Board Question Paper : March 2019
BIOLOGY
Note:
- All questions are compulsory.
- Draw neat, labelled diagrams wherever necessary.
- Question paper consists of 30 questions divided into FOUR sections namely A, B, C and D.
- Section A: contains Q. No. 1 to 4 of multiple choice type of questions carrying one mark each and Q. No. 5 to 8 are very short answer type of questions carrying one mark each.
- Section B: contains Q. No. 9 to 18 of short answer type questions carrying two marks each. Internal choice is provided only to one question.
- Section C: contains Q. No. 19 to 27 of short answer type of questions carrying three marks each. Internal choice is provided only to one question.
- Section D: contains Q. No. 28 to 30 of long answer type of questions carrying five marks each. Internal choice is provided to each question.
- For each MCQ, correct answer must be written along with its alphabet, e.g., (a) ……. / (b) ……. / (c) ……. / (d) ……. etc.
- In case of MCQs, (i.e. Q. No. 1 to 4) evaluation would be done for the first attempt only.
- Start each section on a new page.
- Figures to the right indicate full marks.
SECTION A
(1)
Q.1 As the base sequence present on one strand of DNA decides the base sequence of other strand, this strand is considered as _______
(A) Descending strand
(B) Leading strand
(C) Lagging strand
(D) Complimentary strand
Answer:
(D) Complimentary strand
(1)
Q.2 _______ shows haplo-diploid type of sex-determination.
(A) Pigeon
(B) Honey bee
(C) Parrot
(D) Snake
Answer:
(B) Honey bee
(1)
Q.3 Membrane bound receptors and hormones produce second messengers like _______.
(A) Renin
(B) IP3
(C) ANF
(D) GHRF
Answer:
(B) IP3
(1)
Q.4 During double fertilization second male gamete fuses with _______.
(A) antipodal cell
(B) egg cell
(C) secondary nucleus
(D) synergids
Answer:
(C) secondary nucleus
(1)
Q.5 What is Sinus arrhythmias?
Answer:
Sinus arrhythmia is a normal variation in heart rate in which the heart rate increases during inspiration (breathing in) and decreases during expiration (breathing out). It is commonly seen in healthy individuals.
(1)
Q.6 By which process ammonia is converted into urea in liver?
Answer:
Ammonia is converted into urea in the liver by the Ornithine cycle (also known as the Urea cycle or Krebs-Henseleit cycle).
(1)
Q.7 Give the role of plasmids in bacterial cell.
Answer:
Plasmids provide additional genetic characteristics to the bacterial cell, such as antibiotic resistance (R-plasmids) or the ability to produce toxins. In biotechnology, they are used as vectors to transfer foreign DNA into host cells.
(1)
Q.8 A person is showing symptoms like increased BMR, heart rate, pulse rate, blood pressure and deposition of fats in eye sockets. Name the disease he is suffering from.
Answer:
The person is suffering from Exophthalmic Goiter (also known as Graves' disease), which is a form of Hyperthyroidism.
SECTION B
(2)
Q.9 Define apiculture. Name the products obtained from it.
Answer:
Definition: Apiculture (or beekeeping) is the scientific method of rearing, care, and management of honey bees for the production of honey and other products.
Products obtained:
- Honey
- Beeswax
- Royal Jelly
- Bee Venom (Apitoxin)
(2)
Q.10 Define biofertilizers. Give two types of fungal biofertilizers.
Answer:
Definition: Biofertilizers are preparations containing live or latent cells of efficient strains of nitrogen-fixing, phosphate-solubilizing, or cellulolytic microorganisms used for application to seed, soil, or composting areas to increase soil fertility.
Types of Fungal Biofertilizers (Mycorrhiza):
- Ectomycorrhiza: Fungal hyphae form a mantle on the root surface (e.g., in Pines).
- Endomycorrhiza (VAM): Fungal hyphae penetrate the root cortex cells (e.g., in Orchids, Grasses).
(2)
Q.11 Give the types of blood proteins and human hormones produced by recombinant DNA-technique.
Answer:
1. Blood Proteins:
- Factor VIII (for Hemophilia A treatment)
- Factor IX (for Hemophilia B treatment)
- Erythropoietin (for Anemia)
- Tissue Plasminogen Activator (tPA)
2. Human Hormones:
- Insulin (Humulin - for Diabetes)
- Somatotropin (Human Growth Hormone)
- Somatostatin
(2)
Q.12 Write any two scientific and commercial values of transgenic animals in favour of human being.
