One purple plant, two possible genotypes, and a deceptively simple fraction
A greenhouse student chooses a purple-flowered plant from the offspring of two purple heterozygous parents. The plant might be PP or Pp in the simplified inheritance model. The student has heard that a Pp × Pp cross produces the genotype ratio 1:2:1 and promptly says that the chosen plant has a one-half chance of being Pp. That answer uses the wrong group. We already know the selected plant is purple; the white pp outcome is no longer one of the possibilities. Among the three equally likely purple genotype outcomes, two are Pp. The correct probability under the stated model is two-thirds.
This guide teaches the mathematics behind that distinction. You will draw possibility and tree diagrams, count equally likely outcomes, apply the multiplication and addition rules, distinguish an expected ratio from a short observed run, and explain why four purple offspring in a test cross do not prove that an unknown parent is PP. Every example uses fictional plants and explicitly assumed genetic relationships. The aim is a reusable SEC G3 probability method, not to predict a particular examination question.
Quick route: Identify the sample space; list genotypes and their colours; remove outcomes excluded by the given condition; count favourable outcomes over possible outcomes; use a branching diagram for successive offspring; multiply within a path and add mutually exclusive paths; take the complement for ‘at least one’; check the independence assumption; distinguish theoretical probability from recorded data; explain limits in one sentence.
Where this sits in the official 2027 SEC G3 Mathematics syllabus
The Singapore-Cambridge Secondary Education Certificate (SEC) begins in 2027. This article works within G3 Mathematics syllabus K310, the new G3 code corresponding to the earlier O-Level Mathematics syllabus 4052. The official syllabus covers calculating the probability of single and combined events, using possibility diagrams and tree diagrams, identifying mutually exclusive events, and applying addition or multiplication rules in appropriate situations. We use those core methods in a new, connected setting rather than adding a separate genetics requirement to Mathematics.
K310 Paper 1 carries 90 marks, lasts two hours and fifteen minutes, and accounts for half the final mark. Paper 2 also carries 90 marks, lasts two hours and fifteen minutes, and accounts for the other half; it contains longer problems and a final real-world-context question. There is no separately assessed G3 Mathematics ‘greenhouse probability paper’. The figures in this article are teaching examples and do not imply that the 2027 examination must include biological inheritance.
Read the official 2027 SEAB G3 Mathematics K310 syllabus and the current G3 syllabus directory for the authoritative format. For the scientific assumptions and diagrams behind this example, first consult Learner’s Guide 1178, Biology Greenhouse Lineage.
The model is a rule, not a photograph of nature
Let P be a dominant allele giving purple flowers, and p be a recessive allele giving white flowers. In a simplified complete-dominance model, PP and Pp plants are purple and pp plants are white. Two parents with genotype Pp can each pass P or p to an offspring, with equal probability under the model. Their four equally likely genotype combinations are PP, Pp, Pp and pp.
Probability questions demand that we say what is random and what is fixed. The parental genotypes are stipulated; the allele each parent passes is random within the model; the observed flower colour may become known information that restricts the outcomes being considered. Real plants may involve more genes, environmental conditions, recording errors or selection. Do not smuggle those complications into a classroom model unless the task asks for them, but do not forget the model has assumptions.
A probability is a number between zero and one. Zero means impossible under the model; one means certain under the model. Two-thirds means that in a large collection of comparable cases, about two of every three qualifying possibilities are of the named kind. It never means two-thirds of one plant’s genotype is Pp. The plant has one actual genotype; we are measuring uncertainty about that genotype before it is known.
Rebuild the unconditional sample space from first principles
Draw a two-by-two diagram with the first parent’s gametes P and p across the top and the second parent’s P and p down the side. In the four cells place PP, Pp, Pp and pp. Each cell represents one equally likely combination of parental gametes. Count one PP, two Pp and one pp. This is the source of the 1:2:1 genotype ratio.
Now group outcomes by the displayed flower colour. PP and both Pp cells give purple. Only pp gives white. The chance of purple is therefore three quarters and the chance of white is one quarter. State the denominator: four equally likely gamete combinations. If you write 3:1 without defining whether it is genotype or phenotype, your answer risks becoming wrong even though one familiar number appears in it.
