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By the end of this section, you will be able to:
  • Understand the process of translation and discuss its key factors
  • Describe how the initiation complex controls translation
  • Explain the different ways in which the post-translational control of gene expression takes place

After the RNA has been transported to the cytoplasm, it is translated into protein. Control of this process is largely dependent on the RNA molecule. As previously discussed, the stability of the RNA will have a large impact on its translation into a protein. As the stability changes, the amount of time that it is available for translation also changes.

The initiation complex and translation rate

Like transcription, translation is controlled by proteins that bind and initiate the process. In translation, the complex that assembles to start the process is referred to as the initiation complex    . The first protein to bind to the RNA to initiate translation is the eukaryotic initiation factor-2 (eIF-2)    . The eIF-2 protein is active when it binds to the high-energy molecule guanosine triphosphate (GTP) . GTP provides the energy to start the reaction by giving up a phosphate and becoming guanosine diphosphate (GDP) . The eIF-2 protein bound to GTP binds to the small 40S ribosomal subunit . When bound, the methionine initiator tRNA associates with the eIF-2/40S ribosome complex, bringing along with it the mRNA to be translated. At this point, when the initiator complex is assembled, the GTP is converted into GDP and energy is released. The phosphate and the eIF-2 protein are released from the complex and the large 60S ribosomal subunit binds to translate the RNA. The binding of eIF-2 to the RNA is controlled by phosphorylation. If eIF-2 is phosphorylated, it undergoes a conformational change and cannot bind to GTP. Therefore, the initiation complex cannot form properly and translation is impeded ( [link] ). When eIF-2 remains unphosphorylated, it binds the RNA and actively translates the protein.

Art connection

The eIF2 protein is a translation factor that binds to the small 40S ribosome subunit. When eIF2 is phosphorylated, translation is blocked.
Gene expression can be controlled by factors that bind the translation initiation complex.

An increase in phosphorylation levels of eIF-2 has been observed in patients with neurodegenerative diseases such as Alzheimer’s, Parkinson’s, and Huntington’s. What impact do you think this might have on protein synthesis?

Chemical modifications, protein activity, and longevity

Proteins can be chemically modified with the addition of groups including methyl, phosphate, acetyl, and ubiquitin groups. The addition or removal of these groups from proteins regulates their activity or the length of time they exist in the cell. Sometimes these modifications can regulate where a protein is found in the cell—for example, in the nucleus, the cytoplasm, or attached to the plasma membrane.

Chemical modifications occur in response to external stimuli such as stress, the lack of nutrients, heat, or ultraviolet light exposure. These changes can alter epigenetic accessibility, transcription, mRNA stability, or translation—all resulting in changes in expression of various genes. This is an efficient way for the cell to rapidly change the levels of specific proteins in response to the environment. Because proteins are involved in every stage of gene regulation, the phosphorylation of a protein (depending on the protein that is modified) can alter accessibility to the chromosome, can alter translation (by altering transcription factor binding or function), can change nuclear shuttling (by influencing modifications to the nuclear pore complex), can alter RNA stability (by binding or not binding to the RNA to regulate its stability), can modify translation (increase or decrease), or can change post-translational modifications (add or remove phosphates or other chemical modifications).

The addition of an ubiquitin group to a protein marks that protein for degradation. Ubiquitin acts like a flag indicating that the protein lifespan is complete. These proteins are moved to the proteasome    , an organelle that functions to remove proteins, to be degraded ( [link] ). One way to control gene expression, therefore, is to alter the longevity of the protein.

Multiple ubiquitin groups bind to a protein. The tagged protein is then fed into the hollow tube of a proteasome. The proteasome degrades the protein.
Proteins with ubiquitin tags are marked for degradation within the proteasome.

Section summary

Changing the status of the RNA or the protein itself can affect the amount of protein, the function of the protein, or how long it is found in the cell. To translate the protein, a protein initiator complex must assemble on the RNA. Modifications (such as phosphorylation) of proteins in this complex can prevent proper translation from occurring. Once a protein has been synthesized, it can be modified (phosphorylated, acetylated, methylated, or ubiquitinated). These post-translational modifications can greatly impact the stability, degradation, or function of the protein.