Answer:
1. Study of Disease (Human Models): Transgenic animals are designed to carry genes for specific human diseases (like cancer, cystic fibrosis, Alzheimer's) to understand how genes contribute to disease and to test new treatments.
2. Biological Products (Molecular Farming): Transgenic animals can be used to produce valuable biological products. For example, 'Rosie', the first transgenic cow, produced human protein-enriched milk (alpha-lactalbumin) which is nutritionally more balanced for human babies than natural cow milk.
(2)
Q.13 Define ‘Respiratory Quotient’ (RQ) and calculate the Respiratory Quotient for Carbohydrate.
Answer:
Definition: Respiratory Quotient (RQ) is defined as the ratio of the volume of Carbon dioxide (\(CO_2\)) evolved to the volume of Oxygen (\(O_2\)) consumed during respiration.
$$ RQ = \frac{\text{Volume of } CO_2 \text{ evolved}}{\text{Volume of } O_2 \text{ consumed}} $$
RQ for Carbohydrate:
When carbohydrates are used as respiratory substrate, they are completely oxidized. The equation is:
$$ C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + \text{Energy} $$
$$ RQ = \frac{6CO_2}{6O_2} = 1.0 $$
Therefore, the RQ for Carbohydrate is 1.0.
(2)
Q.14 Light and dark reactions are interdependent – Explain.
Answer:
Light and dark reactions of photosynthesis are interdependent because:
- Product Dependency: The Light reaction (photochemical phase) traps solar energy to produce assimilatory power in the form of ATP and NADPH. These products are absolutely essential for the Dark reaction.
- Utilization: The Dark reaction (biosynthetic phase) utilizes the ATP and NADPH produced in the light reaction to reduce \(CO_2\) into carbohydrates (glucose). Without the products of the light reaction, the dark reaction cannot proceed.
- Regeneration: The Dark reaction regenerates ADP, iP (inorganic phosphate), and NADP+, which are returned to the light reaction to be reused for the synthesis of more ATP and NADPH.
(2)
Q.15 Classify the chromosomes on the basis of position of centromere.
Answer:
Based on the position of the centromere, chromosomes are classified into four types:
- Metacentric: The centromere is situated in the middle of the chromosome. The two arms are nearly equal in length. It appears 'V' shaped during anaphase.
- Sub-metacentric: The centromere is situated slightly away from the middle. One arm is shorter than the other. It appears 'L' shaped during anaphase.
- Acrocentric: The centromere is situated near the end of the chromosome. One arm is very short and the other is very long. It appears 'J' shaped during anaphase.
- Telocentric: The centromere is situated at the tip (proximal end) of the chromosome. It has only one arm and appears rod-shaped ('i' shaped).
(2)
Q.16 Sketch and label structure of male gametophyte in angiosperm.
Answer:
(A diagram should be drawn showing a germinating pollen grain with a pollen tube containing two male gametes, a tube nucleus, and cytoplasm.)
Labels required:
- Exine / Intine (Pollen wall)
- Pollen Tube
- Tube Nucleus
- Male Gametes (Two)
- Cytoplasm
(2)
Q.17 Match the following and rewrite:
| Group ‘A’ | Group ‘B’ |
|---|
| i. Diethyle Carbamacine | a. AIDS |
| ii. Widal test | b. Pneumonia |
| iii. Albendazole | c. Filariasis |
| iv. HAART | d. Typhoid |
| e. Ascariasis |
Answer:
| Group ‘A’ |
Correct Match (Group ‘B’) |
| i. Diethyle Carbamacine |
c. Filariasis |
| ii. Widal test |
d. Typhoid |
| iii. Albendazole |
e. Ascariasis |
| iv. HAART (Highly Active Antiretroviral Therapy) |
a. AIDS |
(2)
Q.18 Complete the following chart and rewrite:
| Agencies | Type of Pollination |
|---|
| i. Water | ………. |
| ii. ………. | Entomophily |
| iii. Bat | ………. |
| iv. ………. | Ornithophily |
OR
Explain outbreeding devices in angiospermic plants.
Answer (Main Question):
| Agencies | Type of Pollination |
|---|
| i. Water | Hydrophily |
| ii. Insects | Entomophily |
| iii. Bat | Chiropterophily |
| iv. Birds | Ornithophily |
Answer (OR Question):
Outbreeding Devices: These are mechanisms that discourage self-pollination and encourage cross-pollination to prevent inbreeding depression.
