501. Which comparison correctly describes spermatogonial mitosis and meiosis II of a secondary spermatocyte?
ⓐ. Both can increase cell number without changing the ploidy present before that division
ⓑ. Both reduce chromosome number from \(46\) to \(23\)
ⓒ. Spermatogonial mitosis produces haploid cells, whereas meiosis II produces diploid cells
ⓓ. Neither process involves cell division
Correct Answer: Both can increase cell number without changing the ploidy present before that division
Explanation: Spermatogonial mitosis begins with a diploid spermatogonium and produces diploid daughter cells, so the \(2n=46\) state is preserved while cell number increases. Meiosis II begins with a haploid secondary spermatocyte produced by meiosis I. It generates two haploid spermatids, so the \(n=23\) condition is preserved during that particular division even though cell number again increases. The key difference is the ploidy state being maintained: diploid during spermatogonial mitosis and haploid during meiosis II. Meiosis I, not meiosis II, is the reductional step that changes a diploid primary spermatocyte into haploid secondary spermatocytes. The comparison separates change in cell number from change in chromosome-set status.
502. Twenty-five spermatids are present in a seminiferous-tubule region. If \(60\%\) complete spermiogenesis and no cell is lost, while spermiation has not yet occurred, which observation is expected?
ⓐ. \(25\) spermatozoa released into the lumen
ⓑ. \(10\) spermatozoa and \(15\) diploid spermatids
ⓒ. \(15\) spermatozoa and \(10\) secondary spermatocytes
ⓓ. \(15\) spermatozoa and \(10\) haploid spermatids
Correct Answer: \(15\) spermatozoa and \(10\) haploid spermatids
Explanation: Spermiogenesis is the differentiation of spermatids into spermatozoa. Sixty percent of \(25\) spermatids is \(0.60\times25=15\), so \(15\) cells complete spermiogenesis. The remaining \(25-15=10\) cells are still spermatids. No meiotic division occurs during spermiogenesis, so both the unchanged spermatids and the newly formed spermatozoa remain haploid with \(23\) chromosomes. The description also states that spermiation has not yet occurred. Spermiation is the subsequent release of formed spermatozoa from their association with Sertoli cells into the seminiferous-tubule lumen; it is distinct from the morphological transformation of spermiogenesis. The expected observation is therefore \(15\) spermatozoa and \(10\) haploid spermatids, without treating the spermatozoa as already released. The calculation connects the percentage to a developmental-stage distinction: \(60\%\) have changed cellular form, while \(40\%\) remain at the spermatid stage, and neither group has changed ploidy. Spermiation would change location, not the \(15:10\) differentiation count established here.
503. Microscopy shows elongated spermatozoa with normal head, middle-piece and tail differentiation still attached to Sertoli cells rather than released into the seminiferous-tubule lumen. Which process is specifically defective?
ⓐ. Meiosis I
ⓑ. Meiosis II
ⓒ. Spermiation
ⓓ. Spermiogenesis
Correct Answer: Spermiation
Explanation: The cells have already acquired the differentiated form of spermatozoa, including the specialised head, middle piece and tail. That observation shows that spermiogenesis, the transformation of spermatids into spermatozoa, has occurred. The abnormality is their continued attachment to Sertoli cells rather than their release into the lumen of the seminiferous tubule. That release step is called spermiation. Meiosis I and meiosis II occurred still earlier, producing secondary spermatocytes and then haploid spermatids respectively. The microscopic evidence localises the defect after differentiation but before normal luminal release. This distinction matters as spermiogenesis and spermiation are consecutive processes with different biological meanings even though their names are similar and both occur late in spermatogenesis. Selective changes in Sertoli, Leydig or germ-cell function can be distinguished by asking whether support, androgen secretion or cell division is altered first.
504. Population P of sperm has normal middle-piece mitochondria but structurally non-functional tails. Population Q has structurally intact tails but severely defective middle-piece mitochondria. Which comparison is most appropriate?
