Developmental Biology: Gametes, Embryos, Differentiation, Metamorphosis and Stem Cells

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

1. In most mammals, including humans, the oocyte released at ovulation is arrested at which stage?

   1. Diplotene of prophase I, the stage at which it has waited since fetal life
   2. Metaphase of the second meiotic division, finished only if a sperm enters
   3. Metaphase of the first meiotic division, with no polar body yet formed
   4. G2 phase after the last oogonial mitosis, before meiosis has begun
   Show answer

   Answer: B — Metaphase of the second meiotic division, finished only if a sperm enters

   The LH surge releases the diplotene arrest, so meiosis I is completed and the first polar body is extruded before ovulation; the secondary oocyte then arrests at metaphase II and completes meiosis II, giving the second polar body, only after a sperm enters. Diplotene is the tempting choice, but that arrest belongs to the primary oocyte inside the follicle before the LH surge, not to the ovulated cell.

2. A sea urchin egg is treated with a drug that blocks its IP₃ receptors and is then inseminated. Which event of fertilisation would fail?

   1. Exocytosis of the sperm acrosomal vesicle on contact with the egg jelly
   2. Binding of the sperm protein bindin to its receptor on the egg envelope
   3. Fusion of the sperm plasma membrane with the egg plasma membrane
   4. Exocytosis of cortical granules and lifting of the fertilisation envelope
   Show answer

   Answer: D — Exocytosis of cortical granules and lifting of the fertilisation envelope

   After fusion, IP₃ opens ER Ca²⁺ channels and the resulting Ca²⁺ wave drives cortical granule exocytosis, which raises the fertilisation envelope (the slow block). With the egg's IP₃ receptors blocked, this step fails. The acrosome reaction is the tempting distractor because it is also a Ca²⁺-triggered exocytosis, but it happens in the sperm, before fusion, and does not use the egg's receptors; bindin binding and membrane fusion also precede the IP₃ step.

3. In a mammal, 250 primary spermatocytes and 120 secondary spermatocytes in a testis, and 80 primary oocytes and 40 secondary oocytes in an ovary, all complete meiosis normally (assume each secondary oocyte is fertilised so that meiosis II is completed). The total number of functional gametes (sperm plus eggs) formed is ____.

   Numerical answer — type the value.

   Show answer

   Answer: 1360

   Each primary spermatocyte gives 4 sperm: 250 × 4 = 1000. Each secondary spermatocyte has only meiosis II left and gives 2: 120 × 2 = 240. Each oocyte, primary or secondary, gives just one egg, the other products being polar bodies: 80 + 40 = 120. Total = 1000 + 240 + 120 = 1360. Multiplying the oocytes by 4, or the secondary spermatocytes by 4, are the usual slips.

4. A spermatid of a diploid mammal contains 3.2 pg of nuclear DNA. The nuclear DNA content, in pg, of a primary spermatocyte of the same animal in prophase I is ____ (to one decimal place).

   Numerical answer — type the value.

   Show answer

   Answer: 12.8

   A spermatid is haploid with unreplicated chromosomes, so it is 1C = 3.2 pg. A primary spermatocyte is diploid and has completed S phase before entering meiosis, so in prophase I it is 4C = 4 × 3.2 = 12.8 pg. Answering 6.4 (2C) forgets the premeiotic S phase; 6.4 pg is the content of a secondary spermatocyte.

5. A researcher wants to follow every somatic cell division from zygote to adult and to know in advance exactly which cells will undergo programmed death. Which model organism suits this best?

   1. Drosophila melanogaster, because balancer chromosomes let lethal mutations be kept
   2. Zebrafish, because its transparent embryos develop outside the mother’s body
   3. Caenorhabditis elegans, because its complete somatic cell lineage is invariant
   4. Xenopus laevis, because its large eggs are easy to inject and to dissect
   Show answer

   Answer: C — *Caenorhabditis elegans*, because its complete somatic cell lineage is invariant

   In C. elegans the same divisions occur in every individual, the whole somatic lineage has been traced, and the 131 cells that die in hermaphrodite development are known by name, which is how the ced cell-death genes were found. Zebrafish is the tempting answer because its embryos are transparent, but its cell lineages vary between individuals, so cell fates cannot be predicted cell by cell. Balancers and injectable eggs are real strengths of the other two, but not for this purpose.