- Unisexuality (Dicliny): The plant bears either male or female flowers, making self-pollination impossible (e.g., Papaya).
- Dichogamy: Anthers and stigma mature at different times.
- Protandry: Anthers mature first (e.g., Sunflower).
- Protogyny: Stigma matures first (e.g., Gloriosa).
- Prepotency: Pollen of other flowers germinates rapidly over the stigma than the pollen from the same flower (e.g., Apple).
- Heterostyly: Anthers and stigma are placed at different levels (e.g., Primrose).
- Self-sterility (Self-incompatibility): Pollen grains fail to germinate on the stigma of the same flower (e.g., Tobacco).
SECTION C
(3)
Q.19 What is Biofortification? Explain selective breeding with suitable example.
Answer:
Biofortification: It is the method of breeding crops with higher levels of vitamins, minerals, or higher protein and healthier fats to improve public health and overcome hidden hunger.
Selective Breeding (Conventional Breeding):
- Selective breeding is a traditional method of plant breeding where parents with desirable traits (like high nutrient content or high yield) are selected and crossed.
- The progeny (offspring) are then selected based on the presence of the desired combination of traits. This process is repeated over generations to stabilize the trait.
- Example: Hybrid maize with twice the amount of amino acids lysine and tryptophan was developed using selective breeding. Wheat variety 'Atlas 66' with high protein content has been used as a donor for improving cultivated wheat.
(3)
Q.20 In the light of Griffith’s experiment, explain the action of two strains of Diplococcus pneumoniae and give his conclusion.
Answer:
Frederick Griffith used two strains of
Diplococcus pneumoniae (Streptococcus pneumoniae):
- S-Strain (Smooth): Virulent, pathogenic, encapsulated (has a mucous coat), causes pneumonia.
- R-Strain (Rough): Non-virulent, non-pathogenic, non-capsulated, does not cause pneumonia.
The Experiment Steps:
- Step 1: Injected Live R-strain into mice → Mice Survived (No disease).
- Step 2: Injected Live S-strain into mice → Mice Died (Pneumonia).
- Step 3: Injected Heat-killed S-strain into mice → Mice Survived.
- Step 4: Injected Heat-killed S-strain + Live R-strain into mice → Mice Died.
Conclusion: Griffith recovered live S-strain bacteria from the dead mice in Step 4. He concluded that some "Transforming Principle" from the heat-killed S-strain had entered the live R-strain bacteria and transformed them into virulent S-strain bacteria by enabling them to synthesize a smooth polysaccharide capsule.
(3)
Q.21 Give scientific reasons:
(A) The pyramid of energy is always upright.
(B) In an ecosystem the energy flow is always unidirectional.
(C) Ozone present in the stratosphere is called as “good ozone”.
Answer:
(A) The pyramid of energy is always upright: According to the second law of thermodynamics and the 10% law, when energy is transferred from one trophic level to the next, only about 10% is stored as biomass. The remaining 90% is lost as heat for metabolic activities. Therefore, energy decreases at each successive trophic level, keeping the base broad and the top narrow.
(B) In an ecosystem the energy flow is always unidirectional: Energy enters the ecosystem from the Sun (captured by producers). It passes to consumers and decomposers. However, energy lost as heat to the environment cannot be reused by plants for photosynthesis. It cannot flow backward from consumers to producers; hence, it is unidirectional.
(C) Ozone present in the stratosphere is called as “good ozone”: This ozone layer acts as a shield absorbing ultraviolet (UV) radiation from the sun. UV rays are highly injurious to living organisms (causing skin cancer, mutation, etc.). Since stratospheric ozone protects life on Earth, it is called "good ozone" (unlike tropospheric ozone, which is a pollutant).
(3)
Q.22 Define ‘reproductive isolation’ and explain two types of reproductive isolation.
Answer:
Definition: Reproductive isolation refers to the mechanisms or barriers that prevent two different species from interbreeding and producing fertile offspring, thereby maintaining the integrity of the species.
Types of Reproductive Isolation (Pre-mating mechanisms):
- Temporal Isolation: Different species breed at different times of the day, different seasons, or different years, preventing them from mating. (e.g., American toad mates in early summer, Fowler's toad in late summer).