Suppose instead that a parent is PP and the other is pp. Their offspring are all Pp. Every cell contains the same genotype and the purple probability is one. In a PP × Pp cross, half the offspring are PP and half Pp, but all are purple. The phrase ‘two purple parents’ alone cannot tell you which diagram to use. The biological premise determines the mathematical sample space.
The information changes: a chosen flower is already purple
In the original four-cell Pp × Pp diagram, the Pp genotype occupies two of four cells. Thus an offspring selected without knowing its colour has probability 2/4, or one-half, of being Pp. The new question says that a purple offspring was selected. The white pp cell is now excluded. Three equally likely cells remain: PP, Pp, Pp. Two contain Pp. The required chance is 2/3.
This is an example of conditional probability, explained through an appropriately reduced sample space. In optional notation we could write P(Pp given purple) = P(Pp and purple)/P(purple) = (1/2)/(3/4) = 2/3. The equation is helpful for more advanced readers, but a labelled diagram and a clear ‘two out of three purple outcomes’ explanation are sufficient to communicate the principle. Do not claim that memorising formal conditional-probability notation is a separate prescribed K310 learning outcome.
The same change gives a one-third chance that the selected purple plant is PP. The two possibilities sum to one, as they must: 2/3 + 1/3 = 1. That final check is simple and powerful. If your reported probabilities for mutually exclusive and exhaustive possibilities do not total one, either the counts or the denominator are wrong.
Why you cannot reduce the denominator without changing the question
Many wrong answers take the original four cells and remove the unwanted white cell from the numerator but not from the denominator. A pupil may write ‘two Pp cells out of four, so 1/2’, even after learning that the chosen flower is purple. Another may count three purple outcomes and claim a 3/4 chance that the selected purple flower is purple. Both confuse the event that generated the condition with the event being asked about.
Write the question in words before calculating: ‘Among the purple offspring, what fraction are expected to be heterozygous?’ The phrase ‘among the purple’ identifies the denominator. In a medical-style or family example this idea would become sensitive and more complex; our fictional greenhouse confines it to an explicit classroom sample space, avoiding predictions about real people’s genotypes.
If a purple offspring was deliberately selected using other information, the simple 2/3 figure may no longer follow. Perhaps a previous test already ruled out one genotype. The question must tell you how the selection was made or clearly imply random selection among all purple outcomes. A good student notices this before using a formula mechanically.
The test cross: what the first branch means
The Biology case now crosses an unknown purple plant U with a white pp tester. If U is PP, every offspring is Pp and purple under the model. If U is Pp, each offspring has probability one-half of Pp purple and one-half of pp white. The diagram for the second hypothesis has two branches at each new offspring: purple and white, each carrying probability 1/2.
The expected result ‘one purple to one white’ does not require the first two offspring to be one of each. It means the long-run chance for each offspring is one-half under the stated independent-trial model. Students often interpret a 1:1 ratio as an alternating colour pattern. That is not how independent outcomes work; two purple in a row are possible, as are two white in a row.
The distinction between a guaranteed outcome and a possible observed run is the central learning objective of this volume. If you are asked to infer the genotype after seeing four purple seedlings, the correct solution must consider what that run would look like under both possible parental genotypes.
Two offspring: a complete tree diagram
Start at one point. Draw a first branch labelled purple, probability 1/2, and another labelled white, probability 1/2. From the end of each first branch draw a second purple and white branch, each again 1/2. There are four equally likely ordered paths: purple–purple, purple–white, white–purple and white–white. Each path has probability (1/2)(1/2) = 1/4, assuming independence.
The probability of exactly one white offspring among the two is 1/4 + 1/4 = 1/2, because purple–white and white–purple are separate mutually exclusive paths that both satisfy the description. The probability of both purple is 1/4. The probability of at least one white is 3/4, since every path except purple–purple contains a white.
