the following categories of control can include the logical mechanisms used to control
access and authenticate users?
A. Administrative
B. Clerical
C. Technical
D. Physical
Solution
Answer: C. Technical
Technical controls can be a software or hardware. They are some logical mechanisms that can be
used to control the access and authenticate users and identified unusual activities and restricted
unauthorized access.
Clerical is nonexistent category towards technical.
Physical controls include locks, guards, alarms and gates.
Admisirative controls are procedurals..
What is it, and how does Chromatic Dispersion Compensation work.pdfeyewatchsystems
What is it, and how does Chromatic Dispersion Compensation work?
Solution
Chromatic dispersion is a phenomenon that is an important factor in fiber optic communications.
It is the result of the different colors, or wavelengths, in a light beam arriving at their destination
at slightly different times. The result is a spreading, or dispersion, of the on-off light pulses that
convey digital information.
Chromatic dispersion is common place, as it is actually what causes rainbows - sunlight is
dispersed by droplets of water in the air. Sir Isaac Newton observed this phenomenon when he
passed sunlight through a prism and saw it diverge into a spectrum of different colors. This
dispersion occurs because different colors, or light frequencies, act slightly differently as they
pass through a medium such as glass. In fiber-based systems, an optical fiber, comprised of a
core and cladding with differing refractive index materials, inevitably causes some wavelengths
of light to travel slower or faster than others.
Chromatic dispersion is due to inherent property of silica fiber. The speed of a light wave is
dependent on the refractive index, , of the medium within which it is propagating. In silica fiber,
as well as many other materials, varies with wavelength. Therefore, different wavelength
channels will travel at slightly different speeds within the fiber. Because, even a pulse of light
from a laser has some spectral width, each of its wavelength components will travel at slightly
different speeds within the fiber. This results in a spreading of the transmission pulse as it travels
through the fiber..
True or FalseDue to developed molecular tools, functional genes c.pdfeyewatchsystems
True or False:
Due to developed molecular tools, functional genes can be identified directly in the clinical and
environmental samples.
Solution
True because the functional genes can be directly identified in samples by the metagenomic
arrays, using special biomarkers characterized by a unique order for the specific gene or gene
sequence, SNPs etc and their molecular structures. This biomarker array is sometimes followed
by the amplification tool and sequencing method for the entire DNA or RNA extracted from the
sample..
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.
Unit 8 - Information and Communication Technology (Paper I).pdfThiyagu K
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
Biological screening of herbal drugs: Introduction and Need for
Phyto-Pharmacological Screening, New Strategies for evaluating
Natural Products, In vitro evaluation techniques for Antioxidants, Antimicrobial and Anticancer drugs. In vivo evaluation techniques
for Anti-inflammatory, Antiulcer, Anticancer, Wound healing, Antidiabetic, Hepatoprotective, Cardio protective, Diuretics and
Antifertility, Toxicity studies as per OECD guidelines
Macroeconomics- Movie Location
This will be used as part of your Personal Professional Portfolio once graded.
Objective:
Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
Acetabularia Information For Class 9 .docxvaibhavrinwa19
Acetabularia acetabulum is a single-celled green alga that in its vegetative state is morphologically differentiated into a basal rhizoid and an axially elongated stalk, which bears whorls of branching hairs. The single diploid nucleus resides in the rhizoid.
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
Instructions for Submissions thorugh G- Classroom.pptxJheel Barad
This presentation provides a briefing on how to upload submissions and documents in Google Classroom. It was prepared as part of an orientation for new Sainik School in-service teacher trainees. As a training officer, my goal is to ensure that you are comfortable and proficient with this essential tool for managing assignments and fostering student engagement.
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.
1) Even the simplest types of infectious agents must have DNA, RNA, .pdf
1. 1) Even the simplest types of infectious agents must have DNA, RNA, and protein.
a) Not all infectious agents have DNA, RNA, and protein. Some viruses have RNA and protein
only, viroids have only protein, and prions are composed only of RNA.
b) Not all infectious agents have DNA, RNA, and protein. Some viruses have DNA and protein
only, viroids have only protein, and prions are composed only of DNA.
c) Not all infectious agents have DNA, RNA, and protein. Some viruses have RNA and protein
only, viroids have only DNA, and prions are composed only of protein.
d) Not all infectious agents have DNA, RNA, and protein. Some viruses have DNA and protein
only, viroids have only RNA, and prions are composed only of protein.
2) A large amount of the DNA in eukaryotic cells has no function and is called "junk DNA".
a) Although some DNA in cells has been called "junk DNA" and have no function, their
concentraition maintains constant to allow other DNA act normally.
b) Although some DNA in cells has been called "junk DNA", it may simply be that we do not
yet understand its function.
c) Although some DNA in cells has been called "junk DNA", they have well known function to
store energy in the cells.
d) Although some DNA in cells has been called "junk DNA", but this is due to their mutation
and they could mutate again to restore their functions.
Solution
1. Even the simplest types of infectious agents must have DNA, RNA, and protein.
ANSWER: The above statement is false. This must be corrected as:
Even the simplest types of infectious agents must have either DNA and protein coat , RNA only
or Protein Coat only.
So the answer is:
D. Not all infectious agents have DNA, RNA, and protein. Some viruses have DNA and protein
only, viroids have only RNA, and prions are composed only of protein.
2.
Junk DNA/Non Coding DNA
The DNA sequences that does not code for a protein, that usually occurs in repetitive sequences
of nucleotides, and does not seem to serve any useful purpose are called Junk DNA.
So the answer is:
B. Although some DNA in cells has been called "junk DNA", it may simply be that we do not
yet understand its function.
