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We saw how to score and decode an HMM-generated sequence in two different ways. Fortunately, the HMM framework enables the learning of these probabilities when provided a set of training data and a set architecture for the model.

When the training data is labelled, estimation of the probabilities is a form of supervised learning. One such instance would occur if we were given a DNA sequence of one million nucleotides in which the CpG islands had all been experimentally annotated and were asked to use this to estimate our model parameters.

In contrast, when the training data is unlabelled, the estimation problem is a form of unsupervised learning. Continuing with the CpG island example, this situation would occur if the provided DNA sequence contained no island annotation whatsoever and we needed to both estimate model parameters and identify the islands.

## Supervised Learning

When provided with labelled data, the idea of estimating model parameters is straightforward. Suppose that you are given a labelled sequence x1, . , xN as well as the true hidden state sequence π1, . , πN. Intuitively, one might expect that the probabilities that maximize the data’s likelihood are the actual probabilities that one observes within the data. This is indeed the case and can be formalized by defining Akl to be the number of times hidden state k transitions to l and Ek(b) to be the number of times b is emitted from hidden state k. The parameters θ that maximize P(x|θ) are simply obtained by counting as follows:

[
egin{aligned}
a_{k l} &=frac{A_{k l}}{sum_{i} A_{k i}}
e_{k}(b) &=frac{E_{k}(b)}{sum_{c} E_{k}(c)}
end{aligned}
]

One example training set is shown in Figure 8.5. In this example, it is obvious that the probability of transitioning from B to P is (egin{equation}
frac{1}{3+1}=frac{1}{4}
end{equation}) (there are 3 B to B transitions and 1 B to P transitions) and the probability of emitting a G from the B state is (egin{equation}
frac{2}{2+2+1}=frac{2}{5}
end{equation}) (there are 2 G's emitted from the B state, 2 C's and 1 A)

Notice, however, that in the above example the emission probability of character T from state B is 0because no such emissions were encountered in the training set. A zero probability, either for transitioning or emitting, is particularly problematic because it leads to an infinite log penalty. In reality, however, the zero probability may merely have arisen due to over-fitting or a small sample size. To rectify this issue and maintain flexibility within our model, we can collect more data on which to train, reducing the possibility that the zero probability is due to a small sample size. Another possibility is to use ’pseudocounts’ instead of absolute counts: artificially adding some number of counts to our training data which we think more accurately represent the actual parameters and help counteract sample size errors.

[egin{equation}
egin{array}{l}
A_{k l}^{*}=A_{k l}+r_{k l}
E_{k}(b)^{*}=E_{k}(b)+r_{k}(b)
end{array}
end{equation}. onumber ]

Larger pseudocount parameters correspond to a strong prior belief about the parameters, reflected in the fact that these pseudocounts, derived from your priors, are comparatively overwhelming the observations, your training data. Likewise, small pseudocount parameters (r << 1) are more often used when our priors are relatively weak and we are aiming not to overwhelm the empirical data but only to avoid excessively harsh probabilities of 0.

## Unsupervised Learning

Unsupervised learning involves estimating parameters based on unlabelled data. This may seem impossible - how can we take data about which we know nothing and use it to ”learn”? - but an iterative approach can yield surprisingly good results, and is the typical choice in these cases. This can be thought of loosely as an evolutionary algorithm: from some initial choice of parameters, the algorithm assesses how well the parameters explain or relate to the data, uses some step in that assessment to make improvements on the parameters, and then assesses the new parameters, producing incremental improvements in the parameters at every step just as the fitness or lack thereof of a particular organism in its environment produces incremental increases over evolutionary time as advantageous alleles are passed on preferentially.

Suppose we have some sort of prior belief about what each emission and transition probability should be. Given these parameters, we can use a decoding method to infer the hidden states underlying the provided data sequence. Using this particular decoding parse, we can then re-estimate the transition and emission counts and probabilities in a process similar to that used for supervised learning. If we repeat this procedure until the improvement in the data’s likelihood remains relatively stable, the data sequence should ultimately drive the parameters to their appropriate values.