ⓐ. Both can lose motility: P lacks effective tail action; Q lacks adequate mitochondrial energy
ⓑ. P should retain normal motility because mitochondria are intact, whereas Q should retain normal motility because its tail is intact
ⓒ. Only P shows impaired motility; mitochondrial energy is not required if Q has an intact tail
ⓓ. Only Q shows impaired motility; tail structure is unnecessary if P has normal mitochondria
Correct Answer: Both can lose motility: P lacks effective tail action; Q lacks adequate mitochondrial energy
Explanation: Sperm motility depends on coordinated function of two different structural components. The middle piece contains numerous mitochondria that provide energy required for movement, whereas the tail provides the major motile apparatus responsible for propulsion. In population P, energy supply can remain available, but a structurally non-functional tail prevents that energy from being converted effectively into movement. In population Q, the tail structure is present, but severely impaired mitochondria reduce the energy support needed for vigorous motility. Both defects can reduce movement through different proximal causes. Neither condition directly changes sperm ploidy or Leydig-cell androgen secretion. The comparison separates the energy-producing role of the middle piece from the movement-producing role associated with the tail. The inference should follow only from the conditions stated, without extending a local reproductive event into unrelated stages.
505. Population P of sperm retains a normal haploid nucleus but lacks functional acrosomal enzymes. Population Q retains a functional acrosome but lacks the normal paternal haploid nucleus. Which comparison is correct?
ⓐ. P cannot move, whereas Q cannot produce mitochondrial energy
ⓑ. P is impaired in oocyte entry; Q cannot provide the paternal haploid nucleus
ⓒ. P retains normal fertilising capacity because its haploid nucleus is intact, whereas Q can compensate for the missing nucleus through its acrosome
ⓓ. P and Q have identical defects because the acrosome and nucleus perform the same role
Correct Answer: P is impaired in oocyte entry; Q cannot provide the paternal haploid nucleus
Explanation: The sperm head contains both the haploid nucleus and the acrosome, but their functions are distinct. Acrosomal secretions assist the fertilising sperm during passage through the oocyte coverings, so population P is primarily impaired at the entry stage despite retaining its paternal chromosomes. Population Q can retain the enzyme-containing acrosomal apparatus and may preserve the entry-assisting function, but without the haploid paternal nucleus it cannot contribute the normal paternal chromosome set required for formation of the diploid zygote. The comparison separates a fertilisation-access function from a genetic contribution. Sharing the same cellular region does not make the two structures functionally interchangeable, and neither defect is primarily a mitochondrial or tail abnormality. Acrosomal assistance, sperm entry, prevention of additional sperm entry and nuclear fusion are related but distinct events around fertilisation.
506. In a semen sample, \(70\%\) of sperm have normal shape and size and \(50\%\) show vigorous motility. Without knowing the exact overlap between these groups, what is the minimum possible percentage that must satisfy both properties?
ⓐ. \(10\%\)
ⓑ. \(50\%\)
ⓒ. \(70\%\)
ⓓ. \(20\%\)
Correct Answer: \(20\%\)
Explanation: The two groups together account for \(70\%+50\%=120\%\) of the sperm population when counted separately. Since only \(100\%\) of the population exists, at least the excess \(20\%\) must belong to both groups. Equivalently, the minimum overlap is:
\[
70\%+50\%-100\%=20\%
\]
The actual overlap could be larger, but it cannot be smaller than \(20\%\). This reasoning goes beyond checking the morphology and motility thresholds independently. It uses the fact that both measurements describe subsets of the same sperm population. The supplied information does not provide an exact overlap value, so the correct conclusion is a mathematical lower bound rather than a claim that exactly \(20\%\) of sperm necessarily satisfy both characteristics. The lower-bound rule can be written generally as \(\max(0,A+B-100\%)\) for two percentages \(A\) and \(B\) measured in the same population. Here the sum exceeds \(100\%\), forcing some overlap. The result is a constraint on shared membership, not a new semen-quality threshold.
507. A semen sample is centrifuged so that intact sperm are separated from the surrounding seminal plasma. The sperm fraction remains morphologically intact. Which change most directly belongs to removal of the fluid fraction rather than alteration of the sperm cells themselves?
ⓐ. Loss of the paternal haploid chromosome set
ⓑ. Conversion of sperm heads into diploid cells
ⓒ. Loss of the plasma membrane that normally encloses each intact sperm cell after ejaculation
ⓓ. Removal of extracellular seminal fluid containing fructose, calcium and enzymes
Correct Answer: Removal of extracellular seminal fluid containing fructose, calcium and enzymes
Explanation: Semen consists of sperm together with seminal plasma. Seminal plasma is the surrounding fluid contributed by male accessory-gland secretions and is described as containing substances such as fructose, calcium and certain enzymes. Centrifugal separation of intact sperm from that fluid removes much of those extracellular components while leaving the sperm cells themselves morphologically present. The paternal haploid nucleus and sperm plasma membrane are parts of each sperm and should not disappear simply because the external fluid is separated. The experiment distinguishes sperm-intrinsic structures from components of semen that reside outside the sperm. It also reinforces why semen and seminal plasma are not synonyms: one contains both cellular and fluid components, whereas the other is only the fluid component.
