Q: About Fatty acids melting point


 

Which fatty acid has the lower melting temperature?

 

a)      18:1D9

b)      18:2D9,12

c)      18:0

d)      18:1D11

e)      18:3D9,12,15

 

 

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Breakable: Mr. Glass and Osteogenesis Imperfecta.


 

 

This movie is called “Unbreakable”, in reference to the character played by Bruce Willis, but from the biochemical point of view, we are more interested in the character of Elijah, played by Samuel L. Jackson: the “Breakable” character, or “Mr. Glass”.

 

Elijah Price: “I have something called Osteogenesis Imperfecta. It’s a genetic disorder. I don’t make a particular protein very well and it makes my bones very low in density… very easy to break.”

 

 

More information, Soon, in a webpage near you!

(Unbreakable: Osteogenesis Imperfecta Case Presentation)

 

 

 

 

USMLE Step I, 2009: Important links


 

Step I Content description on line

http://www.usmle.org/Examinations/step1/step1_content.html

 

Pdf version of Step I content description and sample material 2009

http://download.usmle.org/2009step1.pdf

 

Overview of Step I content in the USMLE Bulletin of Information-2009

http://www.usmle.org/General_Information/bulletin/2009/content.html

 

2009 Orientation Materials

– Multiple choice tutorial and Practice Test items Version FredV1 (current test delivery software)

– Multiple choice tutorial and Practice Test items Versions FredV2 (test delivery software for 2009)

http://www.usmle.org/Orientation/2009/menu.html

 

 

And of course, review frequently the main link:

http://www.usmle.org/

 

Q: About the major rate-limiting step in glycolysis


 

Biochemistry Question CM-18

 

The major rate-limiting step of glycolysis in liver cells is

 

 

a)     The conversion of glucose to glucose 6-phosphate.

 

b)     the conversion of glucose 6-phosphate to fructose 6-phosphate.

 

c)      The conversion of fructose 6-phosphate to fructose 1,6-bisphosphate.

 

d)     the formation of trioses (the aldolase reaction).

 

e)     the conversion of pyruvate to lactate.

 

 

Biochemistry Related News


 

Vitamin B Does Not Slow Cognitive Decline in Alzheimer’s

 

Little-known fat can be a heartbraker

 

Scientists identify role of fatty acids in Alzheimer’s disease

 

Harvard Gene Projct to Reveal DNA Secrets of 10 Volunteers

 

Human Protein Atlas Will Help Pinpoint Disease

 

Scientists Use Light to Conrol Proteins

 

Antioxidants save trauma patients

 

Take your vitamins and double up on D

 

Parkinson linked to Vitamin D

 

School bans Birthday Sweets

 

 

 

 

 

 

About a baby with cataracts


 

(Q:CM-17) Two months after birth, the parents of a baby that was born abroad, detect that the baby does not follow objects with his eyes.  When your assistance is required, you detect cataracts and the presence of a reducing monosaccharide in urine.  The deficit of which of the following enzymes could be causing these signs?

 

a)     Fructokinase

 

b)     Galactokinase

 

c)      Glucokinase

 

d)     Hexokinase

 

e)     Phosphofructokinase

 

f)       Pyruvate kinase

 

 

      The answer here

 

 

How to calculate the energetic balance of the total oxidation of a fatty acid?


 

The total oxidation of a fatty acid comprehends different processes:

 

1.- The activation of the fatty acid

 

2.- Beta-Oxidation

 

3.- Kreb’s Cycle.

 

For being metabolized, a fatty acid should experiment an activation:

 

Fatty acid + CoA + ATP —-à Acyl CoA + AMP + 2(P)

 

The activation of the fatty acid requires 1 molecule of ATP, but since two energy rich bonds are hydrolized (the ATP is hydrolyzed to AMP and 2 (P) ) for energetic balance purposes it is considered that 2 ATP have been consumed in this activation process)

 

The formed acyl CoA will experiment different oxidation reactions. These reaction occurs in the Beta-carbon. That is why the process is called Beta-oxidation

 

Recalling:

 

       CH3 -………………………………..   -CH2 – CH2 – CH2 – CH2 -COOH

(Omega carbon)                                -Delta-Gamma-Beta-Alpha-Carboxyl

 

In Beta-oxidation (a mitochondrial process) the acyl CoA is totally oxidazed to Acetyl groups in form of Acetyl CoA units.

 

How many Acetyl CoA units will be formed?

 

Since the Acetyl group of the acetyl coA is formed by two carbons, we should divide the number of carbons in the acyl group between two.