FAQ

Q: Why does unsupervised learning even work? Or is it magic?

A: Unsupervised learning works because we have the sequence (input data) and this guides every step of the iteration; to go from a labelled sequence to a set of parameters, the later are guided by the input and its annotation, while to annotate the input data, the parameters and the sequence guide the procedure.

For HMMs in particular, two main methods of unsupervised learning are useful.

Expectation Maximization using Viterbi training

The first method, Viterbi training, is relatively simple but not entirely rigorous. After picking some initial best-guess model parameters, it proceeds as follows:

E step: Perform Viterbi decoding to find π

M step: Calculate the new parameters (A_{k l}^{*}, E_{k}(b)^{*}) using the simple counting formalism in supervised learning (Maximization step)

Iteration: Repeat the E and M steps until the likelihood P(x|θ) converges

Although Viterbi training converges rapidly, its resulting parameter estimations are usually inferior to those of the Baum-Welch Algorithm. This result stems from the fact that Viterbi training only considers the most probable hidden path instead of the collection of all possible hidden paths.

Expectation Maximization: The Baum-Welch Algorithm

The more rigorous approach to unsupervised learning involves an application of Expectation Maximization to HMM’s. In general, EM proceeds in the following manner:

Init: Initialize the parameters to some best-guess state

E step: Estimate the expected probability of hidden states given the latest parameters and observed sequence (Expectation step)

M step: Choose new maximum likelihood parameters using the probability distribution of hidden states (Maximization step)

Iteration: Repeat the E and M steps until the likelihood of the data given the parameters converges

The power of EM lies in the fact that P (x|θ) is guaranteed to increase with each iteration of the algorithm. Therefore, when this probability converges, a local maximum has been reached. As a result, if we utilize a variety of initialization states, we will most likely be able to identify the global maximum, i.e. the best parameters θ. The Baum-Welch algorithm generalizes EM to HMM’s. In particular, it uses the forward and backward algorithms to calculate P(x|θ) and to estimate Akl and Ek(b). The algorithm proceeds as follows:

Initialization 1. Initialize the parameters to some best-guess state

Iteration 1. Run the forward algorithm
2. Run the backward algorithm
3. Calculate the new log-likelihood P(x|θ)
4. Calculate Akl and Ek(b)
5. Calculate akl and ek(b) using the pseudocount formulas 6. Repeat until P(x|θ) converges

Previously, we discussed how to compute P(x|θ) using either the forward or backward algorithm’s final results. But how do we estimate Akl and Ek(b)? Let’s consider the expected number of transitions from state k to state l given a current set of parameters θ. We can express this expectation as

[A_{k l}=sum_{t} Pleft(pi_{t}=k, pi_{t+1}=l mid x, heta ight)=sum_{t} frac{Pleft(pi_{t}=k, pi_{t+1}=l, x mid heta ight)}{P(x mid heta)} onumber ]

Exploiting the Markov property and the definitions of the emission and transition probabilities leads to the following derivation:

[egin{aligned}
A_{k l} &=sum_{t} frac{Pleft(x_{1} ldots x_{t}, pi_{t}=k, pi_{t+1}=l, x_{t+1} ldots x_{N} mid heta ight)}{P(x mid heta)}
&=sum_{t} frac{Pleft(x_{1} ldots x_{t}, pi_{t}=k ight) * Pleft(pi_{t+1}=l, x_{t+1} ldots x_{N} mid pi_{t}, heta ight)}{P(x mid heta)}
&=sum_{t} frac{f_{k}(t) * Pleft(pi_{t+1}=l mid pi_{t}=k ight) * Pleft(x_{t+1} mid pi_{t+1}=l ight) * Pleft(x_{t+2} ldots x_{N} mid pi_{t+1}=l, heta ight)}{P(x mid heta)}
Rightarrow A_{k l} &=sum_{t} frac{f_{k}(t) * a_{k l} * e_{l}left(x_{t+1} ight) * b_{l}(t+1)}{P(x mid heta)}
end{aligned}]