6. In an embryo whose early cleavages are synchronous, the first cleavage is completed 90 min after fertilisation and every later cleavage takes 30 min. Assuming all divisions stay synchronous and no cell dies, the number of blastomeres 250 min after fertilisation is ____.

   Numerical answer — type the value.

   Show answer

   Answer: 64

   Cleavages are completed at 90, 120, 150, 180, 210 and 240 min; the next would finish at 270 min, after the time asked. That is 6 cleavages, and n synchronous cleavages give 2ⁿ cells: 2⁶ = 64. A common slip is to divide 250 by 30 and forget that the first cleavage takes 90 min, or to count the 2-cell stage as cleavage zero.

7. Which of the following statements about germ layers and their derivatives in vertebrates are correct?

   1. Melanocytes of the skin and chromaffin cells of the adrenal medulla both arise from the neural crest
   2. The notochord is mesodermal and induces the overlying ectoderm to form the neural plate
   3. In deuterostomes such as sea urchins and vertebrates, the blastopore becomes the mouth
   4. Liver hepatocytes and the lining epithelium of the lungs are derived from mesoderm
   Show answer

   Answer: A — Melanocytes of the skin and chromaffin cells of the adrenal medulla both arise from the neural crest; B — The notochord is mesodermal and induces the overlying ectoderm to form the neural plate

   Neural crest cells migrate to form pigment cells, the adrenal medulla, peripheral neurons and glia, and the notochord (chordamesoderm) induces neural plate formation. In deuterostomes the blastopore becomes the anus and the mouth forms secondarily; mouth-first is the protostome pattern. Hepatocytes and lung epithelium are endodermal outgrowths of the gut tube; only their connective tissue and vessels are mesodermal.

8. A fertilised Xenopus egg is injected with α-amanitin at a dose that blocks RNA polymerase II. What is the expected outcome?

   1. Cleavage proceeds normally to about the blastula stage, then the embryo fails to gastrulate
   2. The first cleavage fails, because each division needs freshly transcribed cyclin mRNA
   3. Development continues normally to a tadpole, since maternal proteins control all early stages
   4. Only animal-pole cells cleave, since vegetal yolky cells need zygotic transcripts to divide
   Show answer

   Answer: A — Cleavage proceeds normally to about the blastula stage, then the embryo fails to gastrulate

   Before the midblastula transition (about 12 cycles) Xenopus cleavage runs on maternal mRNAs and proteins stored in the egg, including cyclin mRNA, so blocking transcription does not stop cleavage. Zygotic transcription becomes essential at the midblastula transition, so the embryo arrests before gastrulation. The tempting option about cyclin is wrong because cyclin is translated from stored maternal mRNA; development cannot reach a tadpole because gastrulation needs zygotic genes.

9. In a tunicate embryo, removing the blastomere pair that inherits the yellow-crescent cytoplasm gives a larva without tail muscle, and the same blastomeres cultured in isolation still make muscle. These results show that tail-muscle fate is specified by

   1. conditional specification, in which neighbouring cells induce muscle after gastrulation
   2. syncytial specification, in which shared nuclei read a gradient of transcription factors
   3. autonomous specification by maternal determinants that segregate into those cells
   4. regulative specification, in which the remaining cells replace any lost blastomeres
   Show answer

   Answer: C — autonomous specification by maternal determinants that segregate into those cells

   A cell that makes its normal product when isolated, and whose removal leaves a gap that others do not fill, is autonomously (mosaic) specified; in tunicates the determinant is maternal macho-1 mRNA in the yellow crescent. Conditional specification is the tempting alternative, but it would predict that isolated blastomeres fail to make muscle without their neighbours. Regulative replacement is ruled out because the larva stays without tail muscle.

10. A bead soaked in Sonic hedgehog protein is implanted beneath the apical ectodermal ridge at the anterior margin of an early chick wing bud. What is the expected result?