508. During fetal development, oogonia form normally but are experimentally prevented from entering meiosis I. No new oogonia are added after birth. Which state is most directly expected later?
ⓐ. A normal prenatal pool of prophase-I-arrested primary oocytes still forms
ⓑ. The primary-oocyte reserve is compromised because oogonia failed to enter prenatal meiosis I
ⓒ. Puberty automatically converts all remaining oogonia into mature ova without meiosis
ⓓ. Graafian follicles form normally because follicular development does not require primary oocytes
Correct Answer: The primary-oocyte reserve is compromised because oogonia failed to enter prenatal meiosis I
Explanation: Normal oogenesis begins during fetal life. Oogonia enter meiosis I and become primary oocytes, which then remain arrested in prophase I for a prolonged period. The experiment allows oogonia to form but blocks precisely the transition that normally establishes the primary-oocyte population. Since the human pattern also states that no new oogonia are added after birth, the missing prenatal transition cannot simply be replaced later by continuous postnatal production of new germ cells. The expected consequence is a severely compromised primary-oocyte reserve. Puberty can act on an existing ovarian follicular pool, but it does not recreate the normal fetal germ-cell history or bypass the meiotic requirements of oogenesis. No new oogonia are added after birth in the stated human pattern, making the prenatal germ-cell pool the source of later primary oocytes.
509. A pubertal ovarian sample contains many cells identified as primary oocytes within follicles, but a second report claims that these cells were newly generated from oogonia after birth. Which evaluation is most appropriate?
ⓐ. Primary-oocyte identification may be correct, but postnatal origin from oogonia conflicts with normal oogenesis
ⓑ. Both the primary-oocyte identification and a postnatal origin from newly generated oogonia are necessarily correct in normal human oogenesis
ⓒ. Primary oocytes exist only after fertilisation, so both reports are wrong
ⓓ. Primary oocytes are formed from secondary oocytes during ovulation
Correct Answer: Primary-oocyte identification may be correct, but postnatal origin from oogonia conflicts with normal oogenesis
Explanation: Primary oocytes are expected to be present in the ovaries at puberty because they were established much earlier during fetal development. Oogonia formed prenatally entered meiosis I and became primary oocytes that remained arrested in prophase I. The normal human pattern is that new oogonia are not added after birth. Observing primary oocytes in a pubertal ovary is entirely consistent, but explaining them as products of newly generated postnatal oogonia conflicts with the developmental sequence. The reasoning separates the identity of the observed cell from the proposed developmental origin of that cell. A correct observation can still be paired with an incorrect developmental explanation if the timing of germ-cell formation is misunderstood.
510. An ovarian follicle is labelled “secondary follicle,” but microscopy shows a distinct fluid-filled antrum and the surrounding theca differentiated into interna and externa. Which interpretation is most appropriate?
ⓐ. The label is correct because every secondary follicle must contain a large antrum
ⓑ. The features fit a tertiary follicle better than a secondary follicle
ⓒ. The structure must be a primary follicle because theca is present
ⓓ. The presence of an antrum proves that ovulation has already occurred
Correct Answer: The features fit a tertiary follicle better than a secondary follicle
Explanation: Secondary follicles develop additional granulosa-cell layers and surrounding thecal tissue, but appearance of the fluid-filled antrum is a defining feature of progression to the tertiary follicle. At this stage, the theca also differentiates into theca interna and theca externa. The combination of a clear antrum with differentiated thecal layers places the observed follicle beyond the ordinary secondary stage. Ovulation occurs later, after further maturation to the Graafian follicle and rupture of that mature follicle. The observation does not indicate that ovulation has already happened. The correct evaluation uses multiple structural markers together and recognises that a stage label should be rejected when its listed microscopic features correspond to a later developmental stage.
511. Meiosis I of a primary oocyte produces a secondary oocyte and a first polar body. If chromosome segregation is normal, which comparison of the two immediate products is correct?