A similar derivation leads to the following expression for Ek(b):

[E_{k}(b)=sum_{i mid x_{i}=b} frac{f_{k}(t) * b_{k}(t)}{P(x mid heta)}]

Therefore, by running the forward and backward algorithms, we have all of the information necessary to calculate P(x|θ) and to update the emission and transition probabilities during each iteration. Because these updates are constant time operations once P(x|θ), fk(t) and bk(t) have been computed, the total time complexity for this version of unsupervised learning is θ(K2NS), where S is the total number of iterations.

FAQ

Q: How do you encode your prior beliefs when learning with Baum-Welch?
A: Those prior beliefs are encoded in the initializations of the forward and backward algorithms Many systems exhibit exponential growth. These systems follow a model of the form (y=y_0e^,) where (y_0) represents the initial state of the system and (k) is a positive constant, called the growth constant. Notice that in an exponential growth model, we have

That is, the rate of growth is proportional to the current function value. This is a key feature of exponential growth. Equation ef involves derivatives and is called a differential equation.

Systems that exhibit exponential growth increase according to the mathematical model

where (y_0) represents the initial state of the system and (k>0) is a constant, called the growth constant.

Population growth is a common example of exponential growth. Consider a population of bacteria, for instance. It seems plausible that the rate of population growth would be proportional to the size of the population. After all, the more bacteria there are to reproduce, the faster the population grows. Figure (PageIndex<1>) and Table (PageIndex<1>) represent the growth of a population of bacteria with an initial population of 200 bacteria and a growth constant of 0.02. Notice that after only 2 hours (120 minutes), the population is 10 times its original size! Figure (PageIndex<1>): An example of exponential growth for bacteria.

Table (PageIndex<1>): Exponential Growth of a Bacterial Population
Time(min) Population Size (no. of bacteria)
10 244
20 298
30 364
40 445
50 544
60 664
70 811
80 991
90 1210
100 1478
110 1805
120 2205

Note that we are using a continuous function to model what is inherently discrete behavior. At any given time, the real-world population contains a whole number of bacteria, although the model takes on noninteger values. When using exponential growth models, we must always be careful to interpret the function values in the context of the phenomenon we are modeling.

Example (PageIndex<1>): Population Growth

Consider the population of bacteria described earlier. This population grows according to the function (f(t)=200e^<0.02t>,) where t is measured in minutes. How many bacteria are present in the population after (5) hours ((300) minutes)? When does the population reach (100,000) bacteria?

There are (80,686) bacteria in the population after (5) hours.

To find when the population reaches (100,000) bacteria, we solve the equation

The population reaches (100,000) bacteria after (310.73) minutes.

Consider a population of bacteria that grows according to the function (f(t)=500e^<0.05t>), where (t) is measured in minutes. How many bacteria are present in the population after 4 hours? When does the population reach (100) million bacteria?

Use the process from the previous example.

There are (81,377,396) bacteria in the population after (4) hours. The population reaches (100) million bacteria after (244.12) minutes.

