Examples of X inactivation in the following topics:
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- The presence of extra X chromosomes in a cell is compensated for by X-inactivation in which all but one X chromosome are silenced.
- In part, this occurs because of a molecular process called X inactivation.
- So how does X-inactivation help alleviate the effects of extra X chromosomes?
- If three X chromosomes are present, the cell will inactivate two of them.
- If four X chromosomes are present, three will be inactivated, and so on.
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- The X-inactivation skew theory, suggesting the female-high tendency, and proposed by Princeton University's Jeff Stewart, has recently been confirmed experimentally in scleroderma and autoimmune thyroiditis.
- Another theory suggests the female-high tendency to get autoimmunity is due to an imbalanced X chromosome inactivation.
- The X-inactivation skew theory, proposed by Princeton University's Jeff Stewart, has recently been confirmed experimentally in scleroderma and autoimmune thyroiditis.
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- When a gene being examined is present on the X chromosome, but not on the Y chromosome, it is said to be X-linked.
- Eye color in Drosophila was one of the first X-linked traits to be identified, and Thomas Hunt Morgan mapped this trait to the X chromosome in 1910.
- Because human males need to inherit only one recessive mutant X allele to be affected, X-linked disorders are disproportionately observed in males.
- Carrier females can manifest mild forms of the trait due to the inactivation of the dominant allele located on one of the X chromosomes.
- Eye color in Drosophila is an example of a X-linked trait
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- Under very high hydrostatic pressure(HHP) of up to 700 MPa, water inactivates pathogens such as E. coli and Salmonella.
- Under very high hydrostatic pressure of up to 700 MPa (100,000 psi), water inactivates pathogens such as Listeria, E. coli and Salmonella.
- The frist reports showed that bacterial spores were not always inactivated by pressure, while vegetative bacteria were usually killed.
- These spores, which caused a lack of preservation in the earlier experiments, were inactivated faster by moderate pressure, but in a manner different from what occurred with vegetative microbes.
- During pascalization, more than 50,000 pounds per square inch (340 MPa) may be applied for around fifteen minutes, leading to the inactivation of yeast, mold, and bacteria.
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- This area is represented by the probability P ( X < x ) .
- The area to the right is then P ( X > x ) = 1 − P ( X < x ) .
- Remember, P ( X < x ) = Area to the left of the vertical line through x.
- P ( X > x ) = 1 − P ( X < x ) = .
- P ( X < x ) is the same as P ( X ≤ x ) and P ( X > x ) is the same as P ( X ≥ x ) for continuous distributions.
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- for $\left | x \right | \leq 1$ (unless $x = -1$).
- Substituting $x − 1$ for $x$, we obtain an alternative form for $\ln(x)$ itself:
- $\ln(x) = (x - 1) - \dfrac{(x - 1)^{2}}{2} + \dfrac{(x - 1)^{3}}{3} - \cdots$
- for $\left | x -1 \right | \leq 1$ (unless $x = 0$).
- Here is an example in the case of $g(x) = \tan(x)$:
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- For $0 < x < \frac{ \pi}{2}$, $\sin x < x < \tan x.$
- $\displaystyle{\lim_{x \to 0} \left ( \frac{x}{\sin x} \right ) = 1}$
- $\displaystyle{\lim_{x \to 0} \left ( \frac{\sin x}{x} \right ) = 1}$
- $\displaystyle{\frac{(1−\cos x)(1+\cos x)}{x(1+\cos x)}=\frac{(1−\cos^2x)}{x(1+\cos x)}=\frac{\sin^2x}{x(1+\cos x)}= \frac{\sin x}{x} \cdot \frac{\sin x}{1+\cos x}}$
- $\displaystyle{\lim_{x \to 0}\left ( \frac{\sin x}{x} \frac{\sin x}{1 + \cos x} \right ) = \left (\lim_{x \to 0} \frac{\sin x}{x} \right ) \left ( \lim_{x \to 0} \frac{\sin x}{1 + \cos x} \right ) = \left (1 \right )\left (\frac{0}{2} \right )= 0}$
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- In this disorder, the Leiden variant of factor V cannot be inactivated by activated protein C, as it would be in a person with normal factor V, resulting in excess clotting.
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- ( x − $\bar{x}$ ) or ( x − µ ) = Deviations from the mean (how far a value is from the mean)
- ( x − $\bar{x}$)2 or ( x − µ )2 = Deviations squared
- $\bar{x} = \frac{\sum{x}}{n} or x = \frac{\sum{f} \cdot x}{n}$$\bar{x} = \frac{\sum{x}}{n} or x = \frac{\sum{f} \cdot x}{n}$
- $s = \sqrt{\frac{\sum(x \bar{x})^2}{n 1}}or s = \sqrt{\frac{\sum{f} \cdot ( x \bar{x})^2}{n-1}}$
- $s = \sqrt{\frac{\sum(x \bar{x})^2}{N}}or s = \sqrt{\frac{\sum{f} \cdot ( x \bar{x})^2}{N}}$
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- If such a phenotype is found then it can be assumed that the insertion has caused the gene relating to that phenotype to be inactivated.
- Synthetic DNA transposon system are constructed to introduce precisely defined DNA sequences into the chromosomes of vertebrate animals for the purposes of introducing new traits and to discover new genes and their functions (e.g. by establishing a loss-of-function phenotype or gene inactivation).
- Insertional inactivation is a technique used in recombinant DNA engineering where a plasmid (such as pBR322) is used to disable the expression of a gene.
- A gene is inactivated by inserting a fragment of DNA into the middle of its coding sequence.
- Any future products from the inactivated gene will not work because of the extra codes added to it.