What is the Difference Between hnRNA and mRNA?

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RNA synthesis takes place in the nucleus in eukaryotes. However, when RNA is formed, it is transferred from the nucleus to the cytoplasm. The physiological environment in the nucleus is quite different than than of the cytoplasm. This physiological difference may harm newly formed RNA in the cytoplasm. In order to prevent RNA from any damage, it is processed first and then transferred to the cytoplasm.

A newly formed RNA before processing is called heterogenous RNA (hnRNA) and RNA after processing is called mRNA (messenger RNA). Processing involves capping and tailing of RNA. Capping is done by adding 5-methyl guanosine, and tailing is done by adding polyA to the 3'-OH using poly A polymerase enzyme.


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What is Chargaff's Rule of DNA?

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Erwin Chargaff was an Australian biochemist who analyzed nucleic acids of many organisms. From his research, he concluded that the number of purine bases (A, G) equals the pyrimidine bases (C, T) in any double stranded DNA of an organism.

It means,

  •  A+G / T+C = 1; 
  • A+T / C+G may vary from organism to organism.
  • % A = % T, and % G = % C

Based on Chargaff's rule, many questions are asked in various competitive exams including AIPMT, CSIR NET, GATE, etc.

Q1. One of the strands of a double stranded DNA has the following number of bases:
A= 3800; T= 2600. What would be the base composition of the double stranded DNA?

Solution:

From the question, it is a double stranded DNA (Chargaff's rule is valid for only double stranded DNA), and only one strand with its base composition is given. So, the total number of bases would be 20k (10k bases on each strand).

Now, A and T bases with their numbers are given on one strand, so A will have T and T will have A on their complementary DNA. It means, on one strand, the total number of A would be 3800+2600 because % A = % T. Similarly, the total number of T would be 2600 + 3800 = 6400

The total number of A and T on both strands would be 6400 + 6400 = 12800

The remaining bases would be G and C = 20,000 - 12800 = 7200

Since % G =  % C; G = 7200/2 = 3600; C = 3600




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What Is Genic Balance Theory?

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C.B. Bridges in 1922 proposed the Genic Balance Theory for sex determination while working on his experimental model, Drosophila.

Like mammals, XX sex chromosome determines female character and XY as males in Drosophila. However, unlike humans, Y chromosome is not responsible for maleness in Drosophila, as Y chromosome is heterochromatic in nature and thus not active in sex determination.

According to Bridges, sex determination in Drosophila depends on the X/A sex index ratio. Following are the sex index ratios for maleness and female characters:

If X/A = 1.0 it means normal female
If X/A = 0.5, it means normal male
If X/A = <0.5, it means super male
If X/A = 0.5 to 1.0, it means intersex
If X/A = >1.0, it means super female

For example, 3A+XXX: X/A = 3/3 = 1.0 => Normal female
                     2A+XO: X/A = 1/2 = 0.5 => Normal male
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What are the different properties of water?

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What is called as the universal solvent. Water has many special properties that enable it to act as a solvent. These features are due to the polar structure of water molecule. Polarity of water molecules gives rise to hydrogen bonds.

The chemical formula of water is H2O. It means, two hydrogen atoms get attached to an oxygen atom. Since oxygen is electronegative in nature, it carries a partial negative charge. Hydrogen atoms, on the other hand, get a partial positive charge. Since the magnitude of these positive and negative charges are equal, water molecule carries no net charge.

However, these partial charges on H2O molecule attract other water molecules. This weak electrostatic attraction between H2O molecules is called HYDROGEN BOND. This hydrogen bond is responsible for many of the unusual physical properties of H2O. Some of the properties are mentioned below:

H2O is an excellent solvent:


The polarity of water molecule makes it an excellent solvent. Hydrogen bonding between water molecules and ions and between water and polar solutes in solution effectively decreases the charged substances and thereby increases their solubility.

H2O has high specific heat


Specific heat is defined as the energy required to raise the temperature of a substance by a specific amount. Unlike other liquids, water requires a relatively large energy input to raise its temperature. This large energy input is essential for plants because it helps buffer temperature fluctuations.

H2O has high latent heat of vaporization


Latent heat of vaporization is defined as the energy required to separate molecules from the liquid phase and move them into the gas phase at constant temperature – a process that occurs during transpiration. Due to high latent heat of vaporization – water enables plants to cool themselves by evaporating water from leaf surfaces.

Cohesive and adhesive properties of water



Cohesive means attraction between similar molecules and adhesive means attraction between different molecules. Due to these properties, water creates surface tension at an air-water interface. Cohesion, adhesion and surface tension give rise to a phenomenon, called CAPILLARY, the movement of H2O along a capillary tube
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AIPMT DISCUSSION QUESTIONS ON GENETIC CODE

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Q1. Which of the following is a characteristic of genetic code?
a)      The genetic code is non overlapping
b)      The genetic code is unambiguous
c)       The genetic code is degenerative
d)      All of the above

Explanation:

The correct answer is‘d’. The genetic code is non overlapping means that the successive triplets are read in order.
The genetic code is unambiguous means each codon or triplet specifies a particular amino acid and only one amino acid.
The genetic code is degenerative means each amino acid can be specified by more than one triplet codon.

Q2. Which of the following is a nonsense codon?
a)      UAG
b)      UGA
c)       UAA
d)      All of the above

Explanation:

The correct answer is‘d’. Out of 64 triplet codons, three codons are called nonsense codons because they terminate the process of translation. UAG is amber, UGA is opal and UAA is ochre.

Q3. Which of the following poly bases could not form any amino acid in Nirenberg’s experiment?
a)      Poly U
b)      Poly A
c)       Poly G
d)      Poly C

Explanation:

The correct answer is ‘c’ i.e., Poly G because it formed triple stranded structure and thus no translation occurred.

