Monohybrid Crosses & Punnett Squares
Learn to build a Punnett square for a one-trait cross, read off 3:1 and 1:2:1 ratios, and turn those ratios into probabilities for any single offspring.
What you'll do in this lesson
A voice-first session with the Crimsora tutor on Monohybrid Crosses & Punnett Squares, then targeted practice and FRQs — with the tutor adapting to where you get stuck.
What this lesson covers
You already know that alleles separate during meiosis so each gamete carries one allele per gene. Now you get a tool that turns that rule into a prediction: the Punnett square. With a simple 2-by-2 grid you can say, before a single seed sprouts, what fraction of offspring should be tall and what fraction short — and what the odds are for one particular baby.
This lesson stays with a single trait (a monohybrid cross). You will set up squares from parent genotypes, count boxes to get genotypic and phenotypic ratios, and then make the mental jump that trips up most students: a 3:1 ratio is a probability statement about each offspring, not a promise about a litter of four. Get that jump right and dihybrid crosses, blood types, and pedigrees all become easier later in the unit.
This lesson stays with a single trait (a monohybrid cross). You will set up squares from parent genotypes, count boxes to get genotypic and phenotypic ratios, and then make the mental jump that trips up most students: a 3:1 ratio is a probability statement about each offspring, not a promise about a litter of four. Get that jump right and dihybrid crosses, blood types, and pedigrees all become easier later in the unit.
What a Monohybrid Cross Is and How to Set One Up
A monohybrid cross follows one gene with two alleles — one dominant, one recessive. Write the dominant allele as a capital letter and the recessive as the same letter lowercase. If purple stem is dominant in tomatoes, the genotypes are (homozygous dominant), (heterozygous), and (homozygous recessive). Both and show the purple phenotype; only is green.
To build the square, first write each parent's genotype, then split it into gametes. Splitting is the whole point: because of segregation, an parent makes two kinds of gametes, half and half . An parent makes only gametes. Put one parent's gametes across the top of a 2-by-2 grid and the other parent's down the left side. Fill each interior box by combining the letter above it with the letter beside it, writing the capital letter first (, not ).
A very common setup error is writing a parent's genotype along the top instead of its gametes — for example heading a single column rather than heading one column and the next column . A related slip shows up when a parent is homozygous: an parent still gets two columns, but both are headed , and those identical headings mean the same allele twice, not two different alleles. Check yourself by confirming that every interior box has exactly two letters, one from each parent, and that all four boxes are filled before you count anything.
To build the square, first write each parent's genotype, then split it into gametes. Splitting is the whole point: because of segregation, an parent makes two kinds of gametes, half and half . An parent makes only gametes. Put one parent's gametes across the top of a 2-by-2 grid and the other parent's down the left side. Fill each interior box by combining the letter above it with the letter beside it, writing the capital letter first (, not ).
A very common setup error is writing a parent's genotype along the top instead of its gametes — for example heading a single column rather than heading one column and the next column . A related slip shows up when a parent is homozygous: an parent still gets two columns, but both are headed , and those identical headings mean the same allele twice, not two different alleles. Check yourself by confirming that every interior box has exactly two letters, one from each parent, and that all four boxes are filled before you count anything.
Reading Genotypic and Phenotypic Ratios Out of the Grid
Once the four boxes are filled, counting gives you two different ratios, and you must say which one you are reporting.
The genotypic ratio counts letter combinations. The phenotypic ratio counts appearances, so it lumps and together. For the boxes are , , , : genotypic ratio and phenotypic ratio dominant to recessive. Those two numbers describe the same square, so a question asking for "the ratio" is incomplete until you decide which is wanted.
Notice the third row. A cross of a heterozygote with a homozygous recessive is called a testcross, and it is how breeders figure out a hidden genotype. If a purple plant is , crossing it with green gives 100% purple offspring; if it is , about half the offspring are green. One green offspring is enough to prove the purple parent carries a recessive allele.
Students often lose track of order in a ratio. Always name the categories: write "3 purple : 1 green," not a bare "3:1," so your teacher and your future self know what each number refers to.
The genotypic ratio counts letter combinations. The phenotypic ratio counts appearances, so it lumps and together. For the boxes are , , , : genotypic ratio and phenotypic ratio dominant to recessive. Those two numbers describe the same square, so a question asking for "the ratio" is incomplete until you decide which is wanted.
| Cross | Boxes | Genotypic ratio | Phenotypic ratio |
|---|---|---|---|
| all | 100% | all dominant | |
| , , , | |||
| , , , | |||
| , , , | all dominant |
Students often lose track of order in a ratio. Always name the categories: write "3 purple : 1 green," not a bare "3:1," so your teacher and your future self know what each number refers to.