Let&rsquos now turn our attention to a financial application: compound interest. Interest that is not compounded is called simple interest. Simple interest is paid once, at the end of the specified time period (usually (1) year). So, if we put (\$1000) in a savings account earning (2%) simple interest per year, then at the end of the year we have

Compound interest is paid multiple times per year, depending on the compounding period. Therefore, if the bank compounds the interest every (6) months, it credits half of the year&rsquos interest to the account after (6) months. During the second half of the year, the account earns interest not only on the initial (\$1000), but also on the interest earned during the first half of the year. Mathematically speaking, at the end of the year, we have

Similarly, if the interest is compounded every (4) months, we have

and if the interest is compounded daily ((365) times per year), we have (\$1020.20). If we extend this concept, so that the interest is compounded continuously, after (t) years we have

Now let&rsquos manipulate this expression so that we have an exponential growth function. Recall that the number (e) can be expressed as a limit:

Based on this, we want the expression inside the parentheses to have the form ((1+1/m)). Let (n=0.02m). Note that as (n&rarr&infin, m&rarr&infin) as well. Then we get

We recognize the limit inside the brackets as the number (e). So, the balance in our bank account after (t) years is given by (1000 e^<0.02t>). Generalizing this concept, we see that if a bank account with an initial balance of (\$P) earns interest at a rate of (r%), compounded continuously, then the balance of the account after (t) years is

Example (PageIndex<2>): Compound Interest

A 25-year-old student is offered an opportunity to invest some money in a retirement account that pays (5%) annual interest compounded continuously. How much does the student need to invest today to have (\$1) million when she retires at age (65)? What if she could earn (6%) annual interest compounded continuously instead?

She must invest (\$135,335.28) at (5%) interest.

If, instead, she is able to earn (6%,) then the equation becomes

In this case, she needs to invest only (\$90,717.95.) This is roughly two-thirds the amount she needs to invest at (5%). The fact that the interest is compounded continuously greatly magnifies the effect of the (1%) increase in interest rate.

Suppose instead of investing at age (25sqrt), the student waits until age (35). How much would she have to invest at (5%)? At (6%)?

Use the process from the previous example.

At (5%) interest, she must invest (\$223,130.16). At (6%) interest, she must invest (\$165,298.89.)

If a quantity grows exponentially, the time it takes for the quantity to double remains constant. In other words, it takes the same amount of time for a population of bacteria to grow from (100) to (200) bacteria as it does to grow from (10,000) to (20,000) bacteria. This time is called the doubling time. To calculate the doubling time, we want to know when the quantity reaches twice its original size. So we have

If a quantity grows exponentially, the doubling time is the amount of time it takes the quantity to double. It is given by

Example (PageIndex<3>): Using the Doubling Time

Assume a population of fish grows exponentially. A pond is stocked initially with (500) fish. After (6) months, there are (1000) fish in the pond. The owner will allow his friends and neighbors to fish on his pond after the fish population reaches (10,000). When will the owner&rsquos friends be allowed to fish?

We know it takes the population of fish (6) months to double in size. So, if (t) represents time in months, by the doubling-time formula, we have (6=(ln 2)/k). Then, (k=(ln 2)/6). Thus, the population is given by (y=500e^<((ln 2)/6)t>). To figure out when the population reaches (10,000) fish, we must solve the following equation:

The owner&rsquos friends have to wait (25.93) months (a little more than (2) years) to fish in the pond.

Suppose it takes (9) months for the fish population in Example (PageIndex<3>) to reach (1000) fish. Under these circumstances, how long do the owner&rsquos friends have to wait?

Use the process from the previous example.

## End-to-End Differentiable Learning of Protein Structure

Predicting protein structure from sequence is a central challenge of biochemistry. Co-evolution methods show promise, but an explicit sequence-to-structure map remains elusive. Advances in deep learning that replace complex, human-designed pipelines with differentiable models optimized end to end suggest the potential benefits of similarly reformulating structure prediction. Here, we introduce an end-to-end differentiable model for protein structure learning. The model couples local and global protein structure via geometric units that optimize global geometry without violating local covalent chemistry. We test our model using two challenging tasks: predicting novel folds without co-evolutionary data and predicting known folds without structural templates. In the first task, the model achieves state-of-the-art accuracy, and in the second, it comes within 1-2 Å competing methods using co-evolution and experimental templates have been refined over many years, and it is likely that the differentiable approach has substantial room for further improvement, with applications ranging from drug discovery to protein design.