Q4. Hargobind Khorana chemically synthesized amino acids:
a)      Cysteine/valine 50:50 percent
b)      Proline/Cysteine 50:50 percent
c)       Phenyalanine/ Proline 50:50 percent
d)      All of the above

Explanation:
The correct answer is ‘a’ i.e., cysteine/valine. He used UGUGUGUGU poly nucleotide bases to synthesize the same.

Q5. Which one represents serine?
a)      CUU, CUC, CUA, and CUG
b)      UAU, UAC, UGU, and UGC
c)       UCU, UCC, UCA, and UCG
d)      UGU, UGC, UGA, and UAG

Explanation:
The correct answer is ‘c’. In order to remember which triplet codon codes for which amino acid, one needs to see the codon chart. If we reference Proline which falls on the second row and second column of the codon chart, we can remember the codons. However, a practice is needed for the same.

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AIPMT DISCUSSION QUESTIONS ON TRANSCRIPTION IN PROKARYOTES

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Q1: What is the correct direction of transcription?
a)      5’à 3’
b)      3’à 5’
c)       5’à 5’
d)      3’à 3’

Explanation:

The correct answer is ‘a’ i.e., transcription always occurs in the 5’ to 3’ direction. 5’ is called the head and 3’ is called the tail of a nucleotide chain. Replication also occurs in the same direction.

 Q2. A transcription unit comprises of:
a)      A promotor, structural gene, and a terminator
b)      A promotor and structural genes
c)       Structural genes and a terminator
d)      A promotor and a terminator

Explanation: 

The correct answer is ‘a’ i.e., a transcription unit comprises of a promoter, the structural gene and a terminator. Promoter is responsible for binding the enzyme RNA polymerase, strcutural genes are expressed to form proteins and a terminator sequence is required to end the process of transcription and translation. 

Q3. Which of the following statement is correct of a Lac operon?
a)      There are three promotor regions present for each gene Z, Y, and A
b)      There is a single promotor region for each gene Z, Y, and A
c)       There could be more than one promotors for the three genes Z, Y, and A
d)      No promotor is required for the three genes Z, Y, and A

Explanation:


The correct answer is ‘b’ i.e., for the Lac operon, only one promotor is required to express the three genes Z, Y, and A. On the other hand, in eukaryotes, each gene has its own promotor region.

Q4. Coding strand in DNA is the one:
a)      Which does not take part in transcription, but its bases will be the same as a new transcript with T replaced by U
b)      Which takes part in transcription, and its bases will the same as a new trascript with T replaced by U
c)       Which does not take part in transcription, but its bases will be the opposite of the new transcript bases
d)      None of the above

Explanation:

The correct answer is ‘a’ because, a coding strand is one that does not take part in transcription. It can also be called as a non template strand. However, since RNA strand does not contain Thymine, it is replaced by Uracil. 

Q5. The transcription start site i.e., +1 region contains:
a)      Mostly Purine bases
b)      Mostly Pyrimidine bases
c)       50 percent purine and 50 percent pyrimidine
d)      75 percent purine and 25 percent pyrmidine

Explanation:

The correct answer is ‘a’. In 90 percent of the cases, the +1 transcription start site contains purine bases especially Adenine.
Q6. Which of the following conserved sequence is called Pribnow box?
a)      TATAAT
b)      TATATA
c)       TTGACG
d)      GCAGAT

Explanation:

The correct answer is ‘a’ i.e., TATAAT a conserved seequence, which is present -10 seequence upstream on the DNA. This conserved sequence is essential for replication as RNA polymerase recognises this conserved sequence to start replication at the transcription start site.

Q7. Which of the following is the function of a sigma factor in transcription?
a)      Sigma factor is responsible for termination
b)      Sigma factor is responsible for elongation of RNA strand
c)       Sigma factor is involved in transcription initiation
d)      Sigma factor is involved in inhibiting the transcription process

Explanation:

The correct answer is ‘c’ i.e., sigma factor is responsible for transcription initiation. Sigma factor when attaches itself to the core enzyme, the combination is called Holoenzyme i.e., RNA polymerase. Sigma factor helps recognize the RNA polymerase to the correct place on the template strand to initiate transcription. Once the transcription process starts, sigma factor is released and the core enzyme gets involved in the elongation of RNA strand.

Q8. What mechanisms are involved in the termination of transcription process?
a)      Rho dependent and rho independent mechanism
b)      Rho dependent mechanism only
c)       Rho independent mechanism only
d)      None

Explanation:


The correct answer is ‘a’ i.e., rho dependent and rho independent mechanisms. The former is called protein based and the latter is called RNA based. 

Q9. Which of the following is true of rho dependent mechanism of termination?
a)      Near the end of the gene, RNA polymerase encounters a run of ‘A’ nucleotides
b)      Near the end of the gene, RNA polymerase encounnters a run of ‘G’ nucleotides
c)       Near the end of the gene, RNA polymerase encounters a run of ‘C’ nucleotides
d)      Near the end of the gene, RNA polymerase encounters a run of ‘U’ nucleotides

Explanation:

The correct answer is ‘b’ i.e., at the end of the gene, RNA polymerase encounters a run of G nucleotides. As a result, the rho protein collides with the polymerase. The interaction with the rho protein releases the mRNA from the termination bubble.

Q10. In RNA based termination, the polymerase enzyme encounters a region rich in:
a)      C-G nucleotides
b)      A-G nucleotides
c)       A-T nucleotides
d)      A-U nucleotides

Explanation:


The correct answer is ‘a’ i.e., the C-G nucleotides. The mRNA folds back on itself and the complementary C-G nucleotides bind together. The result is a hairpin that causes the polymerase to stall as soon as it begins to transcribe a region rich in A-T nucleotide.
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