Q.If a double stranded DNA has 20 per cent of cytosine, calculate the per cent of adenine in the DNA.

The key concept here is the test cross. In Mendelian genetics, when an individual with a dominant phenotype but unknown genotype is crossed with a homozygous recessive individual, the offspring ratios reveal the unknown genotype. Here, we already know the red parent is heterozygous (Rr), so the cross is Rr × rr.

Why does this give a 1:1 ratio? Because the heterozygous parent produces two types of gametes in equal proportion — R and r — while the white parent produces only r gametes. Fertilisation is random, so half the offspring get R from the red parent (making them red, Rr) and half get r (making them white, rr). No dominance complication arises because the white parent contributes only recessive alleles.

Let’s work through the numbers.

1. Write the cross: Red heterozygous = Rr, white = rr. The cross is Rr × rr.

2. Gametes from each parent:

   • Rr produces: ½ R, ½ r.
   • rr produces: all r.

3. Punnett square outcome:

   • R (from red) + r (from white) → Rr (red) — ½ probability.
   • r (from red) + r (from white) → rr (white) — ½ probability.

   So the expected phenotypic ratio is 1 red : 1 white.

4. Apply to 192 offspring:

   • Red = ½ × 192 = 96.
   • White = ½ × 192 = 96.

   The ratio is 96 : 96.

Concept and Intuition

Water movement in and out of cells is governed by water potential ($\Psi_w$). Water always moves from a region of higher water potential to a region of lower water potential. Water potential is a measure of the potential energy of water per unit volume relative to pure water in reference conditions.

Water potential is determined by two main components:

1. Osmotic Potential ($\Psi_s$): Also known as solute potential. This component arises from the presence of dissolved solutes. Solutes reduce the free energy of water, making the osmotic potential always negative (or zero for pure water). A higher concentration of solutes means a more negative osmotic potential.
2. Pressure Potential ($\Psi_p$): This component arises from the physical pressure exerted on water. In plant cells, this is primarily the turgor pressure exerted by the cell contents against the cell wall. Turgor pressure is usually positive, but it can be zero or even negative (tension) under certain conditions.

The relationship between these potentials is given by the formula:

When a cell is placed in a hypertonic solution, the external solution has a higher solute concentration (and thus a more negative osmotic potential) than the cell's cytoplasm. Consequently, water moves out of the cell, causing the cell to lose turgor. If enough water leaves, the protoplast shrinks and pulls away from the cell wall, a process called plasmolysis.

Step-by-step Solution

1. Identify the given information:

   • The cell is placed in a hypertonic solution.
   • The osmotic potential ($\Psi_s$) of the cell is measured as $-0.5$ MPa.

2. Analyze the effect of a hypertonic solution on the cell:

   When a cell is in a hypertonic solution, water moves out of the cell due to osmosis. This causes the cell to lose turgor and become flaccid, or even undergo plasmolysis (where the cell membrane pulls away from the cell wall).

3. Determine the pressure potential ($\Psi_p$) of the cell:

   In a flaccid or plasmolyzed cell, the protoplast is not exerting any outward pressure against the cell wall. Therefore, the turgor pressure is zero. This means the pressure potential ($\Psi_p$) of the cell is $0$ MPa.

4. Calculate the water potential ($\Psi_w$) using the formula:

   We use the formula $\Psi_w = \Psi_s + \Psi_p$.

   Substitute the given osmotic potential and the determined pressure potential:

5. Conclude the water potential: The water potential of the cell is $-0.5$ MPa.

The question asks for the proportions of phenotypes in the $F_1$ generation, not just the green-seeded ones — but the options only list Tall Green and Dwarf Green fractions. That means we need to check which option correctly gives those two fractions from the full cross.

Let’s unpack the genetics first. Height is controlled by one gene: T (tall) is dominant over t (dwarf). Seed colour is controlled by another gene: Y (yellow) is dominant over y (green). The cross is between:

• Parent 1: TtYy (tall, yellow)
• Parent 2: Ttyy (tall, green)

Since the two genes are on different chromosomes (assort independently), we can treat them separately and then combine probabilities.

1. Height cross: Tt × Tt

   This is a monohybrid cross. The offspring genotypes and phenotypes are:

   • TT (tall) — $\frac{1}{4}$
   • Tt (tall) — $\frac{1}{2}$
   • tt (dwarf) — $\frac{1}{4}$ So probability of tall = $\frac{1}{4} + \frac{1}{2} = \frac{3}{4}$, and dwarf = $\frac{1}{4}$.

2. Seed colour cross: Yy × yy

   This is a test cross. Offspring:

   • Yy (yellow) — $\frac{1}{2}$
   • yy (green) — $\frac{1}{2}$

3. Combine using the product rule (since genes assort independently):

   Multiply the probabilities for each combination of traits.

   • Tall Yellow: $\frac{3}{4} \times \frac{1}{2} = \frac{3}{8}$
   • Tall Green: $\frac{3}{4} \times \frac{1}{2} = \frac{3}{8}$
   • Dwarf Yellow: $\frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$
   • Dwarf Green: $\frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$

   The full phenotypic ratio is $3 : 3 : 1 : 1$.