    1. A truncated wing missing its distal parts, because the ridge signal has been disrupted
    2. A mirror-image duplication of the digits along the anterior–posterior axis
    3. A wing with ventral structures on both its dorsal and ventral surfaces
    4. A normal wing, because the posterior ZPA keeps control of digit identity
    Show answer

    Answer: B — A mirror-image duplication of the digits along the anterior–posterior axis

    Sonic hedgehog is the signal of the zone of polarising activity. A second anterior source creates a second posterior-to-anterior gradient, so digits form in mirror image (for example 4-3-2-2-3-4 instead of 2-3-4). Truncation is the tempting distractor, but that follows removal of the AER (loss of FGF), not added Shh; dorsal–ventral identity depends on Wnt7a from the dorsal ectoderm.

11. The corpora allata are surgically removed from an early-instar larva of a holometabolous insect, leaving its brain and prothoracic glands intact. At its next moult the larva is expected to

    1. fail to moult at all, because juvenile hormone is itself the moulting hormone
    2. undergo extra larval moults and grow into a giant larva that never pupates
    3. moult into a larva again, since ecdysone alone is enough to keep larval identity
    4. moult early into a pupa and then emerge as a miniature but complete adult
    Show answer

    Answer: D — moult early into a pupa and then emerge as a miniature but complete adult

    The prothoracic glands still make ecdysone, so moulting continues, but juvenile hormone from the corpora allata is what keeps a moult larval. Without JH the next ecdysteroid pulse gives a pupal moult, and the small insect metamorphoses into a miniature adult. The giant-larva outcome is the tempting distractor, but it is what happens when active corpora allata are added to a last-instar larva, the opposite manipulation.

12. Which of the following statements about anuran (frog) metamorphosis are correct?

    1. T₃, formed from T₄ by deiodinases in target tissues, acts through nuclear thyroid hormone receptors
    2. Resorption of the tail involves apoptosis of tail cells and enzymatic breakdown of matrix
    3. The liver induces urea-cycle enzymes, shifting nitrogen excretion from ammonia towards urea
    4. Tadpoles reared in thiourea metamorphose earlier than controls and become smaller frogs
    Show answer

    Answer: A — T₃, formed from T₄ by deiodinases in target tissues, acts through nuclear thyroid hormone receptors; B — Resorption of the tail involves apoptosis of tail cells and enzymatic breakdown of matrix; C — The liver induces urea-cycle enzymes, shifting nitrogen excretion from ammonia towards urea

    T₃ is the active hormone and regulates transcription through nuclear receptors; it causes tail cells to die by apoptosis while matrix metalloproteinases remove the matrix, and it induces urea-cycle enzymes as the animal moves to land. Thiourea is a goitrogen that blocks thyroid hormone synthesis, so treated tadpoles keep growing as tadpoles and fail to metamorphose; they do not metamorphose early.

13. A Drosophila female homozygous for a loss-of-function bicoid allele is mated to a wild-type male. What will her embryos look like?

    1. Normal, because every embryo receives a wild-type bicoid allele from its father
    2. Missing every alternate segment, because bicoid is expressed in seven stripes
    3. Lacking head and thorax, with posterior terminal structures at both ends
    4. Two-headed, because the posterior Nanos protein now spreads uniformly
    Show answer

    Answer: C — Lacking head and thorax, with posterior terminal structures at both ends

    bicoid is a maternal-effect gene: its mRNA is made by the mother and placed at the anterior pole of the egg before fertilisation, so the embryo's phenotype follows the mother's genotype. With no Bicoid gradient, anterior structures fail and posterior terminal structures form at both ends. The tempting answer, a normal embryo rescued by the paternal allele, ignores that zygotic transcription starts too late to make the gradient. Seven stripes describe pair-rule genes, and Nanos stays posterior.

