Q.Even if a character shows multiple allelism, an individual will only have two alleles for that character. Why?

Step 1 — Read the direction of the arrow in the stem.

The stem says a single gene produces multiple phenotypic expressions. So we need the term for one gene → many traits.

Step 2 — Test each option.

• (A) Incomplete dominance — an allelic interaction at one locus for one trait: the heterozygote shows an intermediate phenotype (e.g. pink Mirabilis jalapa). One trait, not many.
• (B) Pleiotropy — a single gene influences several, often unrelated, phenotypic traits. Classic example: the gene for phenylketonuria causes mental retardation, reduced hair and skin pigmentation together; likewise sickle-cell anaemia affects RBC shape, oxygen transport and organ damage. ✓ This is exactly one gene → many phenotypes. …

Step 1 — The classic example: Mirabilis jalapa (four o'clock plant)

$F_1$: all $Rr$ — and these are pink, not red. The heterozygote does not resemble either parent; it is intermediate. This is incomplete dominance: the dominant allele's product is not enough on its own to give the full phenotype.

Step 2 — Self the $F_1$ and build the Punnett square

Step 3 — Read off the two ratios

Genotypic ratio:

Phenotypic ratio (incomplete dominance):

The two ratios coincide, which is the hallmark of incomplete dominance. (Under complete dominance the genotypic ratio would still be $1:2:1$, but the phenotypic ratio would collapse to $3:1$ because $RR$ and $Rr$ look alike.) …

Step 1 — Set up the cross

Let $R$ = round (dominant), $r$ = wrinkled; $Y$ = yellow (dominant), $y$ = green.

Selfing the $F_1$ gives the familiar $F_2$ phenotypic ratio:

Step 2 — Read the question carefully

It does not ask for the four-class dihybrid ratio. It asks for "the ratio of segregation of round and wrinkled seed traits" — i.e. collapse the yellow/green distinction and look at seed shape alone.

Step 3 — Collapse the $9:3:3:1$

Step 4 — Why this had to be $3:1$ (the concept)

This is Mendel's Law of Independent Assortment: the segregation of one pair of alleles is independent of the segregation of the other pair. Consequently, when you consider seed shape by itself, the dihybrid cross behaves exactly like a monohybrid cross: …

The core idea here is dominance and segregation. Mendel’s first law tells us that when a homozygous dominant parent (say, TT) is crossed with a homozygous recessive parent (tt), the $F_1$ generation inherits one allele from each parent — all offspring are Tt. Because the dominant allele (T) masks the recessive (t), every $F_1$ plant shows the dominant trait. The recessive trait doesn’t vanish; it’s just hidden.

Now, when these $F_1$ plants self-pollinate, the two alleles segregate during gamete formation. The $F_2$ generation gets a mix: TT, Tt, and tt. The dominant trait appears in both TT and Tt plants, while the recessive trait appears only in tt plants. So in $F_2$, both traits are visible — the dominant in about three-fourths of the offspring, the recessive in one-fourth.

Let’s walk through the options:

1. Option (A) says the dominant trait appears in $F_2$ and the recessive appears only in $F_1$. That’s backwards — the recessive is absent in $F_1$, not present. So (A) is wrong.

2. Option (B) says the dominant appears in $F_1$ and the recessive appears in $F_2$. This is partially correct — the dominant does appear in $F_1$, and the recessive does reappear in $F_2$ — but it misses that the dominant also appears in $F_2$. So (B) is incomplete and therefore incorrect. …

1. Evaluate Statement (1)

"Generally a gene regulates a trait, but sometimes one gene has effect on multiple traits."

This is exactly the definition of PLEIOTROPY — a single gene producing multiple phenotypic effects. The mechanism is that the gene affects an early step of a metabolic pathway, and every downstream trait that depends on that pathway is altered.

Standard examples:

• Phenylketonuria — a single gene defect (phenylalanine hydroxylase) causes mental retardation and reduced hair/skin pigmentation.
• Starch synthesis in pea seeds — one gene affects starch grain size and seed shape (round vs wrinkled).

So Statement (1) is CORRECT. ✓

2. Evaluate Statement (2)

"The trait AB-blood group of man is regulated by one dominant allele and another recessive allele. Hence it is co-dominant."

The conclusion (that AB is co-dominance) is right, but the reason given is wrong, so the statement as printed is false. Here is the correct picture.

The ABO gene $I$ has three alleles: $I^A$, $I^B$ and $i$ (multiple allelism).

• $I^A$ and $I^B$ are BOTH DOMINANT over $i$;
• $I^A$ and $I^B$ are CO-DOMINANT with respect to each other — neither masks the other. …

Step 1 — The concept: the ABO blood-group gene

The ABO locus has three alleles — $I^A$, $I^B$ and $I^O$ (often written $i$) — an example of multiple allelism:

• $I^A$ and $I^B$ are CODOMINANT — when both are present, both antigens A and B appear on the RBC surface, giving blood group AB.
• $I^O$ is RECESSIVE to both $I^A$ and $I^B$ (it produces no antigen).

So the genotype → phenotype map is:

Step 2 — Set up the cross

• Husband: $I^A I^B$ → gametes $I^A$ and $I^B$
• Wife: $I^A I^O$ → gametes $I^A$ and $I^O$

Step 3 — The Punnett square

Step 4 — Count the GENOTYPES

All four are distinct ⇒ 4 genotypes (in a 1 : 1 : 1 : 1 ratio).

