Explain the method for the selection of recombinants that makes use of insertional inactivation, with the help of suitable diagram.
**Principle of Insertional Inactivation-** When cloned DNA is inserted into a gene it inactivates its functioning. Sequence/ insert cloned into vectors within genes of antibiotic resistance, leads to loss of function.

**Steps**

1. Host cells (E. coli cells) with recombinant pBR322 plasmid are first plated on solid media (agarose containing) with the antibiotic ampicillin (assume that the insert has been ligated within the tetracycline resistance gene). Colonies from every single cell plate having the plasmid will develop overnight

2. A petri plate containing solid media with antibiotic tetracycline is kept carefully under aseptic conditions (Laminar flow hood)

3. A circular piece of velvet or velvet paper is aligned and pressed onto the colony containing ampicillin plate (master plate). With the same alignment it is pressed onto the tetracycline plate. Overnight only colonies not containing the insert will grow while due to insertional inactivation no colonies will grow which have the insert. The colonies which have the insert can easily be scored off by comparing the two plates.

![](https://www.sarthaks.com/?qa=blob&qa_blobid=14248038838146221627)
method for handling data structures where the first element is processed first and the newest element is processed last. The code is as shown below 

**clc;**

**close** **all;**

**clear** **all;**

**% define n**

**n = 0:18;**

**N = 19;** **% N tap filter**

**M = (N-1)/2;**

**% define window**

**% define the filter**

**h = 1./((n-M)*pi).*sin((n-M)*pi/5).*blackman(19)'.*((n>=0)-(n>=N));**

**h(M+1) = 2/15;** **% this corresponds to n=1 where the denominator is zero**

**% plot the filter**

**figure;**

**stem(n,h,'filled');grid** **on;title('h[n]');**

**xlabel('n');ylabel('stem height');**

**% plot the frequency response of the filter**

**figure;**

**freqz(h,1);grid** **on;**

  


  


![](/image?k=f31b88da6411d9811f191c8b336bed5e)![](/image?k=d351b5006ccd60c46820c64292935f2f) Assuming you're joining on columns with no duplicates, which is a very common case:

* An inner join of A and B gives the result of A intersect B, i.e. the inner part of a [Venn diagram](http://en.wikipedia.org/wiki/Venn_diagram) intersection.
* An outer join of A and B gives the results of A union B, i.e. the outer parts of a Venn diagram union.

**Examples**

Suppose you have two tables, with a single column each, and data as follows:


```
A    B
-    -
1    3
2    4
3    5
4    6

```
Note that (1,2) are unique to A, (3,4) are common, and (5,6) are unique to B.

**Inner join**

An inner join using either of the equivalent queries gives the intersection of the two tables, i.e. the two rows they have in common.


```
select * from a INNER JOIN b on a.a = b.b;
select a.*, b.*  from a,b where a.a = b.b;

a | b
--+--
3 | 3
4 | 4

```
**Left outer join**

A left outer join will give all rows in A, plus any common rows in B.


```
select * from a LEFT OUTER JOIN b on a.a = b.b;
select a.*, b.*  from a,b where a.a = b.b(+);

a |  b
--+
1 | null
2 | null
3 |    3
4 |    4

```
**Right outer join**

A right outer join will give all rows in B, plus any common rows in A.


```
select * from a RIGHT OUTER JOIN b on a.a = b.b;
select a.*, b.*  from a,b where a.a(+) = b.b;

a    |  b
+
3    |  3
4    |  4
null |  5
null |  6

```
**Full outer join**

A full outer join will give you the union of A and B, i.e. all the rows in A and all the rows in B. If something in A doesn't have a corresponding datum in B, then the B portion is null, and vice versa.


```
select * from a FULL OUTER JOIN b on a.a = b.b;

 a   |  b
+
   1 | null
   2 | null
   3 |    3
   4 |    4
null |    6
null |    5

