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Genetics - Genes

Grade 11IBBiology

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

🔑Concepts

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

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

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.

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.

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.

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.

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.

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:GAGmRNA:GAGAmino Acid:GluDNA_{Normal}: GAG \rightarrow mRNA: GAG \rightarrow Amino \ Acid: Glu

DNASickle:GTGmRNA:GUGAmino 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:

3,200,000,000170,000=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.

Genes - Revision Notes & Key Diagrams | IB Grade 11 Biology