Step 5 — Count the PHENOTYPES

Apply the dominance rules from Step 1:

• $I^A I^A \rightarrow$ A
• $I^A I^O \rightarrow$ A (since $I^O$ is recessive)
• $I^A I^B \rightarrow$ AB (codominance) …

The key here is understanding the genetics of the ABO blood group system. This system is controlled by a single gene with three alleles: $I^A$, $I^B$, and $i$. The $I^A$ and $I^B$ alleles are codominant with each other (both are expressed when present together), and both are dominant over the recessive $i$ allele.

When a parent is described as "heterozygous for A", it means they have one $I^A$ allele and one $i$ allele — their genotype is $I^A i$, and their blood group is A. Similarly, "heterozygous for B" means the genotype is $I^B i$, and the blood group is B.

The question asks about the possible blood groups of children when one parent is $I^A i$ and the other is $I^B i$. Since each parent contributes one allele to the child, we can work through all possible combinations.

1. List the possible gametes from each parent.

   The parent with $I^A i$ can produce two types of gametes: one carrying $I^A$ and one carrying $i$.

   The parent with $I^B i$ can produce two types of gametes: one carrying $I^B$ and one carrying $i$.

2. Set up a Punnett square to find all possible offspring genotypes.

   The four equally likely combinations are:

   • $I^A$ from first parent + $I^B$ from second parent → genotype $I^A I^B$ → blood group AB (both codominant alleles expressed).
   • $I^A$ from first parent + $i$ from second parent → genotype $I^A i$ → blood group A (dominant $I^A$ over $i$).
   • $i$ from first parent + $I^B$ from second parent → genotype $I^B i$ → blood group B (dominant $I^B$ over $i$).
   • $i$ from first parent + $i$ from second parent → genotype $i i$ → blood group O (no A or B antigens). …

Step 1 — Set up the cross (Mendel's dihybrid experiment).

Round (R) and yellow (Y) are dominant; wrinkled (r) and green (y) recessive.

Step 2 — Self the $F_1$.

$RrYy \times RrYy$ gives the classic $F_2$ phenotypic ratio (independent assortment — the genes are on different chromosomes):

Total parts $= 9+3+3+1 = 16$.

Step 3 — Scale to 64 progeny.

Each "part" $= \dfrac{64}{16} = 4$ plants.

The key idea: Morgan’s famous experiment used two X‑linked genes — body colour (yellow vs brown) and eye colour (white vs red). In Drosophila, females have two X chromosomes, males have one X and one Y. When a yellow‑bodied, white‑eyed female (both recessive traits) is crossed with a brown‑bodied, red‑eyed male (both dominant traits), the $F_1$ females are heterozygous for both genes, and the $F_1$ males inherit their X from the mother. When these $F_1$ flies are intercrossed, the $F_2$ offspring show a huge excess of parental combinations and very few recombinants — because the two genes are very close together on the X chromosome.

Let’s walk through the cross step by step.

1. Parental generation

   Female: yellow body, white eyes — both recessive. Genotype: $X^{yw} X^{yw}$ (where $y$ = yellow, $w$ = white).

   Male: brown body, red eyes — both dominant. Genotype: $X^{YW} Y$ (where $Y$ = brown/red wild‑type allele, and the Y chromosome carries no alleles for these genes).

2. $F_1$ generation

   • All daughters get one $X^{yw}$ from mother and one $X^{YW}$ from father → $X^{yw} X^{YW}$ (brown body, red eyes — both dominant traits expressed).
   • All sons get the $X^{yw}$ from mother and the Y from father → $X^{yw} Y$ (yellow body, white eyes — both recessive).

   So $F_1$ females are dihybrid (heterozygous for both genes), and $F_1$ males are hemizygous recessive.

3. Intercrossing $F_1$

   Cross: $X^{yw} X^{YW}$ (female) $\times$ $X^{yw} Y$ (male).

   The female produces four types of eggs, but because the two genes are linked on the same X chromosome, the majority are parental types ($X^{yw}$ and $X^{YW}$) and only a small fraction are recombinant types ($X^{yW}$ and $X^{Yw}$). The male produces two types of sperm: $X^{yw}$ and Y.

4. Recombination frequency …

The question is about a population that is not evolving — no mutation, no migration, no selection, random mating, and a large population size. Under those ideal conditions, the Hardy–Weinberg principle tells us that allele frequencies stay constant from generation to generation, and genotype frequencies are simply the product of the allele probabilities.

If the frequency of allele $A$ is $p$ and the frequency of allele $a$ is $q$, then:

• $p + q = 1$ (only two alleles at this locus).
• The three genotypes appear with frequencies:
  • $AA$: $p^2$
  • $aa$: $q^2$
  • $Aa$: $2pq$

The factor of 2 in $2pq$ comes from the two ways a heterozygote can form: the $A$ can come from the mother and $a$ from the father, or vice versa.

Here we are given $p = 0.6$ and $q = 0.4$. Notice that $0.6 + 0.4 = 1$, so the numbers are consistent.

1. Write the formula for heterozygote frequency

   The expected frequency of $Aa$ individuals is $2pq$.

2. Substitute the given values

3. Calculate

4. Convert to percentage $0.48 \times 100\% = 48\%$ …