Order matters when counting paths but not when naming a final count. Exactly one white has two arrangements. Exactly zero white has one arrangement. Both white has one arrangement. If you forget the second arrangement, your answer for exactly one white becomes half the correct value. The tree makes this mistake visible before you commit it to a final line.
Three offspring: count every qualifying arrangement
For three successive offspring under Pp × pp, there are eight equally likely ordered paths. Each has probability 1/8. The number of white offspring can be zero, one, two or three. Zero white occurs in just PPP, where P now stands informally for a purple outcome rather than the allele in the word-only path list. To avoid ambiguity in formal working, use ‘Pu’ for purple and ‘Wh’ for white rather than reusing P.
The count distribution is: zero whites, one path, probability 1/8; exactly one white, three paths, probability 3/8; exactly two whites, three paths, probability 3/8; three whites, one path, probability 1/8. These four probabilities sum to eight eighths. This total-one check detects omitted arrangements.
If an examiner asks for at least two white among three, include exactly two and exactly three: 3/8 + 1/8 = 1/2. If the examiner asks for more purple than white, include exactly zero white and exactly one white: 1/8 + 3/8 = 1/2. A short verbal rephrasing of the event can prevent an entire page of unnecessary arithmetic.
Four purple offspring: the informative non-proof
Under the heterozygous-parent hypothesis U = Pp, every test-cross offspring has a 1/2 chance of being purple. Four purple in four independent trials therefore have probability (1/2)^4 = 1/16. Under U = PP, the same four-purple run has probability 1, because all offspring are predicted purple. The observation is much more likely under PP than under Pp, but it is not logically impossible under Pp.
It would be incorrect to announce ‘there is now a 1/16 probability that U is Pp’. The value 1/16 measures the probability of a particular observed run if U were Pp. It is not the probability of that genotype after seeing the run. To find a posterior probability one would need further assumptions about how likely PP and Pp were before testing, along with a more advanced updating method. G3 candidates should not swap these two probabilities as if they were the same.
A compact examination response is: ‘The four-purple outcome is compatible with PP, but it can also occur with probability 1/16 if U is Pp, so the genotype cannot be proved from the four observations alone.’ If the stem assumes a pure-breeding parent, the answer changes. The mathematical discipline is to obey the given conditions, not the pattern that seems most attractive.
The complement: at least one white
Long trees can become untidy. For four test-cross offspring under Pp × pp, the chance of at least one white is one minus the chance of no white. No white means all four purple, probability 1/16. Thus at least one white has probability 15/16. The complement route is efficient because all the many qualifying paths are gathered in one calculation.
For six offspring, the chance of no white under the Pp hypothesis is (1/2)^6 = 1/64. The chance of at least one white is 63/64. This is a prediction made before conducting six independent observations. It does not say that any particular batch is required to contain white. A rare outcome can occur without the laws of probability failing.
For n independent trials with probability one-half of purple each time, the mathematical pattern is that all-purple has probability (1/2)^n, and at least one white has probability 1 − (1/2)^n. Treat this as a helpful generalisation after you understand the short tree; the ability to build and interpret a suitable possibility diagram matters more than memorising an isolated expression.
A changed probability: the model must change with it
Suppose a fictional greenhouse selection model instead states that each offspring is purple with probability 0.6 and white with probability 0.4. This is no longer the equal-likelihood inheritance case and should not be described as a simple Pp × pp genetic cross. It is an independent probability exercise with supplied outcome chances. For three successive offspring, the chance all three are purple is 0.6 × 0.6 × 0.6 = 0.216.
The chance at least one is white is 1 − 0.216 = 0.784. The chance that exactly one is white is the sum of three ordered arrangements, each with probability 0.4 × 0.6 × 0.6 = 0.144, giving 0.432. This demonstrates how the same diagram structure survives when the numerical labels change. The biological interpretation must not be carried over unchanged.
If a test stem says outcomes are dependent, multiplication by the same probability at every branch is invalid. For example, drawing coloured counters without replacement changes the composition of the remaining group. In that case, calculate the second-branch chance from the updated counts. Words such as ‘with replacement’, ‘without replacement’ and ‘independent’ are mathematical instructions, not background decoration.