Keywords: biophysics co-evolution deep learning geometric deep learning homology modeling machine learning protein design protein folding protein structure prediction structural biology.

## Action Potential

A nerve impulse is a sudden reversal of the electrical gradient across the plasma membrane of a resting neuron. The reversal of charge is called an action potential . It begins when the neuron receives a chemical signal from another cell or some other type of stimulus . If the stimulus is strong enough to reach threshold , an action potential will take place is a cascade along the axon.

This reversal of charges ripples down the axon of the neuron very rapidly as an electric current, which is illustrated in the diagram below (Figure 8.4.2). A nerve impulse is an all-or-nothing response depending on if the stimulus input was strong enough to reach threshold. If a neuron responds at all, it responds completely. A greater stimulation does not produce a stronger impulse. Figure 8.4.2 An action potential speeds along an axon in milliseconds. Sodium ions flow in and cause the action potential, and then potassium ions flow out to reset the resting potential.

In neurons with a myelin sheath on their axon, ions flow across the membrane only at the nodes between sections of myelin. As a result, the action potential appears to jump along the axon membrane from node to node, rather than spreading smoothly along the entire membrane. This increases the speed at which the action potential travels.

## A Level Biology - Gel Electrophoresis

Check Where this Lesson fits into your Exam Specification!

00:22 What is Gel Electrophoresis?

01:00 How does Gel Electrophoresis work?

02:50 Small DNA fragments migrate further an quicker.

03:49 Watch VNTRs (Genetic fingerprinting next)

04:14 Exam style Q and A (1).

04:40 Exam style Q and A (2).

★ AQA A Level Biology Specification Reference: - 3.8.4.1 Recombinant DNA technology (A-level only). Recombinant DNA technology involves the transfer of fragments of DNA from one organism, or species, to another. Since the genetic code is universal, as are transcription and translation mechanisms, the transferred DNA can be translated within cells of the recipient (transgenic) organism. Interpret information relating to the use of recombinant DNA technology.

★ CIE A Level Biology Specification Reference : - 19.1 Principles of genetic Technology: a) define the term recombinant DNA. b) explain that genetic engineering involves the extraction of genes from one organism, or the synthesis of genes, in order to place them in another organism (of the same or another species) such that the receiving organism expresses the gene product.

★ Edexcel A Level Biology (Biology A &ndash Salters-Nuffield) Specification Reference : - N/A

★ Edexcel A Level Biology (Biology B) Specification Reference: - Topic 7: Modern Genetics. 7.4 Gene technology: Understand how recombinant DNA can be produced.

★ OCR A Level Biology (Biology A) Specification Reference : - 6.1.3 Manipulating genomes (the principles of genetic engineering - Recombinant DNA).

★ OCR A Level Biology (Biology B) Specification Reference : - 5.1.3 Gene technologies.

★ WJEC A Level Biology Specification Reference: - 7. Application of reproduction and genetics.

★ AQA A Level Biology Specification Reference: - 3.8.4.1 Recombinant DNA technology (A-level only). Recombinant DNA technology involves the transfer of fragments of DNA from one organism, or species, to another. Since the genetic code is universal, as are transcription and translation mechanisms, the transferred DNA can be translated within cells of the recipient (transgenic) organism. Fragments of DNA can be produced by: Conversion of mRNA to complementary DNA (cDNA), using reverse transcriptase.

★ CIE A Level Biology Specification Reference: - 19.1 Principles of genetic Technology: h) explain the role of reverse transcriptase in genetic engineering.

★ OCR A Level Biology (Biology A) Specification Reference: - 6.1.3 Manipulating genomes (f) the principles the principles of genetic engineering. To include the isolation of genes from one organism and the placing of these genes into another organism using suitable vectors. NOTE: cDNA and using reverse transcriptase is not specifically mentioned in the specification!

