Identify the parts of the pistil labeled A, B, C, and D in the diagram, and determine the most common abiotic pollinating agent.

The figure shows a diagram of a pistil, the female reproductive organ of a flower.

• A points to the stigma, the receptive tip for pollen.
• B points to the style, the stalk connecting the stigma to the ovary.
• C points to the ovary, the swollen basal part containing ovules.
• D points to the thalamus (or receptacle), the part of the flower stalk where the parts of the flower are attached.

Comparing with the options:

• A. A: Thalamus, B: Style, C: Ovary, D: Stigma (Incorrect order)
• B. A: Style, B: Ovary, C: Stigma, D: Thalamus (Incorrect order)
• C. A: Stigma, B: Style, C: Ovary, D: Thalamus (Correct)
• D. A: Ovary, B: Stigma, C: Thalamus, D: Style (Incorrect order)

The final answer is $\boxed{\text{C}}$

The question asks for the most common abiotic (non-living) pollinating agent.

• A. Anemophily: Pollination by wind. Wind is an abiotic agent.
• B. Hydrophily: Pollination by water. Water is an abiotic agent.
• C. Pollination by bees: Bees are biotic agents.
• D. Pollination by ants: Ants are biotic agents. Between wind and water, wind (anemophily) is a far more common abiotic pollinating agent for plants than water (hydrophily).

The final answer is $\boxed{\text{A}}$

A human female with blood group 'A' has specific antigens on her red blood cells and specific antibodies in her serum.

• Blood group A means that Antigen A is present on the surface of her red blood cells.

• To prevent agglutination (clumping) if exposed to other blood types, her body produces antibody-anti-b (or anti-B antibody) in her blood serum.

• A. Antibody-anti-b in the red blood cells and antigen A in the serum: Incorrect, antibodies are in serum, antigens on red blood cells.

• B. Antigen A in the red blood cells and antibody-anti-b in the serum: Correct.

• C. Antigen B in the red blood cells and antibody-anti-a in the serum: This describes blood group B.

• D. Antigen A in the red blood cells and antibody-anti-a in the serum: Incorrect, a person with blood group A does not produce anti-A antibodies.

The final answer is $\boxed{\text{B}}$

A Mendelian monohybrid cross typically refers to a cross between two heterozygous individuals for a single trait (e.g., $Aa \times Aa$).
The genotypes produced are $AA$, $Aa$, and $aa$.
The ratio of these genotypes is $1:2:1$.

• A. $3:1$: This is the phenotypic ratio for a dominant-recessive monohybrid cross.
• B. $1:2:1$: This is the genotypic ratio for a monohybrid cross.
• C. $1:1$: This is the ratio for a test cross (e.g., $Aa \times aa$).
• D. $3:2$: Not a standard Mendelian ratio.

The final answer is $\boxed{\text{B}}$

A person with blood group 'O' has the genotype $ii$. This means they inherited an $i$ allele from each parent.
The mother has blood group 'A'. For her to pass an $i$ allele to her child, her genotype must be heterozygous $I^A i$.
The father has blood group 'B'. For him to pass an $i$ allele to his child, his genotype must be heterozygous $I^B i$.
Therefore, both the mother and the father must be heterozygous for their respective blood groups.

• A. Mother is homozygous for 'A' blood group ($I^A I^A$) and father is heterozygous for 'B' ($I^B i$). If the mother is $I^A I^A$, she cannot pass an $i$ allele. Incorrect.
• B. Mother is heterozygous for 'A' blood group ($I^A i$) and father is homozygous for 'B' ($I^B I^B$). If the father is $I^B I^B$, he cannot pass an $i$ allele. Incorrect.
• C. Both mother and father are heterozygous for 'A' and 'B' blood group, respectively. Mother ($I^A i$) and Father ($I^B i$). This combination can produce an $ii$ (blood group O) child. Correct.

The final answer is $\boxed{\text{C}}$