Observed percentages are not theoretical probabilities
A class grows 24 offspring from Pp × Pp and records 17 purple and 7 white. The observed purple proportion is 17/24, about 70.8%. The theoretical model predicts 3/4, or 75%. The difference is 4.2 percentage points approximately. None of those calculations by itself establishes that the genetic model is false. A finite sample can differ from the expected ratio.
A poor answer replaces the observed 17:7 with 18:6 because ‘3:1 is correct’. That erases the actual evidence. Another poor answer writes that the theoretical probability has changed to 17/24. The model probability remains 3/4 under its assumptions. Experimental proportion is a summary of what happened in one sample, not a replacement for the underlying theoretical rule.
Make a small table in your rough work if needed: observed purple 17, white 7, total 24; predicted purple 18, white 6 if you are translating a 3:1 prediction into expected counts for 24 trials. The expected counts are not promises of exact outcomes. The observation–expectation distinction is a bridge between G3 Mathematics statistics and G3 Biology inheritance.
Unequal sample sizes: do not average the two percentages
Suppose Group A records 3 white flowers out of 6, giving 50%, while Group B records 6 white flowers out of 18, giving about 33.3%. A student takes the simple mean of these percentages and claims the pooled white rate is 41.7%. That is not the fraction of white flowers in the combined sample because the groups have unequal sizes.
The combined white count is 3 + 6 = 9. The combined total is 6 + 18 = 24. Thus the pooled white fraction is 9/24, or 37.5%. The large group contributes three times as many plants and should not receive the same weight as the small group. The cleanest method is often to add raw counts before computing a percentage.
If the question asks for the mean of the two group’s percentages as separately defined units, 41.7% could be the desired statistic. But it should never be mislabeled as the rate among all 24 plants. Ask, ‘What exactly is the denominator?’ This question repairs errors in science tables, survey reports, public claims and ordinary household budgeting.
A word problem with a hidden condition
A greenhouse selects one purple plant at random from the F2 offspring of a Pp × Pp cross. It is crossed with pp. Given the original simple genotype model, what is the probability that the first test-cross offspring is white? First identify the conditional genotype possibilities for the selected purple plant: probability 1/3 PP and 2/3 Pp. If PP, the chance of white is zero; if Pp, the chance is 1/2.
Under those explicitly stated selection and independence assumptions, the combined probability of white on the first test cross is (1/3)(0) + (2/3)(1/2) = 1/3. A possibility tree can show the two genotype branches and the outcome branches. This is an extension connecting two already learnt structures; it should not be presented as proof that complex two-stage conditional calculations will appear as a compulsory 2027 SEC question.
A common wrong attempt multiplies 2/3 by 1/2 and then divides by 3/4 again. That counts the purple selection condition twice. Once you reduce the genotype possibilities to the group of selected purple offspring, do not reapply the original condition unless a later observation genuinely changes the sample space.
More evidence arrives: what can be concluded?
After three test-cross offspring, a selected purple plant has produced two purple and one white. In the simple model a PP plant cannot pass p to make a pp white offspring with a pp tester. Therefore the existence of a reliably recorded white offspring identifies the selected plant as Pp. This is a deductive conclusion about the simplified mechanism. It is not computed from the overall purple percentage.
There is a difference between calculating the probability of a possible outcome and drawing a logical inference from an impossible outcome under one hypothesis. Under PP the chance of white is zero, so a white offspring excludes PP if the records and assumptions are correct. Under Pp white is possible. Do not use a frequency comparison when a straightforward impossibility argument answers the question.
Suppose the tester plant’s genotype is not confirmed. Now the exclusion is not as clean. The relevant mathematical branch labels have changed because the cross is not fully specified. A careful learner requests the missing premise rather than forcing the earlier calculation onto an unsuitable problem.
Build a reliable tree-diagram page
Write a clear event key before drawing: Pu means purple flower, Wh means white flower. State the model in the margin: U = Pp crossed with pp, independent offspring, each colour with probability 1/2. Each set of branches from a node must sum to one. Label each branch rather than writing a loose fraction beside the entire tree.