★ OCR A Level Biology (Biology B) Specification Reference: - 5.1.3 Gene technologies (b) To include the role of reverse transcriptase - example of human protein to include insulin.

★ Edexcel A Level Biology (Biology A &ndash Salters-Nuffield) N/A

★ Edexcel A Level Biology (Biology B) N/A

★ AQA A Level Biology Specification Reference: - 3.8.4 Gene technologies. Fragments of DNA can be amplified by in vitro and in vivo techniques - The principles of the polymerase chain reaction ( PCR ) as an in vitro method to amplify DNA fragments.

★ CIE A Level Biology Specification Reference: - 19.1 Principles of genetic technology. c) describe the principles of the polymerase chain reaction ( PCR ) to clone and amplify DNA (the role of Taq polymerase (the DNA Polymerase enzyme from the heat tolerant bacterium Thermus aquaticus) should be emphasised).

★ Edexcel A Level Biology (Biology A &ndash Salters-Nuffield) Specification Reference: - Topic 6: Immunity, Infection and Forensics. 6.4 Know how DNA can be amplified using the polymerase chain reaction ( PCR ).

★ Edexcel A Level Biology (Biology B) Specification Reference: - Topic 7: Modern Genetics. 7.1 Using gene sequencing. ii Understand how PCR can be used to amplify DNA samples, and how these samples can be used: e.g. to predict the amino acid sequence of proteins and possible links to genetically determined conditions, using gene sequencing / in forensic science, to identify criminals and to test paternity, using DNA profiling.

★ OCR A Level Biology (Biology A) Specification Reference: - 6.1.3 Manipulating genomes. (d) the principles of the polymerase chain reaction ( PCR ) and its application in DNA analysis.

★ OCR A Level Biology (Biology B) Specification Reference: - 5.1.3 Gene technologies. The principles and uses of the Polymerase Chain Reaction (PCR). To include the use of PCR in amplifying DNA, the role of primers and Taq polymerase ((the DNA Polymerase enzyme from the heat tolerant bacterium Thermus aquaticus ) in PCR.

★ WJEC A Level Biology Specification Reference: - 7. Application of reproduction and genetics. (d) the use of PCR and electrophoresis to produce a genetic fingerprint the forensic use of genetic fingerprinting.

★ AQA A Level Biology Specification Reference: - 3.8.4.3 Genetic fingerprinting (A-level only). An organism&rsquos genome contains many variable number tandem repeats (VNTRs). The probability of two individuals having the same VNTRs is very low. The technique of genetic fingerprinting in analysing DNA fragments that have been cloned by PCR, and its use in determining genetic relationships and in determining the genetic variability within a population. The use of genetic fingerprinting in the fields of forensic science, medical diagnosis, animal and plant breeding. Students should be able to: Explain the biological principles that underpin genetic fingerprinting techniques. Interpret data showing the results of gel electrophoresis to separate DNA fragments. Explain why scientists might use genetic fingerprinting in the fields of forensic science, medical diagnosis, animal and plant breeding.

★ CIE A Level Biology Specification Reference: - 19.1 Principles of genetic technology . Describe and explain how gel electrophoresis is used to analyse proteins and nucleic acids, and to distinguish between the alleles of a gene (limited to the separation of polypeptides and the separation of DNA fragments cut with restriction endonucleases). 19.2 Genetic technology applied to medicine outline the use of PCR and DNA testing in forensic medicine and criminal investigations.

★ Edexcel A Level Biology (Biology A &ndash Salters-Nuffield) Specification Reference: - Topic 6: Immunity, Infection and Forensics. 6.3 Know how DNA profiling is used for identification and determining genetic relationships between organisms (plants and animals).

★ Edexcel A Level Biology (Biology B) Specification Reference: - 7.1 Using gene sequencing. ii Understand how PCR can be used to amplify DNA samples, and how these samples can be used: in forensic science, to identify criminals and to test paternity, using DNA profiling.

