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Evolution - A Brief Account of Evolution

Grade 12CBSEBiology

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

🔑Concepts

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About 20002000 million years ago (mya), the first cellular forms of life appeared on Earth, some of which had the ability to release O2O_2 through mechanisms similar to photosynthesis.

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Invertebrates were formed and became active around 500500 mya, followed by jawless fish evolving around 350350 mya.

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Seaweeds and few plants existed probably around 320320 mya. Plants were the first organisms to colonize land.

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In 19381938, a Coelacanth (lobefin) was caught in South Africa. Lobefins were the ancestors of modern-day frogs and salamanders (amphibians).

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Amphibians evolved into reptiles, which lay thick-shelled eggs that do not dry up in the sun, unlike those of amphibians.

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Giant ferns (Pteridophytes) were present but fell to form coal deposits slowly.

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Some reptiles moved back into water to evolve into fish-like reptiles, such as Ichthyosaurs, approximately 200200 mya.

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The largest land reptiles were dinosaurs, with Tyrannosaurus rex being about 2020 feet in height and possessing dagger-like teeth.

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About 6565 mya, the dinosaurs suddenly disappeared from Earth, possibly due to climatic changes or a meteor hit.

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The first mammals were like shrews. Mammals were more intelligent and sensed/avoided danger better than other animals.

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Due to continental drift, South American mammals joined North American fauna, while Australian marsupials survived because of a lack of competition from other mammals.

📐Formulae

p2+2pq+q2=1p^2 + 2pq + q^2 = 1

p+q=1p + q = 1

Frequency of homozygous dominant=p2\text{Frequency of homozygous dominant} = p^2

Frequency of heterozygous=2pq\text{Frequency of heterozygous} = 2pq

Frequency of homozygous recessive=q2\text{Frequency of homozygous recessive} = q^2

💡Examples

Problem 1:

Calculate the time gap between the origin of the first cellular forms of life (20002000 mya) and the extinction of dinosaurs (6565 mya).

Solution:

Time gap = 2000 mya−65 mya=1935 mya2000 \text{ mya} - 65 \text{ mya} = 1935 \text{ mya}.

Explanation:

Using simple subtraction for the timeline: 2000−651935\begin{array}{r} 2000 \\ - 65 \\ \hline 1935 \end{array}

The result shows that cellular life existed for 19351935 million years before dinosaurs went extinct.

Problem 2:

If the frequency of a recessive allele qq in a population is 0.40.4, find the frequency of the heterozygous individuals in a population at equilibrium.

Solution:

Given q=0.4q = 0.4. Since p+q=1p + q = 1, then p=1−0.4=0.6p = 1 - 0.4 = 0.6. The frequency of heterozygotes is 2pq=2×0.6×0.4=0.482pq = 2 \times 0.6 \times 0.4 = 0.48.

Explanation:

According to the Hardy-Weinberg principle, the frequency of the heterozygous genotype is represented by the term 2pq2pq in the expansion of (p+q)2(p+q)^2.