For a path, multiply along the sequence because every choice on that path must happen. For a combined event, add the probabilities of distinct qualifying paths because any one of those alternatives can satisfy the question. Check whether the paths are mutually exclusive before adding. This method is easier to audit than a formula pasted onto an unexplained number.
When an event is described as ‘at least one’, ‘none’, ‘exactly two’ or ‘more than half’, translate the phrase into explicit qualifying paths. Underline the word exactly. Students lose marks by finding at least two when the question requires exactly two, not by lacking enough formulae.
Practice set: five exam-style tasks
Task 1. For a Pp × Pp cross, find the chance of white, purple and heterozygous Pp among all offspring. Explain whether the three answers are disjoint events. Answer: white = 1/4; purple = 3/4; Pp = 1/2. They are not a partition into three disjoint groups because Pp is a subset of purple; adding all three would double-count some outcomes.
Task 2. From the purple offspring of the same cross, choose one at random. What is the chance that it is homozygous dominant? Answer: one of the three qualifying purple genotype cells is PP, so 1/3. State the reduced sample space; do not merely write ‘one quarter’ from the original diagram.
Task 3. For an independent Pp × pp test cross, calculate the probability that in three offspring exactly two are white. Answer: WWP, WPW or PWW are three separate equally likely paths; each has probability 1/8, giving 3/8. Here W and P in the path list stand for white and purple outcomes, not the alleles.
Task 4. In four independent Pp × pp outcomes, calculate the probability at least one offspring is white. Answer: the all-purple path has probability 1/16; its complement is 15/16. Do not count three, seven or fifteen paths without explaining the event.
Task 5. Group A has two white among eight plants. Group B has seven white among 28 plants. Find the overall white fraction. Answer: 9/36 = 1/4 = 25%. Both groups individually happen to have 25%, but a future example with unequal rates should still be pooled from counts, not a simple mean of percentages.
A nine-mark mock question with a full model answer
A class studies Pp × Pp offspring. One purple individual, U, is randomly selected from the purple offspring. (a) State the probability that U is Pp and explain your denominator. [2] (b) If U is Pp and is crossed with pp, draw a tree for three independently produced offspring and find the chance of at least two whites. [3] (c) A separate tester cross produces four purple and no white offspring. Explain why the observation does not prove that its purple parent was PP. [2] (d) From 30 plants, 21 are purple and 9 white. Find the observed purple percentage and compare it with the theoretical 75%. [2]
Model answer (a). Purple limits the original four genotype outcomes to PP, Pp, Pp, so two of three equally likely qualifying cases are heterozygous: 2/3. (b). Under Pp × pp each branch is 1/2; exactly two white gives three paths at 1/8 each, and three whites gives one path at 1/8. Total 4/8 = 1/2. (c). Pp × pp can also produce four purple by chance, with probability 1/16. (d). The observed purple rate is 21/30 = 70%, which is 5 percentage points below the 75% model prediction.
This mock allocation is educational, not an actual SEAB mark scheme. The task assesses four different moves: conditioning, combined-event computation, limits of evidence and conversion of counts into percentages. Repeating the same arithmetic under different headings would be poor preparation. A useful paper offers distinct jobs that reveal distinct misconceptions.
How to check without a calculator
For two or three offspring, small fractions can be checked mentally. Four equally likely two-offspring paths each have probability 1/4; eight three-offspring paths each have 1/8. For 1/16, recognise that halving four times gives 1/2, 1/4, 1/8, 1/16. Simplify fractions at the end so the denominator still tells the story during working.
For percentages, 3/4 is 75%, 1/4 is 25%, 1/2 is 50% and 2/3 is approximately 66.7%. The last number is an approximation, not exactly 67%. If you round an intermediate result too early and then use it in further calculations, you can accumulate small errors. Keep fractions exact until the final requested precision.
Check an answer against the problem: probabilities cannot be negative or exceed one; four probabilities covering all cases should sum to one; a conditional event must use the correct restricted group; an observed rate must use actual counts. These checks are inexpensive and catch many errors faster than restarting the whole question.