Art connections

[link] An increase in phosphorylation levels of eIF-2 has been observed in patients with neurodegenerative diseases such as Alzheimer’s, Parkinson’s, and Huntington’s. What impact do you think this might have on protein synthesis?

[link] Protein synthesis would be inhibited.

Questions & Answers

show that the set of all natural number form semi group under the composition of addition
Nikhil Reply
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Dominic
explain and give four Example hyperbolic function
Lukman Reply
_3_2_1
felecia
⅗ ⅔½
felecia
_½+⅔-¾
felecia
The denominator of a certain fraction is 9 more than the numerator. If 6 is added to both terms of the fraction, the value of the fraction becomes 2/3. Find the original fraction. 2. The sum of the least and greatest of 3 consecutive integers is 60. What are the valu
SABAL Reply
1. x + 6 2 -------------- = _ x + 9 + 6 3 x + 6 3 ----------- x -- (cross multiply) x + 15 2 3(x + 6) = 2(x + 15) 3x + 18 = 2x + 30 (-2x from both) x + 18 = 30 (-18 from both) x = 12 Test: 12 + 6 18 2 -------------- = --- = --- 12 + 9 + 6 27 3
Pawel
2. (x) + (x + 2) = 60 2x + 2 = 60 2x = 58 x = 29 29, 30, & 31
Pawel
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Ifeanyi
on number 2 question How did you got 2x +2
Ifeanyi
combine like terms. x + x + 2 is same as 2x + 2
Pawel
x*x=2
felecia
2+2x=
felecia
×/×+9+6/1
Debbie
Q2 x+(x+2)+(x+4)=60 3x+6=60 3x+6-6=60-6 3x=54 3x/3=54/3 x=18 :. The numbers are 18,20 and 22
Naagmenkoma
Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 3 times the number of marbles Mark has, how many does each boy have to sell if the total number of marbles is 113?
mariel Reply
Mark = x,. Don = 3x + 1 x + 3x + 1 = 113 4x = 112, x = 28 Mark = 28, Don = 85, 28 + 85 = 113
Pawel
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Harshika Reply
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Harshika
find the subring of gaussian integers?
Rofiqul
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Shirley Reply
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Mark
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Abdullahi Reply
find the value of 2x=32
Felix Reply
divide by 2 on each side of the equal sign to solve for x
corri
X=16
Michael
Want to review on complex number 1.What are complex number 2.How to solve complex number problems.
Beyan
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Mark
16
Makan
x=16
Makan
use the y -intercept and slope to sketch the graph of the equation y=6x
Only Reply
how do we prove the quadratic formular
Seidu Reply
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Darius
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Opoku
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Tric Reply
4
Trista
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Sidiki Reply
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Mark
Solve for the first variable in one of the equations, then substitute the result into the other equation. Point For: (6111,4111,−411)(6111,4111,-411) Equation Form: x=6111,y=4111,z=−411x=6111,y=4111,z=-411
Brenna
(61/11,41/11,−4/11)
Brenna
x=61/11 y=41/11 z=−4/11 x=61/11 y=41/11 z=-4/11
Brenna
Need help solving this problem (2/7)^-2
Simone Reply
x+2y-z=7
Sidiki
what is the coefficient of -4×
Mehri Reply
-1
Shedrak
A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place.
Kimberly Reply
Jeannette has $5 and $10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
August Reply
What is the expressiin for seven less than four times the number of nickels
Leonardo Reply
How do i figure this problem out.
how do you translate this in Algebraic Expressions
linda Reply
why surface tension is zero at critical temperature
Shanjida
I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason
s.
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
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Source:  OpenStax, Gene expression (gpc). OpenStax CNX. Mar 13, 2014 Download for free at http://cnx.org/content/col11638/1.1
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