ⓐ. They have different chromosome numbers because their cytoplasmic volumes differ
ⓑ. The polar body remains diploid while the secondary oocyte becomes haploid
ⓒ. Both contain \(23\) chromosomes, although their cytoplasmic volumes are highly unequal
ⓓ. The secondary oocyte contains \(46\) chromosomes because it receives most of the cytoplasm
Correct Answer: Both contain \(23\) chromosomes, although their cytoplasmic volumes are highly unequal
Explanation: Meiosis I is a reductional division, so both immediate products receive a haploid chromosome complement of \(23\) chromosomes when segregation is normal. The pronounced inequality in oogenesis concerns cytoplasmic distribution, not chromosome number. Most cytoplasm is retained by the secondary oocyte, while the first polar body receives very little. This asymmetry allows one major female gametic cell to preserve a large cytoplasmic reserve while still completing meiotic chromosome reduction. The larger size of the secondary oocyte does not make it diploid, and the small size of the polar body does not imply that it contains fewer than \(23\) chromosomes. Ploidy and cytoplasmic volume are separate properties that must be interpreted independently.
512. A tertiary follicle contains an antrum and a primary oocyte. The oocyte completes meiosis I, and the follicle then undergoes further normal maturation. Which combination is expected just before ovulation?
ⓐ. A diploid ovum without a zona pellucida inside a fully formed corpus luteum
ⓑ. A Graafian follicle containing a zona-associated secondary oocyte
ⓒ. A primary follicle containing four equivalent haploid ova
ⓓ. An implanted blastocyst surrounded by theca interna
Correct Answer: A Graafian follicle containing a zona-associated secondary oocyte
Explanation: The tertiary follicle is characterised by its antrum and differentiated surrounding tissues. During further maturation, the primary oocyte completes meiosis I and produces the large haploid secondary oocyte together with the first polar body. Continued follicular development produces the mature Graafian follicle. The secondary oocyte is associated with the zona pellucida and is the cell released when the Graafian follicle ruptures during ovulation. It has not yet completed meiosis II under the normal sequence; that completion is triggered after fertilisation is initiated. The corpus luteum forms from the ruptured follicle after ovulation. The sequence coordinates follicular maturation, meiotic stage and oocyte-associated structure immediately before release.
513. Following normal meiosis I in a primary oocyte, let \(C_s\) be the chromosome number of the secondary oocyte and \(C_p\) the chromosome number of the first polar body. Which relation is correct even though the two cells differ greatly in size?
ⓐ. \(C_s:C_p=2:1\)
ⓑ. \(C_s:C_p=1:2\)
ⓒ. \(C_s+C_p=23\)
ⓓ. \(C_s:C_p=1:1\)
Correct Answer: \(C_s:C_p=1:1\)
Explanation: Meiosis I in a primary oocyte separates homologous chromosome sets. Both the secondary oocyte and the first polar body are therefore haploid, with \(23\) chromosomes each. If \(C_s\) denotes the chromosome number of the secondary oocyte and \(C_p\) that of the first polar body, then \(C_s:C_p=23:23=1:1\). Their very different sizes do not alter this chromosome-number relation. Oogenesis uses highly unequal cytokinesis, so most cytoplasm is retained in the secondary oocyte and only a small amount enters the polar body. Cytoplasmic volume and chromosome number are separate properties: unequal partitioning of cytoplasm can occur while chromosome segregation gives each product the same haploid count. The symbolic ratio represents chromosome distribution, not cell size or functional importance. A \(2:1\) or \(1:2\) chromosome-number ratio would incorrectly treat cytoplasmic asymmetry as though it changed how homologous chromosomes were separated during normal meiosis I.
514. Which sequence correctly follows the secondary oocyte from its release through the earliest fertilisation-related events involving the zona pellucida?
ⓐ. Graafian-follicle rupture releases the zona-associated secondary oocyte → sperm interacts with the zona pellucida → fertilisation-related entry events can follow
ⓑ. Sperm interacts with the zona pellucida → graafian-follicle rupture releases the zona-associated secondary oocyte → fertilisation-related entry events can follow
ⓒ. Graafian-follicle rupture releases the zona-associated secondary oocyte → fertilisation-related entry events can follow → sperm interacts with the zona pellucida
ⓓ. Fertilisation-related entry events can follow → sperm interacts with the zona pellucida → graafian-follicle rupture releases the zona-associated secondary oocyte
Correct Answer: Graafian-follicle rupture releases the zona-associated secondary oocyte → sperm interacts with the zona pellucida → fertilisation-related entry events can follow
Explanation: The mature Graafian follicle ruptures during ovulation and releases the secondary oocyte, which remains associated with the zona pellucida. If viable sperm reach the appropriate oviductal region at the correct time, sperm can interact with this zona during fertilisation. That interaction participates in the sequence leading to successful sperm entry and the subsequent block to additional sperm entry. The zona is not simply discarded at ovulation, nor does it arise only after zygote formation. Meiosis I has already been completed before the secondary oocyte is released. The correct order links follicular rupture, oocyte-associated structure and fertilisation without confusing ovulation itself with sperm entry or nuclear fusion.