- Ethological (Behavioral) Isolation: Members of two populations have different mating rituals or courtship behaviors. If the female does not recognize the courtship display of the male, mating does not occur.
(Other types include Mechanical Isolation, Habitat Isolation, Gametic Mortality, etc.)
(3)
Q.23 Name the connecting link between glycolysis and TCA cycle and explain it.
Answer:
The connecting link between Glycolysis and TCA (Krebs) cycle is the
Acetylation of Pyruvate (or Link Reaction).
Explanation:
- The end product of glycolysis is Pyruvate (3-carbon), which is produced in the cytoplasm.
- Pyruvate enters the mitochondrial matrix.
- It undergoes oxidative decarboxylation. One molecule of \(CO_2\) is removed, and NAD+ is reduced to NADH + \(H^+\).
- The remaining 2-carbon fragment (acetyl group) combines with Coenzyme-A (CoA) to form Acetyl Co-A.
Reaction:
$$ \text{Pyruvate} + \text{NAD}^+ + \text{CoA} \xrightarrow[\text{Pyruvate Dehydrogenase}]{\text{Mg}^{++}} \text{Acetyl-CoA} + \text{NADH} + \text{H}^+ + \text{CO}_2 $$
(3)
Q.24 Explain internal structure of kidney with the help of suitable diagram.
Answer:
(Diagram of L.S. of Kidney showing Cortex, Medulla, Pyramids, Pelvis, Ureter)
Explanation:
- Capsule: The kidney is covered by a tough, fibrous connective tissue layer called the renal capsule.
- Cortex: The outer dark red region is called the Cortex. It contains Malpighian bodies, PCT, and DCT of nephrons.
- Medulla: The inner pale red region is the Medulla. It is divided into conical masses called Renal Pyramids (6 to 20 in number). It contains the Loop of Henle and collecting ducts.
- Columns of Bertini: The extensions of the cortex into the medulla between the pyramids are called Renal Columns of Bertini.
- Pelvis: The broad funnel-shaped space near the hilum is called the Renal Pelvis. The tips of pyramids (Renal Papillae) open into calyces which lead to the pelvis and finally the ureter.
(3)
Q.25 Explain the mechanism of reflex action with the help of a suitable diagram.
Answer:
(Diagram showing Receptor, Sensory Neuron, Spinal Cord/Interneuron, Motor Neuron, Effector Muscle)
Mechanism of Reflex Action:
It is a sudden, involuntary, and instantaneous response to a stimulus. The path traveled by the impulse is called the Reflex Arc.
- Receptor: Receives the stimulus (e.g., skin touching a hot object).
- Sensory Neuron (Afferent): Transmits the impulse from the receptor to the spinal cord via the dorsal root.
- Association Neuron (Interneuron): Located in the spinal cord, it processes the information and transfers it to the motor neuron.
- Motor Neuron (Efferent): Carries the impulse from the spinal cord to the effector organ via the ventral root.
- Effector: The muscle or gland that responds (e.g., muscle contraction to withdraw hand).
(3)
Q.26 Define pollution. “Industries are pouring poison in water”– Explain.
Answer:
Definition: Pollution is an undesirable change in the physical, chemical, or biological characteristics of air, water, or land that is harmful to human life and other living organisms.
"Industries are pouring poison in water" – Explanation:
- Industrial effluents often contain toxic substances like heavy metals (Mercury, Lead, Cadmium), organic compounds, acids, and alkalis.
- When released into water bodies without treatment, these substances kill aquatic life (fish, plants).
- Bioaccumulation: Toxins like mercury and DDT enter the food chain and their concentration increases at successive trophic levels (Biomagnification), eventually poisoning humans (e.g., Minamata disease).
- High organic load from industries like sugar mills increases Biological Oxygen Demand (BOD), reducing dissolved oxygen and suffocating aquatic life. Thus, untreated industrial waste acts as poison for the water ecosystem.
(3)
Q.27 With the help of a suitable diagram, describe ultra structure of the cell organelle, which is essential for photosynthesis.
OR
During photosynthesis “O2 is evolved from water molecule and not from CO2”. Give the experimental proof given by Robert Hill.
Answer (Main Question - Chloroplast):
(Diagram showing Double membrane, Stroma, Granum, Thylakoids, Intergranal lamellae)
The organelle essential for photosynthesis is the
Chloroplast.
- Envelope: It is bounded by a double membrane (outer and inner).