What to do when an exam question feels unfamiliar
First identify the random experiment: choosing a plant, producing an offspring, selecting a colour from a group or repeating independent trials. Second list the outcomes that are possible at the stage of the experiment the question asks about. Third write any information already known. Fourth sketch a diagram. Only then calculate.
This route is especially useful when the same problem mixes Biology words and Mathematics requirements. You do not need to know more genetics than the model supplies. If the question states a complete-dominance cross, use it. If it gives probabilities directly without genetics, use those probabilities. Marks come from interpreting the question accurately, not from adding irrelevant science vocabulary.
Make the final sentence answer in the units requested. If the answer is a probability, give a fraction, decimal or percentage as directed. If it asks for an expected number out of 40, multiply the theoretical probability by 40 and label it as an expectation, not a guaranteed count. If it asks for a conclusion, state what the calculation does and does not imply.
A short three-day revision plan
Day one: visualise the sample space
Work the Pp × Pp four-cell diagram twice, once asking for all offspring and once asking only about purple offspring. Explain aloud why the denominator changes from four to three. Then replace the flower labels with fictional tokens so you know the mathematics is transferable. Ask a parent or tutor to give a new condition such as ‘not white’ and test whether the diagram still works.
Day two: branch, multiply, then add
Draw a three-level tree for independent Pp × pp offspring. Find exactly one white, exactly two whites and at least one white. Do not read answers until you have named every qualifying ordered path. Recreate the ‘at least one’ result with a complement and verify both methods agree. Correct the first mistake by drawing, not by memorising the numerical result.
Day three: interpret the evidence
Take two short classroom data sets with unequal totals. Pool the counts correctly and compare the combined experimental percentage with the theoretical chance. Write one cautious conclusion and one sentence that explains why an all-purple test-cross run does not prove PP. Timed accuracy grows from clear reasoning; simply making the calculation faster does not improve a mistaken denominator.
Frequently asked questions
Why is Pp one-half among all offspring but two-thirds among purple offspring?
The original Pp × Pp square has four equally likely cells, two of which are Pp. Knowing the chosen offspring is purple excludes the white pp cell. Two of the three remaining cells are Pp. The numerator stays two while the relevant denominator changes from four to three.
Is the probability of six purple offspring under a Pp × pp cross zero?
No. It is (1/2)^6 = 1/64 if successive offspring are modelled as independent. The result is unusual but possible. The probability that at least one of six is white is 63/64.
Does a 3:1 expected ratio mean every group of four must have three purple?
No. Each new offspring is a random outcome in the simplified model. The theoretical ratio describes the chances, not a fixed order or fixed quota in every small group.
May I use formal conditional-probability notation in G3 Mathematics?
Clear notation can help if you understand it, but do not mistake this article’s optional extension for an assurance about a particular 2027 question. The official K310 syllabus and your teacher’s instructions determine the examinable content. A labelled possibility diagram is an excellent way to show the same reasoning transparently.
What if the group has 100% purple seedlings?
Report the observed fraction exactly as recorded, but do not automatically replace the genetic model. A short sample containing only purple does not prove the underlying white probability was zero. Check the cross, count, context and whether the conclusions are being overstated.
Why do we pool counts rather than average percentages?
Pooling answers the question about all individuals together. Groups of different sizes contribute different numbers of individuals. A simple average of group percentages treats each group equally, which is a different statistic.
Reading route and next examination skill
The previous Biology inheritance volume 1178 explains the genotype logic. For general branching diagrams, read Mathematics probability trees in volume 0178. The PSLE Mathematics reasoning bridge helps readers who are building the habit of showing each step, while the Sengkang G3 SEC overview explains where Mathematics fits alongside English and Science.
Once the maths is reliable, the next challenge is language. A greenhouse report might accurately record seventeen purple flowers but publish a misleading headline saying that the result ‘proves’ every purple plant is genetically pure. English comprehension requires the reader to separate evidence, inference, qualifiers and persuasive language. The next volume takes that step without turning a comprehension paper into a Biology test.