515. Four primary spermatocytes and four primary oocytes are each followed through their normal complete gametogenic outcomes, with fertilisation available to permit completion of female meiosis II. Considering only the major functional gametes produced from these primary meiocytes, what is the male-to-female functional-gamete ratio?
ⓐ. \(1:1\)
ⓑ. \(2:1\)
ⓒ. \(4:1\)
ⓓ. \(8:1\)
Correct Answer: \(4:1\)
Explanation: Each primary spermatocyte completes meiosis I and meiosis II to produce four spermatids, which differentiate through spermiogenesis into four spermatozoa. Four primary spermatocytes therefore yield \(4\times4=16\) functional male gametes. Oogenesis follows a different product pattern. One primary oocyte retains most of the cytoplasm in a single principal female gametic lineage, while polar bodies are small meiotic products rather than equivalent functional ova. With fertilisation available to permit completion of meiosis II, four primary oocytes yield four major female gametes. The male-to-female functional-gamete ratio is consequently \(16:4=4:1\). This ratio does not arise from a difference in haploid chromosome number; both sperm and ova are haploid. It arises from the contrast in meiotic cytokinesis and functional-product number: spermatogenesis produces four equivalent gametic products from each primary spermatocyte, whereas oogenesis preserves one major functional gamete from each primary oocyte.
516. A menstrual-cycle graph has cycle day on the x-axis and relative endometrial thickness on the y-axis. The endometrium is similarly thick on day \(13\) and day \(21\). At day \(13\), a mature follicle and high estrogen-associated activity are present; at day \(21\), a corpus luteum and high progesterone are present. Which interpretation best explains the equal thickness?
ⓐ. Equal endometrial thickness proves that both observations belong to the same ovarian phase and are supported by the same hormone
ⓑ. Day \(13\) represents luteal maintenance, whereas day \(21\) represents follicular proliferation
ⓒ. Equal thickness can reflect follicular proliferation on day \(13\) and progesterone-supported maintenance on day \(21\)
ⓓ. Endometrial thickness is unrelated to ovarian structures or ovarian hormones
Correct Answer: Equal thickness can reflect follicular proliferation on day \(13\) and progesterone-supported maintenance on day \(21\)
Explanation: Endometrial thickness alone does not uniquely identify a menstrual-cycle phase. During the follicular phase, developing ovarian follicles are associated with increasing estrogen secretion, while the endometrium regenerates and proliferates after menstrual shedding. Around day \(13\) of a typical cycle, a mature follicle together with high estrogen-associated activity fits a proliferative pre-ovulatory state. By day \(21\), ovulation has already occurred and the ruptured follicle has formed a corpus luteum. Progesterone from the corpus luteum supports maintenance of the developed endometrium during the luteal phase. The graph can show similar endometrial thickness at two times even though the ovarian structure, dominant hormonal relation and physiological meaning of that thickness differ. Phase identification requires integrating uterine, ovarian and hormonal evidence. Ovarian and uterine events are coordinated, yet the follicle, corpus luteum and endometrium respond at different points in the cycle.
517. Fertilisation does not occur, the corpus luteum degenerates and progesterone concentration falls normally. The endometrium and its blood vessels are experimentally prevented from breaking down. Which outcome is most directly expected?
ⓐ. Menstrual flow is reduced or absent despite the usual ovarian-hormonal sequence
ⓑ. Ovulation must immediately recur while progesterone remains high and the endometrium cannot break down
ⓒ. A diploid zygote forms even though fertilisation did not occur
ⓓ. The corpus luteum is restored by the intact endometrium
Correct Answer: Menstrual flow is reduced or absent despite the usual ovarian-hormonal sequence
Explanation: In an unfertilised cycle, degeneration of the corpus luteum lowers progesterone support for the endometrium. Under normal conditions, this loss of support is followed by breakdown of the endometrial lining and its blood vessels, producing menstrual flow. The experiment preserves the upstream ovarian sequence but blocks the uterine tissue response that normally generates the visible menstrual phase. A fall in progesterone can occur without producing the usual amount of menstrual shedding if the endometrium is artificially prevented from breaking down. This separates the hormonal trigger from the responding uterine tissue. Menstruation is not simply another name for corpus-luteum degeneration; it is the downstream uterine consequence of losing luteal support in a non-pregnant cycle. Preventing the tissue breakdown interrupts that final step while earlier ovarian events can still occur.