- Stroma (Matrix): The colorless, proteinaceous ground substance inside. It contains enzymes for the Dark reaction, DNA, RNA, and ribosomes (70S).
- Thylakoids: Flattened sac-like structures present in the stroma. The membrane of thylakoids contains photosynthetic pigments (Chlorophyll).
- Grana: Thylakoids are stacked like coins to form Grana. This is the site of the Light reaction.
- Intergranal Lamellae (Stroma Lamellae): These connect different grana.
Answer (OR Question - Hill's Reaction):
Robert Hill's Experiment (1937):
- Robert Hill suspended isolated chloroplasts from spinach leaves in water which was free of \(CO_2\).
- He added a hydrogen acceptor (like ferric salts or hemoglobin) to the suspension.
- On illuminating the suspension, he observed that oxygen bubbles were evolved and the hydrogen acceptor was reduced (Ferric to Ferrous).
- Since there was no \(CO_2\) present in the mixture, the oxygen must have come from the water ($H_2O$).
- Conclusion: This proved that the source of \(O_2\) evolved during photosynthesis is the photolysis of water, not carbon dioxide.
SECTION D
(5)
Q.28 Explain with help of a suitable diagram conducting system of human heart.
OR
Give reasons:
(A) Lymphatic vessels are milky in appearance.
(B) Monocytes are called scavengers.
(C) The wall of left ventricle is thicker than right ventricle.
(D) Valves are present in the veins.
(E) Pulmonary veins carry oxygenated blood.
Answer (Main Question):
The conducting system consists of specialized cardiac muscle fibers that initiate and conduct cardiac impulses.
(Diagram showing SA Node, AV Node, Bundle of His, Purkinje Fibers)
- SA Node (Sino-atrial Node): Located in the wall of the right atrium near the opening of the superior vena cava. It acts as the "Pacemaker" because it generates the impulse for heart contraction.
- AV Node (Atrio-ventricular Node): Located in the lower left corner of the right atrium near the inter-atrial septum. It receives the impulse from the SA node.
- Bundle of His (AV Bundle): Arises from the AV node and divides into right and left branches running down the interventricular septum.
- Purkinje Fibers: Fine fibers arising from the bundle branches that spread into the walls of the ventricles. They convey the impulse to the ventricular muscles causing contraction.
Answer (OR Question):
- Lymphatic vessels are milky: The lymph vessels originating from the intestine (lacteals) absorb fats. The presence of these absorbed fats gives the lymph a milky appearance, hence they are called lacteals or look milky.
- Monocytes are scavengers: Monocytes are large phagocytic WBCs. They engulf and destroy damaged cells, dead tissue, and cellular debris at the site of infection, effectively "cleaning" the area.
- Left ventricle wall is thicker: The left ventricle has to pump blood to all parts of the body (systemic circulation) against high pressure, whereas the right ventricle only pumps to the lungs nearby. The thick muscular wall provides the force required.
- Valves in veins: Blood pressure in veins is very low, and they carry blood against gravity towards the heart. Valves prevent the backflow of blood.
- Pulmonary veins carry oxygenated blood: By definition, veins carry blood towards the heart. Pulmonary veins bring blood from the lungs (where oxygenation occurs) to the left atrium; hence, they carry oxygenated blood.
(5)
Q.29 Which phenomenon gives 2:1 ratio instead of 3:1 ratio? Describe with graphical representation.
OR
A pea plant homozygous for yellow round seed is crossed with its recessive parents. Calculate the phenotypic and genotypic ratio with the help of checker board.
Answer (Main Question):
The phenomenon is
Lethal Genes. In certain cases, a gene in the homozygous condition causes the death of the organism.
Example: Coat color in Mice or Sickle Cell Anemia.
Sickle Cell Anemia Example:
- \(Hb^A\): Normal gene (Dominant)
- \(Hb^S\): Sickle cell gene (Recessive/Co-dominant)
Cross between two Carriers (Sickle-cell trait):
Parents: Carrier (\(Hb^A Hb^S\)) x Carrier (\(Hb^A Hb^S\))
| Gametes |
\(Hb^A\) |
\(Hb^S\) |
| \(Hb^A\) |
\(Hb^A Hb^A\) (Normal) |
\(Hb^A Hb^S\) (Carrier) |
| \(Hb^S\) |
\(Hb^A Hb^S\) (Carrier) |
\(Hb^S Hb^S\) (Sickle cell Anaemic - Dies) |
Result:
- Genotypes produced: 1 \(Hb^A Hb^A\) : 2 \(Hb^A Hb^S\) : 1 \(Hb^S Hb^S\)
- Since \(Hb^S Hb^S\) is lethal and the individual dies, the surviving ratio is modified.
