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Sexual Reproduction in Flowering Plants Set-4 MCQ Online Practice Test | Class 12 | Quiz Tweek

Sexual Reproduction in Flowering Plants Set-4

Practice Important MCQs for Sexual Reproduction in Flowering Plants — Biology, Class 12.

Sexual reproduction in flowering plants is a high-scoring topic because it links meiosis, gametogenesis, double fertilization, ploidy changes, and fertilization-triggered endosperm development. Board as well as competitive exams frequently ask about chromosome numbers (egg, sperm, zygote, PEN), compatibility/incompatibility mechanisms, and how apomixis or ploidy imbalance leads to seed failure. Hence, mastering the logic of double fertilization and endosperm dosage is essential.

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Q1. In a flowering plant with somatic chromosome number $2n=24$ , what will be the chromosome numbers in the zygote and the primary endosperm nucleus (PEN) immediately after double fertilization?

(A)

Zygote $12$ , PEN $24$

(B)

Zygote $24$ , PEN $36$

(C)

Zygote $24$ , PEN $48$

(D)

Zygote $12$ , PEN $36$

Q2. A pollen donor is heterozygous $S_1S_2$ and the recipient pistil is $S_1S_3$ . Which self‑incompatibility (SI) system yields 50% of the donor pollen grains compatible and which yields 0% compatible?

(A)

Gametophytic SI → 0% compatible; Sporophytic SI → 50% compatible

(B)

Both gametophytic and sporophytic SI → 50% compatible

(C)

Gametophytic SI → 100% compatible; Sporophytic SI → 0% compatible

(D)

Gametophytic SI → 50% compatible; Sporophytic SI → 0% compatible

Q3. A pollen tube grows at constant rate $v$ (mm h^{-1}). The generative cell divides into two sperm after $T_d$ hours from germination; each sperm remains viable for $T_v$ hours after formation. If style length is $L$ mm, successful double fertilization requires $\displaystyle\frac{L}{v}\le T_d+T_v$ . Given $L=12\ \text{mm}$ , $T_d=2\ \text{h}$ and $T_v=6\ \text{h}$ , the minimum $v$ (mm h^{-1}) required is:

(A)

$1.5\ \mathrm{mm\;h^{-1}}$

(B)

$0.75\ \mathrm{mm\;h^{-1}}$

(C)

$2.0\ \mathrm{mm\;h^{-1}}$

(D)

$3.0\ \mathrm{mm\;h^{-1}}$

Q4. Assertion (A): Apomixis produces seeds that are genetically identical to the maternal parent.
Reason (R): In apomictic species both the embryo and the endosperm always develop without any fertilization, so pollination is never required.

(A)

Both A and R are true and R is the correct explanation of A

(B)

Both A and R are true but R is not the correct explanation of A

(C)

A is true but R is false

(D)

A is false but R is true

Q5. Due to meiotic failure a plant produces diploid sperm cells ( $2n$ ) instead of haploid ( $n$ ). If such a diploid sperm fertilizes a normal haploid egg ( $n$ ) and the two polar nuclei ( $n+n$ ), what will be the chromosome numbers of the zygote and the primary endosperm nucleus (PEN), and the likely consequence for seed development?

(A)

Zygote $2n$ , PEN $3n$ — normal seed development

(B)

Zygote $3n$ , PEN $4n$ — abnormal embryo/endosperm balance; likely seed abortion or defective seed

(C)

Zygote $n$ , PEN $3n$ — viable seed but sterile adult

(D)

Zygote $4n$ , PEN $6n$ — viable and fully fertile seed

Q6. In a typical angiosperm double fertilisation, an unreduced pollen grain delivers a diploid sperm ( $2n$ ) which fertilises a normal haploid egg ( $n$ ) and also fertilises the central cell containing two haploid polar nuclei ( $n + n$ ). What will be the ploidy of the resulting embryo and the primary endosperm nucleus?

(A)

Embryo $2n$ , endosperm $3n$

(B)

Embryo $3n$ , endosperm $4n$

(C)

Embryo $4n$ , endosperm $5n$

(D)

Embryo $3n$ , endosperm $3n$

Q7. A mutant embryo sac contains only a single haploid polar nucleus in the central cell (instead of the usual two), while egg and pollen are normal haploids. After fertilisation by a normal sperm ( $n$ ), which of the following describes the expected ploidy of the primary endosperm nucleus, the maternal:paternal genome ratio in it, and the likely effect on seed development?

(A)

Endosperm $3n$ ; maternal:paternal $2:1$ ; endosperm development normal

(B)

Endosperm $1n$ ; maternal:paternal $1:0$ ; maternal-only endosperm, seed aborts

(C)

Endosperm $2n$ ; maternal:paternal $2:0$ ; maternal-excess endosperm, abnormal development

(D)

Endosperm $2n$ ; maternal:paternal $1:1$ ; deviation from normal $2:1$ ratio, likely disturbed endosperm development and seed failure

Q8. A plant shows these floral features: numerous small green inconspicuous flowers, no nectar or showy petals, exposed stamens and feathery stigmas, and a pollen:ovule ratio greater than $10{,}000$ . Which pollination mechanism is most consistent with these traits?

