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Genetics - Genes

Grade 11IBBiology

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

🔑Concepts

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A gene is a heritable factor that consists of a length of DNADNA and influences a specific characteristic.

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A gene occupies a specific position on a type of chromosome called the locus (plural: loci).

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Alleles are alternative forms of a gene that code for the variations of a specific trait. They differ from each other by only one or a few bases.

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New alleles are formed by mutation. A base substitution mutation is the most common, where one base in the sequence of a gene is replaced by a different base.

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The genome is the whole of the genetic information of an organism. The human genome consists of approximately 3.2×1093.2 \times 10^9 base pairs.

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Sickle cell anemia is a genetic disease caused by a base substitution mutation on the HBBHBB gene which codes for the β\beta-globin polypeptide of hemoglobin.

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In sickle cell anemia, the DNADNA sequence changes from GAGGAG to GTGGTG on the coding strand, resulting in the mRNAmRNA codon changing from GAGGAG to GUGGUG.

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The mutation in sickle cell anemia causes the amino acid Glutamic AcidGlutamic \ Acid to be replaced by ValineValine at the sixth position of the polypeptide chain.

📐Formulae

DNANormal:GAG→mRNA:GAG→Amino Acid:GluDNA_{Normal}: GAG \rightarrow mRNA: GAG \rightarrow Amino \ Acid: Glu

DNASickle:GTG→mRNA:GUG→Amino Acid:ValDNA_{Sickle}: GTG \rightarrow mRNA: GUG \rightarrow Amino \ Acid: Val

Genome Size (Homo sapiens)≈3.2×109 bpGenome \ Size \ (Homo \ sapiens) \approx 3.2 \times 10^9 \text{ bp}

Gene Number (Homo sapiens)≈21,000Gene \ Number \ (Homo \ sapiens) \approx 21,000

💡Examples

Problem 1:

Calculate the difference in the number of base pairs between the genome of T2 phageT2 \ phage (1.7×105 bp1.7 \times 10^5 \text{ bp}) and the human genome (3.2×109 bp3.2 \times 10^9 \text{ bp}).

Solution:

3,200,000,000−170,000=3,199,830,000 bp3,200,000,000 - 170,000 = 3,199,830,000 \text{ bp}

Explanation:

The human genome is significantly larger than viral genomes. In scientific notation, the difference is approximately 3.19983×109 bp3.19983 \times 10^9 \text{ bp}.

Problem 2:

Identify the change in the 6th6^{th} amino acid of the hemoglobin β\beta-chain in a patient with Sickle Cell Anemia.

Solution:

Glutamic Acid (polar)→Valine (non-polar)Glutamic \ Acid \text{ (polar)} \rightarrow Valine \text{ (non-polar)}

Explanation:

The replacement of a hydrophilic (polar) amino acid with a hydrophobic (non-polar) amino acid causes the hemoglobin molecules to polymerize at low oxygen concentrations, distorting the red blood cell into a sickle shape.