krit.club logo

Biology: Life and Systems - Genetics and DNA Structure

Grade 8IB

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

🔑Concepts

DNA (Deoxyribonucleic Acid) is the molecule that carries the genetic instructions for life. It is structured as a double helix, which resembles a twisted ladder.

The basic building block of DNA is the nucleotide. Each nucleotide consists of three parts: a phosphate group (PO43PO_4^{3-}), a deoxyribose sugar (C5H10O4C_5H_{10}O_4), and a nitrogenous base.

There are four types of nitrogenous bases: Adenine (AA), Thymine (TT), Cytosine (CC), and Guanine (GG).

Complementary Base Pairing Rule: In the DNA double helix, Adenine always pairs with Thymine (ATA-T) via two hydrogen bonds, and Cytosine always pairs with Guanine (CGC-G) via three hydrogen bonds.

Genes are specific sequences of DNA that code for proteins. Humans have approximately 20,00020,000 to 25,00025,000 genes packaged into 2323 pairs of chromosomes (4646 in total).

Alleles are different versions of the same gene, represented by letters (e.g., BB for dominant and bb for recessive).

Genotype refers to the genetic makeup (e.g., BbBb), while Phenotype refers to the physical expression of the trait (e.g., Brown eyes).

📐Formulae

%A=%T\%A = \%T

%G=%C\%G = \%C

%A+%T+%G+%C=100%\%A + \%T + \%G + \%C = 100\%

Probability of Genotype=Specific OutcomesTotal Possible Outcomes\text{Probability of Genotype} = \frac{\text{Specific Outcomes}}{\text{Total Possible Outcomes}}

💡Examples

Problem 1:

If a sample of DNA is analyzed and found to contain 18%18\% Guanine (GG), calculate the percentage of Thymine (TT) in the sample.

Solution:

G=18%    C=18%G = 18\% \implies C = 18\% G+C=18%+18%=36%G + C = 18\% + 18\% = 36\% A+T=100%36%=64%A + T = 100\% - 36\% = 64\% T=64%2=32%T = \frac{64\%}{2} = 32\%

Explanation:

According to Chargaff's rule, the amount of Guanine equals Cytosine, and Adenine equals Thymine. By subtracting the sum of GG and CC from 100%100\%, we find the total for AA and TT, which is then divided by 22 to find the individual percentage for Thymine.

Problem 2:

Use a Punnett Square to determine the probability of an offspring having a recessive phenotype if both parents are heterozygous (BbBb).

Solution:

The possible genotypes are: 1×BB1 \times BB, 2×Bb2 \times Bb, and 1×bb1 \times bb. Probability of recessive phenotype (bbbb) = 14=25%\frac{1}{4} = 25\%

Explanation:

In a cross between two heterozygous parents (Bb×BbBb \times Bb), the recessive trait only appears when the offspring inherits the recessive allele from both parents (bbbb).

Genetics and DNA Structure - Revision Notes & Key Formulas | IB Grade 8 Science