- Ratio: 1 Normal : 2 Carriers i.e., 2:1.
Answer (OR Question):
Note: The question states "Homozygous (YYRR) crossed with recessive parents (yyrr)". This produces F1 hybrids. Usually, a 5-mark question implies finding the F2 ratio (Dihybrid Cross) or a Dihybrid Test Cross. Below is the solution for the Dihybrid Cross (F2 generation) as per standard board pattern.
1. Parents: Yellow Round (YYRR) x Green Wrinkled (yyrr)
2. F1 Generation: All Yellow Round (YyRr)
3. Selfing F1: YyRr x YyRr
Gametes: YR, Yr, yR, yr
Checker Board (F2 Generation):
| ♂ / ♀ |
YR |
Yr |
yR |
yr |
| YR |
YYRR (Yellow Round) |
YYRr (Yellow Round) |
YyRR (Yellow Round) |
YyRr (Yellow Round) |
| Yr |
YYRr (Yellow Round) |
YYrr (Yellow Wrinkled) |
YyRr (Yellow Round) |
Yyrr (Yellow Wrinkled) |
| yR |
YyRR (Yellow Round) |
YyRr (Yellow Round) |
yyRR (Green Round) |
yyRr (Green Round) |
| yr |
YyRr (Yellow Round) |
Yyrr (Yellow Wrinkled) |
yyRr (Green Round) |
yyrr (Green Wrinkled) |
Phenotypic Ratio: 9 Yellow Round : 3 Yellow Wrinkled : 3 Green Round : 1 Green Wrinkled (9:3:3:1)
Genotypic Ratio: 1:2:1:2:4:2:1:2:1
(5)
Q.30 After puberty human female shows cyclic changes in her reproductive system. Explain structural and hormonal changes in the uterus.
OR
Give reasons:
(A) Scrotal sac serves as thermoregulator.
(B) Corpus luteum gets converted into corpus albicans in absence of fertilization.
(C) Missing of menses is the first indication of pregnancy.
(D) Surgical sterilization is a permanent method of birth control.
(E) Human egg is microlecithal.
Answer (Main Question):
The cyclic changes in the uterus constitute the Menstrual Cycle. It has three phases:
- Menstrual Phase (Bleeding Phase): (Days 1-5)
- Hormones: Progesterone and Estrogen levels fall sharply due to degeneration of corpus luteum.
- Structure: The endometrium breaks down. Blood vessels rupture, causing bleeding. The unfertilized egg and tissue debris are discharged.
- Proliferative Phase (Follicular Phase): (Days 5-13)
- Hormones: FSH stimulates follicle development which secretes Estrogen. Estrogen stimulates repair.
- Structure: The endometrium regenerates. It becomes thicker (3-5mm) and vascular. Glands become active.
- Secretory Phase (Luteal Phase): (Days 15-28)
- Hormones: LH causes ovulation and formation of Corpus Luteum. Corpus Luteum secretes large amounts of Progesterone.
- Structure: Endometrium becomes maximum thickened, highly vascular, and glandular (uterine milk) to prepare for implantation. If fertilization does not occur, it degenerates, leading back to the menstrual phase.
Answer (OR Question):
- Thermoregulator: The testes need a temperature 2-3°C lower than body temperature for spermatogenesis. The scrotum hangs outside the body and its muscles contract or relax to adjust the distance from the body, maintaining this optimal temperature.
- Corpus Albicans: If fertilization does not occur, the egg dies. The high levels of LH drop. The Corpus Luteum degenerates and becomes a white scar tissue called Corpus Albicans.
- Missing Menses: During pregnancy, the corpus luteum persists and secretes progesterone, which maintains the endometrium. Hence, the shedding (menstruation) stops. This amenorrhea is the first sign.
- Permanent Method: In surgical sterilization (Vasectomy/Tubectomy), the gamete transport path is cut and tied. This makes gamete transport impossible permanently, and reversal is very difficult.
- Microlecithal: Human eggs develop inside the mother's uterus and receive nutrition via the placenta. Therefore, they do not need large stored food (yolk). Hence, they contain very little yolk (Microlecithal).