``` <p>concepts and reason</p>
<p>the concept used to solve this problem is the formal charge calculation of the oxygen atoms present in the ozone molecule.</p>
<p>fundamentals</p>
<p>a charge that is assigned to an atom, which is present in the molecule, is represented as a formal charge. in two bonded atoms, a formal charge assumes that the electrons are equally shared between them. the formal charge is calculated using the following formula:</p>
<p>formal charge \(=\mathrm{v}-\mathrm{n}-\frac{\mathrm{b}}{2}\)</p>
<p>here, \(\mathrm{v}\) is the number of valence electrons, \(\mathrm{n}\) is the number of non-bonding electrons and \(\mathrm{b}\) is the number of bonding electrons.</p>
<p> </p>
<p>the structure is as follows:</p>
<p>calculate the formal charge for the circled oxygen in ozone molecule as follows:</p>
<p><img src="https://img.wizedu.com/images/5af6d1f54822b4dfb8fe89d91ef2011c5fe3020815aa4.png" alt="" width="88" /></p>
<p>formal charge \(=\mathrm{v}-\mathrm{n}-\frac{\mathrm{b}}{2}\)</p>
<p>\(\begin{aligned}=& 6-6-\frac{2}{2} \\ &=-1 \end{aligned}\)</p>
<p>explanation | common mistakes | hint for next step</p>
<p>the marked oxygen atom has 6 valence electrons. it contains 6 non-bonding electrons. the marked oxygen atom is bonded with one oxygen atom. therefore, the bonding electron of oxygen is \(2 .\) substitute the values in a formal charge formula to determine the formal charge of oxygen (marked).</p>
<p> </p>
<p>the structure is as follows:</p>
<p><img src="https://img.wizedu.com/images/39493056a8515708b79350d1976e17b35fe30216f0473.png" alt="" width="88" /></p>
<p>calculate the formal charge for the circled oxygen in ozone molecule as follows:</p>
<p>formal charge \(=\mathrm{v}-\mathrm{n}-\frac{\mathrm{b}}{2}\)</p>
<p>\(=6-2-\frac{6}{2}\)</p>
<p>\(=+1\)</p>
<p>explanation | common mistakes | hint for next step</p>
<p>the oxygen atom in the middle has 6 valence electrons and has 2 non-bonding electrons. the central oxygen atom is bonded with 1 double bond and 1 single bond. so, there is a total of 3 bonds containing 6 electrons (bonding electrons). the formal charge of the middle oxygen atom can be determined by substituting the values in a formal charge formula.</p>
<p> </p>
<p>the structure is as follows:</p>
<p><img src="https://img.wizedu.com/images/b1122ce7723aa4ab1f9ae5c9e457299e5fe3022ca77a5.png" alt="" width="88" /></p>
<p>calculate the formal charge for the circled oxygen in ozone molecule as follows:</p>
<p>formal charge \(=\mathrm{v}-\mathrm{n}-\frac{\mathrm{b}}{2}\)</p>
<p>\(=6-4-\frac{4}{2}\)</p>
<p>\(=0\)</p>
<p>the formal charge of each oxygen atom in \(\mathrm{o}\_{3}\) molecule are labeled as follows:</p>
<p><img src="https://img.wizedu.com/images/1de5ee227194f6d8c7fb9fb0e76160f25fe3024117fde.png" alt="" width="120" /></p>
<p>the oxygen atom present on the right side contains 6 valence electrons which have 2 non-bonded electrons. the marked oxygen is bonded with 1 double bond with a central oxygen atom. so, the number of bonding electrons is \(4 .\) the formal charge of the corner oxygen atom (right corner) can be determined by substituting the values in the formal charge formula.</p>
<p>the formal charge of each oxygen atom in \(\mathrm{o}\_{3}\) molecule are labeled as follows:</p>
<p><img src="https://img.wizedu.com/images/1de5ee227194f6d8c7fb9fb0e76160f25fe3024117fde.png" alt="" width="120" /></p> the block diagram in the [boron datasheets](https://docs.particle.io/datasheets/cellular/boron-datasheet/#block-diagram) shows this

[[image]](https://aws1.discourse-cdn.com/business7/uploads/particle/original/3x/e/e/ee67f4ef8137993dea8cf5954316868091b0d7a9.png "image")

and these are not the wire pins.

but you may want to try wire1 instead. [image] nirgrahamuk:

observe({

fv <- req(input$filter\_var)

td <- req(tabledata())

updateselectizeinput(inputid = "filter\_vals",

choices = pull(td,fv))