14. A Drosophila larva has denticle belts for only half the normal number of segments, with every other segment-width region missing. The mutation most likely lies in

    1. fushi tarazu, a pair-rule gene expressed in seven stripes
    2. Krüppel, a gap gene expressed in one broad central band
    3. engrailed, a segment-polarity gene expressed in 14 stripes
    4. Ultrabithorax, a homeotic gene of the bithorax complex
    Show answer

    Answer: A — *fushi tarazu*, a pair-rule gene expressed in seven stripes

    Pair-rule genes are expressed in seven stripes, one per two segments, so loss of one removes alternate segment-width regions and halves the segment count; fushi tarazu means “too few segments” in Japanese. A gap mutant such as Krüppel loses a block of adjacent segments; a segment-polarity mutant such as engrailed keeps the segment number but replaces part of each with a mirror image; a homeotic mutant keeps the number but changes segment identity.

15. Which of the following statements about the amphibian (Spemann) organizer are correct?

    1. It secretes chordin and noggin, which bind BMP and allow nearby ectoderm to become neural
    2. A grafted dorsal lip induces a second axis whose neural tube is built mostly from host cells
    3. Neural plate forms where BMP signalling in the ectoderm is strongest
    4. β-catenin is degraded on the future dorsal side after cortical rotation, marking that side as the organizer
    Show answer

    Answer: A — It secretes chordin and noggin, which bind BMP and allow nearby ectoderm to become neural; B — A grafted dorsal lip induces a second axis whose neural tube is built mostly from host cells

    The organizer works by secreting BMP antagonists (chordin, noggin, follistatin); ectoderm shielded from BMP becomes neural. In the Spemann–Mangold graft the pigment difference showed that the secondary neural tube came largely from host tissue, proving induction. BMP signalling instructs epidermis, so neural plate forms where BMP is lowest. Cortical rotation stabilises, not degrades, β-catenin on the dorsal side, which sets up the Nieuwkoop centre and then the organizer.

16. Which set of transcription factors did Takahashi and Yamanaka use in 2006 to reprogram mouse fibroblasts into induced pluripotent stem cells?

    1. Oct3/4, Nanog, Pax6 and MyoD
    2. Sox2, Gata1, Hnf4 and c-Myc
    3. Pdx1, Ngn3, MafA and Klf4
    4. Oct3/4, Sox2, Klf4 and c-Myc
    Show answer

    Answer: D — Oct3/4, Sox2, Klf4 and c-Myc

    The 2006 mouse study screened 24 candidate genes and found that four, Oct3/4, Sox2, Klf4 and c-Myc, were enough to give pluripotent cells. Each distractor mixes a pluripotency factor with lineage-specifying factors (Pax6 for eye and neural, MyoD for muscle, Gata1 for erythroid, Hnf4 for liver, Pdx1, Ngn3 and MafA for pancreatic β-cells), which drive cells towards particular differentiated fates rather than back to pluripotency.

17. Mouse embryonic stem cells injected into a host blastocyst contribute to every tissue of the resulting chimera, including its germ line, but essentially not to the trophoblast-derived parts of the placenta. The best description of these cells is

    1. totipotent, because they can make every adult cell type including germ cells
    2. pluripotent, because they form all embryonic lineages but not trophoblast
    3. multipotent, because they are limited to lineages of a single germ layer
    4. unipotent, because in the chimera they behave like germline stem cells
    Show answer

    Answer: B — pluripotent, because they form all embryonic lineages but not trophoblast

    Contributing to all three germ layers and the germ line, but not to trophoblast, is the definition of pluripotency, matching the inner cell mass from which ES cells come. Totipotent is the tempting answer because every adult tissue is formed, but totipotency also requires the extraembryonic trophoblast, which only the zygote and early blastomeres can make. Contribution to tissues of all germ layers rules out multipotent and unipotent.

18. An asynchronously dividing population of embryonic cells has a cell-cycle time of 24 h. Of 2,000 cells scored, 150 are in mitosis. Assuming that the fraction of cells in a phase equals the fraction of the cycle spent in it, the duration of M phase in minutes is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 108

    Mitotic index = 150 ÷ 2000 = 0.075. Duration of M = 0.075 × 24 h = 1.8 h = 1.8 × 60 = 108 min. Forgetting to convert hours to minutes gives 1.8, the usual slip.