518. Follicular development is experimentally suppressed so severely that no mature Graafian follicle is present near mid-cycle. An artificial LH pulse identical in magnitude to a normal LH surge is then produced. Which prediction is most appropriate?
ⓐ. A secondary oocyte must be released because an LH surge alone is sufficient regardless of follicular maturity
ⓑ. The artificial LH pulse directly produces a new mature Graafian follicle within the uterus
ⓒ. Ovulation occurs from a primary follicle because any ovarian follicle can respond identically to an LH surge
ⓓ. Normal ovulation is unlikely because the LH surge requires a mature Graafian follicle capable of rupture
Correct Answer: Normal ovulation is unlikely because the LH surge requires a mature Graafian follicle capable of rupture
Explanation: The LH surge is the decisive hormonal event that induces rupture of a mature Graafian follicle during normal ovulation. Its action, however, occurs within a developmental context: the follicle must first have progressed through earlier stages and reached appropriate maturity. In the experiment, the hormonal signal is recreated but the mature follicular structure on which that signal normally acts is absent. The artificial LH peak cannot by itself substitute for the missing developmental history of the follicle. Primary or incompletely developed follicles are not equivalent to a mature Graafian follicle simply because LH concentration rises. This changed-condition experiment demonstrates that ovulation depends on both a competent ovarian structure and the appropriate mid-cycle hormonal signal. A normal signal cannot guarantee the downstream event when its required biological substrate has not formed.
519. An ovarian section taken after the middle of a menstrual cycle shows a well-developed corpus luteum but no intact mature Graafian follicle. No information about menstrual bleeding is provided. Which inference is best supported?
ⓐ. Fertilisation must already have occurred
ⓑ. Ovulation has occurred earlier in the current cycle
ⓒ. The primary oocyte is still arrested in a primary follicle
ⓓ. Menopause has necessarily begun
Correct Answer: Ovulation has occurred earlier in the current cycle
Explanation: The corpus luteum develops from the follicular structure that remains after rupture of the mature Graafian follicle during ovulation. Observing a corpus luteum provides retrospective evidence that the follicle has already undergone the ovulatory transition in that cycle. The absence of an intact mature Graafian follicle is consistent with this interpretation, since that follicle has already ruptured and changed into a post-ovulatory structure. The observation does not establish that fertilisation occurred; ovulation can occur in cycles in which the released secondary oocyte is never fertilised. It also does not indicate menopause, which is a reproductive-life boundary involving cessation of menstrual cycles. The ovarian structure itself permits inference about a preceding local event, but it does not by itself prove subsequent fertilisation, implantation or pregnancy.
520. Ovulation occurs normally, but the newly formed corpus luteum is experimentally removed. Progesterone is then supplied so that its circulating level remains similar to that of a normal luteal phase, and the endometrium remains maintained. What does this experiment most strongly support?
ⓐ. Progesterone normally triggers formation of the mature Graafian follicle after ovulation
ⓑ. The corpus luteum maintains the endometrium only through direct physical contact with the uterus
ⓒ. Progesterone can mediate corpus-luteum support of the luteal endometrium
ⓓ. Endometrial maintenance is independent of ovarian hormonal influence
Correct Answer: Progesterone can mediate corpus-luteum support of the luteal endometrium
Explanation: After ovulation, the ruptured follicle forms the corpus luteum, which becomes an important source of progesterone. Progesterone supports maintenance of the developed endometrium during the luteal phase. The experiment removes the corpus luteum but replaces one of its major hormonal outputs, and the uterine lining remains maintained. This result separates the physical presence of the ovarian structure from the hormonal signal through which an important part of its uterine effect is produced. The observation supports progesterone as a functional mediator between corpus-luteum activity and endometrial maintenance. It does not show that every corpus-luteum function has been replaced, nor does progesterone recreate the earlier Graafian follicle. The evidence specifically concerns the hormone-dependent maintenance relation tested by the experimental replacement. Loss of luteal progesterone support promotes endometrial breakdown in an unfertilised cycle, linking ovarian change to the next menstruation.