 

Miristic acid (14 carbons): 14 carbons /2 = 7 Acetil CoA

 

Palmitic acid (16 carbons): 16 carbons/2 = 8 Acetyl CoA

 

In order to be completed degraded to Acetyl CoA, the fatty acid in form of acyl CoA should experiment several “rounds” in the Beta-oxidation process. En each round, an Acetyl CoA is released and, a NADH.H+  and a FADH2 are produced.

 

 

We know already how many acetyl Coa are formed from each fatty acid: n/2, where n  is the number of carbons.

 

Now, how many rounds are necessary for converting all the fatty acid to acetyl CoA?

 

Since in the last round we already obtain two Acetyl CoA, the number of necessary rounds is (n/2) -1

 

Observe:

 

Miristic acid (14 carbons)

 

 1st round:

Produce one acyl CoA of 12 carbons and one Acetyl CoA + NAD H.H+  +  FADH2

 

2nd round:

Produce one acyl CoA of 10 carbons and one Acetyl CoA + NAD H.H+  +  FADH2

 

3rd round:

Produce one acyl CoA of 8 carbons and one Acetyl CoA + NAD H.H+  +  FADH2

 

4th round:

Produce one acyl CoA of 6 carbons and one Acetyl CoA + NAD H.H+  +  FADH2

 

5th round:

Produce one acyl CoA of 4 carbons and one Acetyl CoA + NAD H.H+  +  FADH2

 

6th round:

Produce one acyl CoA of 2 carbons and one Acetyl CoA + NAD H.H+  +  FADH2

 

But the acyl CoA of 2 carbons is already an Acetyl CoA, so it is the last round! 

 

So, after 4 rounds, all the Miristic acid has been converted to Acetyl CoA.

 

Then, what we have obtained as a result of the beta-oxidation of Miristic acid?

 

7 acetyl CoA

And 6 NADH.H+ y 6 FADH2

 

The acetyl CoA units are oxidized up to CO2 and H2O in the Krebs Cycle.

 

In terms of  ATP, the yielding depends on the kind of yielding that is used for reduced cofactors:

 

a) If during your course it is considered that each NADH.H+ yields 2.5 ATP and each FADH2 yields 1.5 ATP, then

7 acetyl CoA x 10 ATP/AcetylCoA in the Krebs Cycle: 70 ATP

6  NADH.x 2.5 ATP/NADH = 11 ATP

6 FADH2 x 1.5 ATP/FADH2 = 9 ATP

Minus 2 ATP used in the activation (Miristic acid to miristyl CoA)

70+11+9-2 = 88 ATP

 

b) If during your course it is considered that each NADH yields 3 ATP and each FADH2 yields 2 ATP, then

7 acetyl CoA x 12 ATP/AcetylCoA in the Krebs Cycle: 84 ATP

6 NADHx 3 ATP/NADH = 18 ATP

6 FADH2 x 2 ATP/FADH2 = 12 ATP

Minus 2 ATP used in the activation (Miristic  acid to Miristyl coA)

84+18+12-2 = 112 ATP

 

How to calculate the energetic balance of any fatty acid?

 

Step 1.- Number of Carbons/2 = Number of Acetyl CoA formed.

 

Step 2.- Number of rounds in the Beta-oxidation necessary for converting the whole fatty acid to Acetyl Co A units:  Number of Acetyl CoA minus 1 [(n/2)-1]

 

Step 3 (Option a) If you consider that each NADH yields 2.5 ATP and each FADH2 yields 1.5 ATP then multiply the number of rounds times 4 and multiply the number of Acetyl CoA times 1o

 

OR

 

Step 3 (Option b).-  If you consider that each NADH yields 3 ATP and each FADH2 yields 2 ATP then multiply the number of rounds times 5 and multiply the number of Acetyl CoA times 12.

 

Step 4.- Take two ATP that were used for the activation of the Fatty Acid

 

 

Example:

 

Fatty acid with 12 Carbons

 

– Option (a):

 

Step 1:  12/2 = 6 Acetyl CoA

 

Step 2: 6-1 = 5 rounds

 

Step 3 (Option a): (5 x 4) + (6 x10)

 

Step 4 = -2

 

Total: 78 ATP

 

– Option (b):

 

Step 1:  12/2 = 6 Acetyl CoA

 

Step 2: 6-1 = 5 rounds

 

Step 3 (Option b): (5 x 5) + (6 x12)

 

Step 4 = -2

 

Total: 95 ATP

 

I would like now that the reader calculates the energetic balance of the total oxidation of a fatty acid with 18 carbons. Assume that the oxidation of each NADH.H+ yields 3 ATP and that each FADH2 yields 2 ATP.

 

I am looking forward to read your answers in the comments!