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Biology - Reproduction in Plants and Humans

Grade 7IGCSE

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

🔑Concepts

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Sexual reproduction involves the fusion of two haploid (nn) gametes to form a diploid (2n2n) zygote.

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Asexual reproduction involves only one parent and produces genetically identical offspring (clones) without the fusion of gametes.

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In plants, the male reproductive part is the stamen (consisting of the anther and filament), and the female part is the carpel/pistil (consisting of the stigma, style, and ovary).

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Pollination is the transfer of pollen grains from the anther to the stigma, which can be facilitated by wind or insects.

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The human male reproductive system produces sperm and the hormone testosterone (C19H28O2C_{19}H_{28}O_2).

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The human female reproductive system produces ova (eggs) and hormones such as estrogen (C18H24O2C_{18}H_{24}O_2) and progesterone (C21H30O2C_{21}H_{30}O_2).

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The menstrual cycle is a monthly cycle of approximately 2828 days where the uterine lining thickens and is shed if fertilization does not occur.

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Fertilization in humans occurs in the oviduct (Fallopian tube) when a sperm nucleus fuses with an egg nucleus.

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The placenta is a specialized organ that allows for the exchange of nutrients, oxygen (O2O_2), and waste products (CO2CO_2, urea) between the mother and the fetus via diffusion.

📐Formulae

Haploid Gamete (n)+Haploid Gamete (n)→Diploid Zygote (2n)\text{Haploid Gamete } (n) + \text{Haploid Gamete } (n) \rightarrow \text{Diploid Zygote } (2n)

Magnification=Measured size of imageActual size of specimen\text{Magnification} = \frac{\text{Measured size of image}}{\text{Actual size of specimen}}

C6H12O6+6O2→6CO2+6H2O+Energy (ATP)C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + \text{Energy (ATP)} (Aerobic respiration providing energy for fetal growth)

💡Examples

Problem 1:

Compare the characteristics of wind-pollinated and insect-pollinated flowers.

Solution:

Insect-pollinated flowers have large, brightly colored petals and nectar; wind-pollinated flowers have small, dull petals and long filaments with feathery stigmas.

Explanation:

Insect-pollinated flowers need to attract pollinators using visual and chemical signals. Wind-pollinated flowers, such as grasses, have feathery stigmas to provide a large surface area to catch pollen grains drifting in the air (vwindv_{wind}).

Problem 2:

Describe the change in the thickness of the uterine lining during the menstrual cycle and the role of hormones.

Solution:

Days 1−51-5: Lining sheds (menstruation). Days 6−146-14: Lining thickens due to estrogen. Days 15−2815-28: Lining is maintained by progesterone.

Explanation:

If fertilization does not occur, the levels of progesterone drop, leading to the breakdown of the lining. This cycle is approximately T=28T = 28 days.

Problem 3:

A pollen grain is viewed under a microscope. The image size is 44 mm and the actual size is 0.050.05 mm. Calculate the magnification.

Solution:

Magnification =×80= \times 80

Explanation:

Using the formula M=IAM = \frac{I}{A}, we calculate M=4 mm0.05 mm=80M = \frac{4\text{ mm}}{0.05\text{ mm}} = 80.

Reproduction in Plants and Humans - Revision Notes & Key Formulas | IGCSE Grade 7 Science