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Genetics - Molecular Biology and Biotechnologies

Grade 9IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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DNA Structure: Deoxyribonucleic acid (DNA) is a double helix consisting of two polynucleotide chains. Each nucleotide is composed of a phosphate group, a deoxyribose sugar, and one of four nitrogenous bases: Adenine (AA), Thymine (TT), Cytosine (CC), and Guanine (GG).

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Base Pairing Rules: According to Chargaff's rules, Adenine always pairs with Thymine via two hydrogen bonds (A=TA = T), and Cytosine always pairs with Guanine via three hydrogen bonds (G≡CG \equiv C).

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The Central Dogma: Genetic information flows from DNADNA to RNARNA to Protein. This involves two main stages: Transcription (DNA to mRNAmRNA) and Translation (mRNAmRNA to protein).

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Genetic Engineering: The process of modifying an organism's genome using biotechnology. This involves using restriction enzymes to 'cut' DNA at specific sequences and DNA ligase to 'paste' or join DNA fragments.

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Polymerase Chain Reaction (PCR): A laboratory technique used to amplify specific segments of DNADNA, creating millions of copies from a very small initial sample.

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Gel Electrophoresis: A method used to separate DNADNA fragments based on their size and charge. Since DNADNA is negatively charged, it moves toward the positive electrode; smaller fragments move faster and further through the gel matrix.

📐Formulae

%A+%G=%T+%C=50%\%A + \%G = \%T + \%C = 50\%

ncodons=nnucleotides3n_{\text{codons}} = \frac{n_{\text{nucleotides}}}{3}

Number of DNA molecules after n PCR cycles=2n\text{Number of DNA molecules after } n \text{ PCR cycles} = 2^n

💡Examples

Problem 1:

A sample of double-stranded DNADNA is analyzed and found to contain 22%22\% Cytosine (CC). Calculate the percentage of Thymine (TT) in this sample.

Solution:

  1. According to Chargaff's rule, %C=%G\%C = \%G, so %G=22%\%G = 22\%.
  2. The sum of CC and GG is: 22%+22%=44%22\% + 22\% = 44\%.
  3. The remaining percentage for AA and TT is: 100%−44%=56%100\% - 44\% = 56\%.
  4. Since %A=%T\%A = \%T, the percentage of Thymine is: 56%2=28%\frac{56\%}{2} = 28\%

Explanation:

In double-stranded DNA, the amount of Cytosine equals Guanine, and Adenine equals Thymine. By subtracting the known C+GC+G content from 100%100\%, we find the total A+TA+T content, which is then halved to find the specific percentage of Thymine.

Problem 2:

A segment of mRNAmRNA consists of 450450 nucleotides. How many amino acids will be present in the protein synthesized from this mRNAmRNA (ignoring the stop codon)?

Solution:

Each amino acid is coded for by a triplet of nucleotides called a codon. Number of amino acids=4503=150\text{Number of amino acids} = \frac{450}{3} = 150

Explanation:

Because the genetic code is a triplet code, every 33 nucleotides on the mRNAmRNA strand correspond to 11 amino acid in the polypeptide chain.

Problem 3:

Calculate the total number of DNADNA copies produced from a single template molecule after 55 cycles of PCR.

Solution:

The formula for DNADNA amplification is 2n2^n, where nn is the number of cycles. 25=2×2×2×2×2=322^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32

Explanation:

In each cycle of PCR, the amount of DNADNA doubles. Therefore, after 55 cycles, the original molecule has been replicated 3232 times.