4. Match with the options

   The question only lists the green-seeded phenotypes in the options. From above:

   • Tall Green = $\frac{3}{8}$
   • Dwarf Green = $\frac{1}{8}$

   That corresponds exactly to option (A).

Concept & Intuition

Gel electrophoresis separates DNA fragments by size using an electric field. DNA is a nucleic acid with a sugar-phosphate backbone — the phosphate groups are negatively charged at neutral pH. Because opposite charges attract, DNA moves toward the positive electrode (the anode). The reason given claims DNA fragments are positively charged, which is the exact opposite of the truth. So the assertion stands, but the reason is wrong.

Step-by-step reasoning

1. Understand the assertion (A):

   In gel electrophoresis, DNA fragments are placed in a gel matrix and an electric current is applied. Because DNA is negatively charged, it migrates toward the anode (positive terminal). Smaller fragments move faster, so they separate by size. This statement is correct.

2. Understand the reason (R):

   The reason claims DNA fragments are positively charged molecules. In reality, the phosphate backbone of DNA gives it a strong negative charge at physiological pH. Therefore, this statement is incorrect.

3. Evaluate the relationship:

   Since (R) is false, it cannot be the correct explanation of (A). The only option that matches “(A) is correct but (R) is not correct” is option (C).

The question tests your understanding of mitosis and the physical state of chromosomes during metaphase. Let’s break it down.

Why metaphase is ideal for observing chromosomes

During metaphase, chromosomes have reached their highest level of condensation (supercoiling). They are thick, short, and distinct — easy to see under a microscope. They also line up at the metaphase plate (the equatorial plane of the cell), which makes them neatly arranged and countable. This is why cytogeneticists routinely use metaphase spreads for karyotyping.

What the Reason says — and why it’s wrong

The Reason claims that in metaphase, chromosomes are "scattered in the cytoplasm." That is incorrect. In metaphase, chromosomes are not scattered; they are precisely aligned at the equator. The nuclear envelope has broken down, so they are in the cytoplasm, but they are not randomly dispersed — they are held by spindle fibres and arranged in a single plane. "Scattered" implies disorder, which is the opposite of what happens.

Now, step by step:

1. Evaluate the Assertion (A): "Suitable stage to observe chromosome is metaphase." This is true. The high condensation and equatorial alignment make metaphase the best stage for counting and studying chromosome morphology.

2. Evaluate the Reason (R): "In metaphase chromosomes are scattered in cytoplasm." This is false. Chromosomes are in the cytoplasm (because the nuclear membrane has dissolved), but they are not scattered — they are aligned at the metaphase plate.

3. Determine the relationship: Since R is false, it cannot be a correct explanation of A. The correct option is therefore the one where A is true and R is false.

When we cross two pure individuals—one homozygous dominant, one homozygous recessive—we are performing what Mendel called a monohybrid cross between true-breeding lines. The key word here is "pure": it tells us both parents are homozygous, so every gamete from the grey parent carries the dominant allele and every gamete from the black parent carries the recessive allele.

Let $G$ represent the dominant allele (grey body) and $g$ the recessive allele (black body).

The parental genotypes are:

• Pure grey-bodied: $GG$
• Pure black-bodied: $gg$

Now we trace what happens at fertilization.

1. Gamete formation: The grey parent ($GG$) produces only $G$ gametes. The black parent ($gg$) produces only $g$ gametes.

2. F₁ offspring genotype: Every offspring receives one $G$ from the grey parent and one $g$ from the black parent, giving genotype $Gg$ in 100% of the progeny.

3. F₁ offspring phenotype: Since grey is dominant, every $Gg$ individual expresses the grey body color. There are no $gg$ individuals in this generation, so no black-bodied flies appear.

4. The ratio: All offspring are grey-bodied, none are black-bodied. The phenotypic ratio is grey : black = $1 : 0$.

The key concept here is the rediscovery of Mendel’s laws in 1900. Mendel published his findings on inheritance in 1866, but they were largely ignored. Thirty-four years later, three scientists working independently—Hugo de Vries (Netherlands), Carl Correns (Germany), and Erich von Tschermak (Austria)—each arrived at similar conclusions and, while reviewing literature, found Mendel’s original paper. They are credited with the rediscovery, which launched modern genetics.

The question presents four unnamed scientists (A, B, C, D) and asks which pair rediscovered Mendel’s results. Without the key naming the letters, we must rely on standard historical knowledge: the three rediscoverers are de Vries, Correns, and Tschermak. Any two of them form a correct pair. The options given are:

• (A) A, B
• (B) B, C
• (C) A, D
• (D) A, C

Since the question is from a typical biology exam, the standard mapping is:

• A = Hugo de Vries
• B = Carl Correns
• C = Erich von Tschermak
• D = (often a distractor, e.g., William Bateson, who actually championed Mendel’s work later but did not rediscover it)

Thus, the pair of rediscoverers is A and C (de Vries and Tschermak) or A and B (de Vries and Correns) or B and C (Correns and Tschermak). Among the options, only (D) A, C matches a valid pair of rediscoverers.