★ OCR A Level Biology (Biology A) Specification Reference: - 6.1.3 Manipulating genomes. The principles of DNA sequencing and the development of new DNA sequencing techniques. The principles of DNA profiling and its uses, to include forensics and analysis of disease risk. the principles of the polymerase chain reaction (PCR) and its application in DNA analysis. the principles and uses of electrophoresis for separating nucleic acid fragments or proteins.

★ OCR A Level Biology (Biology B) Specification Reference: - 5.1.3 Gene technologies. The principles and uses of the Polymerase Chain Reaction (PCR). The nature and use VNTRs (variable number tandem repeats) in human genome studies. To include forensics, disease pre-disposition, ethnic migration, paternity testing, selection for clinical trials.

★ WJEC A Level Biology Specification Reference : - 7. Application of reproduction and genetics. This topic covers gene technology and its applications, including the sequencing of genomes, the use of PCR and recombinant DNA technology. Learners should be able to demonstrate and apply their knowledge and understanding of: the use of PCR and electrophoresis to produce a genetic fingerprint the forensic use of genetic fingerprinting.

★ AQA A Level Biology Specification Reference : - 3.8.4.3 Genetic fingerprinting (A-level only). Students should be able to: interpret data showing the results of gel electrophoresis to separate DNA fragments.

★ CIE A Level Biology Specification Reference: - 19.1 Principles of genetic Technology: d) describe and explain how gel electrophoresis is used to analyse proteins and nucleic acids, and to distinguish between the alleles of a gene (limited to the separation of polypeptides and the separation of DNA fragments cut with restriction endonucleases)

★ Edexcel A Level Biology (Biology A &ndash Salters-Nuffield) Specification Reference: - Topic 6: Immunity, Infection and Forensics. Use gel electrophoresis to separate DNA fragments of different length.

★ Edexcel A Level Biology (Biology B) Specification Reference: - N/A. Note Electrophoresis is NOT specifically mentioned in the spec. but understanding it is essential for Topic 7: Modern Genetics. 7.1 Using gene sequencing - in particular in forensic science, to identify criminals and to test paternity, using DNA profiling.

★ OCR A Level Biology (Biology A) Specification Reference: - 6.1.3 Manipulating genomes (e) the principles and uses of electrophoresis for separating nucleic acid fragments or proteins

★ OCR A Level Biology (Biology B) Specification Reference: - 5.1.3 Gene technologies (d) the principles and uses of agarose gel electrophoresis.

★ WJEC A Level Biology Specification Reference: - 7. Application of reproduction and genetics: (d) the use of PCR and electrophoresis to produce a genetic fingerprint the forensic use of genetic fingerprinting.

## Creating Coherence

Careful writers use transitions to clarify how the ideas in their sentences and paragraphs are related. These words and phrases help the writing flow smoothly. Adding transitions is not the only way to improve coherence, but they are often useful and give a mature feel to your essays. Table 8.3 “Common Transitional Words and Phrases” groups many common transitions according to their purpose.

Table 8.3 Common Transitional Words and Phrases

 Transitions That Show Sequence or Time after before later afterward before long meanwhile as soon as finally next at first first, second, third soon at last in the first place then Transitions That Show Position above across at the bottom at the top behind below beside beyond inside near next to opposite to the left, to the right, to the side under where Transitions That Show a Conclusion indeed hence in conclusion in the final analysis therefore thus Transitions That Continue a Line of Thought consequently furthermore additionally because besides the fact following this idea further in addition in the same way moreover looking further considering…, it is clear that Transitions That Change a Line of Thought but yet however nevertheless on the contrary on the other hand Transitions That Show Importance above all best especially in fact more important most important most worst Transitions That Introduce the Final Thoughts in a Paragraph or Essay finally last in conclusion most of all least of all last of all All-Purpose Transitions to Open Paragraphs or to Connect Ideas Inside Paragraphs admittedly at this point certainly granted it is true generally speaking in general in this situation no doubt no one denies obviously of course to be sure undoubtedly unquestionably Transitions that Introduce Examples for instance for example Transitions That Clarify the Order of Events or Steps first, second, third generally, furthermore, finally in the first place, also, last in the first place, furthermore, finally in the first place, likewise, lastly

