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Biology: Life and Systems - Genetics and DNA Structure

Grade 8IB

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

🔑Concepts

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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.

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The basic building block of DNA is the nucleotide. Each nucleotide consists of three parts: a phosphate group (PO43−PO_4^{3-}), a deoxyribose sugar (C5H10O4C_5H_{10}O_4), and a nitrogenous base.

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

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

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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).

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

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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).