(A)

Anemophily (wind pollination)

(B)

Entomophily (insect pollination)

(C)

Hydrophily (water pollination)

(D)

Ornithophily (bird pollination)

Q9. In a cross where the maternal parent is diploid ( $2n$ ) and the paternal parent is tetraploid ( $4n$ ), the female gamete is haploid ( $n$ ) and the male gamete from the $4n$ parent is diploid ( $2n$ ). What are the ploidies of embryo and primary endosperm nucleus, what is the maternal:paternal genome ratio in the endosperm, and which explanation best accounts for frequent seed failure in such crosses?

(A)

Embryo $2n$ , endosperm $3n$ , ratio $2:1$ — balanced dosage; seeds viable

(B)

Embryo $3n$ , endosperm $5n$ , ratio $2:3$ — paternal excess causes overproliferation of endosperm

(C)

Embryo $3n$ , endosperm $4n$ , ratio $1:1$ — deviation from normal $2:1$ maternal:paternal ratio disrupts endosperm development leading to seed abortion

(D)

Embryo $4n$ , endosperm $3n$ — embryo ploidy mismatch is the primary cause of abortion

Q10. The first zygotic division in angiosperms is normally asymmetric, producing a small apical cell (embryo proper) and a large basal cell (suspensor). If a mutation causes the first division to be symmetric (two equal-sized daughter cells), which developmental consequence is most likely?

(A)

Polarity will be re-established by later divisions and a normal suspensor will form

(B)

Loss of zygote polarity resulting in absence or malformation of the suspensor and disrupted apical–basal patterning of the embryo

(C)

Only suspensor will differentiate while the embryo proper fails to form

(D)

Endosperm development will be directly inhibited while embryo and suspensor develop normally

Q11. In a flowering plant species with somatic chromosome number $2n=14$ , during normal double fertilization one male gamete fuses with the egg cell and the other fuses with the two polar nuclei. What will be the chromosome numbers of the zygote and the primary endosperm nucleus (PEN) respectively?

(A)

Zygote $2n=14$ ; PEN $3n=21$

(B)

Zygote $n=7$ ; PEN $3n=21$

(C)

Zygote $2n=14$ ; PEN $2n=14$

(D)

Zygote $3n=21$ ; PEN $2n=14$

Q12. A genetic defect produces pollen grains that contain only one functional sperm nucleus instead of the usual two. A pollen tube from such a pollen grain reaches an embryo sac and releases the single sperm. Which of the following outcomes is most likely?

(A)

Only the central cell (polar nuclei) is fertilized and endosperm develops while egg remains unfertilized, resulting in an endosperm-only, non-viable seed.

(B)

The single sperm splits and fertilizes both the egg and the central cell, producing both zygote and endosperm; normal seed develops.

(C)

Only the egg is fertilized forming a zygote, but because endosperm is absent or insufficient the embryo cannot develop normally and the seed is typically non-viable.

(D)

The single sperm fuses with a synergid triggering autonomous endosperm development and embryo formation proceeds normally.

Q13. A diploid seed parent with somatic chromosome number $2n=14$ is pollinated by an autotetraploid pollen parent ( $4n=28$ ) that produces unreduced pollen grains ( $2n=14$ ). After normal double fertilization, what are the chromosome numbers of the zygote and the primary endosperm nucleus (PEN) respectively?

(A)

Zygote $2n=14$ ; PEN $3n=21$

(B)

Zygote $3n=21$ ; PEN $4n=28$

(C)

Zygote $3n=21$ ; PEN $3n=21$

(D)

Zygote $2n=14$ ; PEN $4n=28$

Q14. Assertion (A): Viability of seeds resulting from interploidy crosses depends more on the genomic ratio in the endosperm than on the ploidy level of the embryo.
Reason (R): Endosperm supplies nourishment to the developing embryo and requires a specific maternal:paternal genomic ratio (typically $2:1$ ); deviation from this ratio disrupts endosperm development and commonly causes seed abortion.

(A)

Both A and R are true but R does not explain A

(B)

A is true but R is false

(C)

A is false but R is true

(D)

Both A and R are true and R explains A

Q15. Species X has somatic chromosome number $2n=16$ ( $n=8$ ). Line A shows diplospory (an apomictic route) in which an unreduced embryo sac is formed and the embryo develops parthenogenetically from the unreduced egg cell ( $2n$ ), while the central cell requires fertilization by a normal reduced sperm from a sexual diploid pollen donor (pseudogamy). If Line A is pollinated by a normal sexual diploid (pollen nucleus $n=8$ ), what will be the chromosome numbers of the embryo and the primary endosperm nucleus (PEN) respectively?

(A)

Embryo $3n=24$ ; PEN $3n=24$

(B)

Embryo $2n=16$ ; PEN $5n=40$

(C)

Embryo $2n=16$ ; PEN $3n=24$

(D)

Embryo $4n=32$ ; PEN $6n=48$

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