After Maria revised for unity, she next examined her paragraph about televisions to check for coherence. She looked for places where she needed to add a transition or perhaps reword the text to make the flow of ideas clear. In the version that follows, she has already deleted the sentences that were off topic.

Many writers make their revisions on a printed copy and then transfer them to the version on-screen. They conventionally use a small arrow called a caret (^) to show where to insert an addition or correction. ### Exercise 2

2. Now return to the first draft of the essay you wrote in Section 8 “Writing Your Own First Draft” and revise it for coherence. Add transition words and phrases where they are needed, and make any other changes that are needed to improve the flow and connection between ideas.

## Budget

How much money do you think the training will cost? The type of training performed will depend greatly on the budget. If you decide that web-based training is the right delivery mode, but you don’t have the budget to pay the user fee for the platform, this wouldn’t be the best option. Besides the actual cost of training, another cost consideration is people’s time. If employees are in training for two hours, what is the cost to the organization while they are not able to perform their job? A spreadsheet should be developed that lists the actual cost for materials, snacks, and other direct costs, but also the indirect costs, such as people’s time.

## Bachelors Degree in Biology Education

### Learning Outcomes for Bachelors Degree in Biology Education

• Demonstrate an overall knowledge of biology needed to teach in the secondary education system.
• Pass the State Praxis Test Scores
• Students who gain employment in secondary education will demonstrate skill and knowledge in pedagogy.

### About UVU's BS in Biology Education

Graduates with a bachelor´s degree in Biology Education will be qualified to obtain state licensure to teach at the secondary level. The degree fulfills the requirements for the Biological Science Composite Major endorsement. If the student completes additional chemistry and earth science courses, then the student may also qualify for other teaching endorsements in the sciences.

Current employment opportunities for graduates from Biology Education programs are strong, particularly for those who also have a chemistry endorsement or integrated science endorsement in addition to the biological science composite endorsement.

The following information is from "Supply and Demand Needs of K-12 Education in the State of Utah 2005-2006" published by the Utah System of Higher Education:

1. Teacher demand in Utah is critically outstripping the supply of new teachers being provided by Utah colleges and universities.
2. About half of all attrition in 2005-2006 was due to retirement reflecting the relatively large number of educators becoming eligible for retirement in Utah. Data suggest this will continue for the next decade.
3. If attrition and growth continue at current rates (which they are likely to do) without a parallel increase in newly trained educators (which isn´t currently in the works), Utah will face a severe teacher shortage crisis unlike anything it has ever experienced within just a year or two.
4. Because Utah´s colleges of education aren´t producing sufficient numbers of new educators, Utah school districts are relying more heavily upon former teachers, retirees, out-of-state recruiting, and to a lesser extent alternative certification programs to meet their unmet teacher needs. The supply of former teachers and retirees is restricted and, anecdotally, districts are reporting that this is having the effect of significantly reducing their substitute reserves. If this were a water study, the conclusion would be that the water table is shrinking and that underground water is drying up quickly.
5. Utah school districts are especially facing severe difficulty in finding and hiring teachers in specific areas of special education (severely handicapped and mild/moderate), math, science, and early childhood education (K-3). With increased math and science requirements being introduced into Utah´s secondary schools, the demand for math and science teachers will grow even more rapidly
6. Although more stable with regards to long-term retention, the ranks of Utah´s non-teaching professional educators showed a significant turnover with 8.4% leaving in 2005-2006. Nearly two-thirds were attributable to retirement which again reflects the aging population of Utah´s educators

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