})

observe({

fv <- as.character(req(input$filter\_var))

td <- req(table… using the long division method: we are going to do the following steps:step 1: determine the classification of the compoundstep 2: write the formula of the compound[readmore]step 1: determine the classification of the compoundrecall the definition of the following:ionic compound - usually made up of a metal and a non-metalcovalent compound - composed of more than one nonmetal togetheracid - substance whose molecules yield hydrogen ions (h+) when dissolved in waterhydrate - substance that contains water or its constituent elementsfor cobaltic iodidecobaltic – metaliodide – nonmetalmetal + nonmetal = ionicstep 2: write the formula of the compoundthe rules for naming ionic compounds are as follows:1. the metal keeps its name and is named and written first.– for transition metals, we use roman numerals to denote its charge.2. the nonmetal keeps its base name but has its ending changed to –ide.for cobaltic iodidecobaltic  → transition metal → +2 charge → co2+iodide → group 7a → -1 charge → i-criss-cross the charges and make them the subscript of the other ioncobaltic iodide → coi2answer: cobaltic iodide is an ionic compound with a formula of coi2 to answer this question we have to take into account several aspects. 

the first of these is the assessment of the male factor. when we speak of oligoazoospermia we refer to a low concentration of sperm in the semen. we consider an altered concentration when there are less than 15 million spermatozoa per milliliter (ml) of ejaculate. 

but although 15 million per ml is the reference parameter for talking about oligoazoospermia, when we assess the seminogram we must take into account the degree of deviation from that normality. that is, a concentration of 13 million/ml, where we would speak of a mild oligoazoospermia, is not the same as 1 million/ml, where we would speak of a severe oligoazoospermia. in addition we must also assess the other parameters of the study of the semen (total volume, motility). 

in order to group all the important parameters of the seminogram into a single value that will help us decide which assisted reproduction technique to use, the concept of msc (mobile sperm count) is established. this tells us the number of spermatozoa with good motility in each milliliter of ejaculate after the training of the semen in the laboratory. 

it is considered that in order to perform an artificial insemination (ai), the msc must be over 3 million spermatozoa. 

therefore, as far as the male factor is concerned, it will depend on the severity of the oligoazoospermia and how it influences the msc whether the technique of choice is an ai or, on the contrary, we have to opt for a more complex technique, such as in vitro fertilization (ivf). 

the second aspect to take into account is the female factor. in this case we are presented with a woman with a polycystic ovary syndrome (pcos). pcos is an endocrine pathology that is often difficult to achieve ovulation. when a woman is unable to ovulate, she will not be able to become pregnant and the clinic that most women with pcos consult for will appear, which is the absence of menstruation. 

thus, what we must do with this patient is to give her medication to help her achieve, first, follicular development and, later, ovulation. once ovulation is achieved, the semen sample should be added either by ai or by directed intercourse. one option or the other will depend on the quality of the semen. 

to achieve ovulation in a patient with pcos we perform an "ovulation induction" treatment. for this we have several therapeutic options. as first-line drugs we would have the drugs for oral administration (clomiphene citrate and letrozole) and, as second-line treatment, we would have the drugs for subcutaneous administration (gonadotropins). 

even so, before considering ovulation induction treatment in a patient with pcos, we should assess the patient's age. the main reason for this is that ovulation induction treatments end with ai or directed intercourse. it is known that women over 38 years of age do not benefit from these techniques due to poor egg quality. in such cases, the first option should be an ivf. 

therefore, and in response to the initial question, the most likely treatment of choice for this couple is ai with ovulation induction, but before indicating such a technique we must take into account other parameters such as emr and the age of the patient. 讓戰將來回答~~

圖片參考：http://www.molecularstation.com/images/western-blot.gif

 0.05%tbst
1. pour 3 l of purified water into a 4-l beaker and begin stirring.
2. measure 400ml 10x tbs in a 500-ml graduated cylinder and add to water.
3. add bsa and 2ml tween 20 to the beaker. rinse all of the tween 20 from the pipette tip into the solution.
4. transfer solution to a graduated cylinder and adjust to final volume 4l with purified water.
5. return solution to beaker and stir until thoroughly mixed.
以上是以 4l 為配置體積, 若你用不到那麼多可依比例減少配置體積, 而 2%bsa 的配法則依此要領, 將你們實驗室中用的 bsa 加入即可, 因為不知道你們是用 powder 還是 liquid 的 bsa~

圖片參考：http://www1.qiagen.com/products/genesilencing/allstarrnaicontrols/images/western.jpg

希望以上有幫到您~~