From Ratios to Probability for a Single Offspring
A Punnett square really answers a probability question. Each box represents an equally likely fertilization event, so the fraction of boxes showing a genotype is the probability that any one offspring has it. For , each offspring independently has a chance of , a chance of , and a chance of the dominant phenotype. Those same numbers can be written as 25%, 50%, and 75%.
Here is where the most persistent misconception lives. A ratio does not mean that in a family of four offspring exactly three will be purple. Fertilization is like flipping a coin: four flips can easily give four heads. Ratios are expected values that get closer to the prediction as the number of offspring grows, which is why Mendel counted thousands of pea plants instead of four.
A second trap: previous offspring do not change the odds for the next one. If two heterozygous parents already have three children with the recessive trait, the fourth child still has a chance of showing it. Each fertilization uses a fresh pair of gametes, so the events are independent.
If you need the chance of a combination across two offspring, multiply the separate probabilities. Two offspring both from is . To convert an expected ratio into predicted counts, multiply the fraction by the total: out of 60 offspring from , expect about purple.
Here is where the most persistent misconception lives. A ratio does not mean that in a family of four offspring exactly three will be purple. Fertilization is like flipping a coin: four flips can easily give four heads. Ratios are expected values that get closer to the prediction as the number of offspring grows, which is why Mendel counted thousands of pea plants instead of four.
A second trap: previous offspring do not change the odds for the next one. If two heterozygous parents already have three children with the recessive trait, the fourth child still has a chance of showing it. Each fertilization uses a fresh pair of gametes, so the events are independent.
If you need the chance of a combination across two offspring, multiply the separate probabilities. Two offspring both from is . To convert an expected ratio into predicted counts, multiply the fraction by the total: out of 60 offspring from , expect about purple.
Working Backward: Deducing Parent Genotypes from Offspring
Many homework problems reverse the process — you are given phenotypes of parents and offspring and asked for genotypes. Two rules crack almost all of them.
First, an individual showing the recessive phenotype must be homozygous recessive, so its genotype is fully known and it can only donate a recessive allele. Second, if any offspring shows the recessive phenotype, then each parent contributed a recessive allele, so each parent carries at least one .
So if two purple-stemmed tomatoes produce a green-stemmed offspring, both parents must be . Neither could be , because has no recessive allele to give. Conversely, if a purple plant crossed with a green plant produces 100% purple offspring over many seeds, the purple parent is very likely — though "likely" matters, since a small sample from an parent could by chance contain no green plants.
A useful habit is to write what you know with a blank: a purple plant of unknown genotype is . Then use the offspring evidence to fill the blank. Students go wrong by assuming that a dominant phenotype means homozygous dominant, or by assuming that carrier parents must produce visible carriers — carriers of a recessive allele look exactly like homozygous dominant individuals, which is precisely why testcrosses and pedigrees exist. This backward reasoning is the same skill you will use when tracing recessive conditions through a family tree later in the unit.
First, an individual showing the recessive phenotype must be homozygous recessive, so its genotype is fully known and it can only donate a recessive allele. Second, if any offspring shows the recessive phenotype, then each parent contributed a recessive allele, so each parent carries at least one .
So if two purple-stemmed tomatoes produce a green-stemmed offspring, both parents must be . Neither could be , because has no recessive allele to give. Conversely, if a purple plant crossed with a green plant produces 100% purple offspring over many seeds, the purple parent is very likely — though "likely" matters, since a small sample from an parent could by chance contain no green plants.
A useful habit is to write what you know with a blank: a purple plant of unknown genotype is . Then use the offspring evidence to fill the blank. Students go wrong by assuming that a dominant phenotype means homozygous dominant, or by assuming that carrier parents must produce visible carriers — carriers of a recessive allele look exactly like homozygous dominant individuals, which is precisely why testcrosses and pedigrees exist. This backward reasoning is the same skill you will use when tracing recessive conditions through a family tree later in the unit.
Key terms
- Monohybrid cross.
- A cross that follows the inheritance of a single gene with two alleles, one dominant and one recessive.
- Punnett square.
- A grid that combines the possible gametes of two parents to show all equally likely offspring genotypes and their proportions.
- Genotype.
- The specific pair of alleles an individual carries for a gene, such as , , or .
- Phenotype.
- The observable trait produced by a genotype; and share the same phenotype when one allele is completely dominant.
- Homozygous.
- Having two identical alleles for a gene ( or ); such an individual produces only one kind of gamete for that gene.
- Heterozygous.
- Having two different alleles for a gene (); produces two kinds of gametes in equal numbers.
- Testcross.
- A cross between an individual showing the dominant phenotype and a homozygous recessive individual, used to reveal whether the dominant individual is heterozygous.
- Expected ratio.
- The proportion of offspring categories predicted by a Punnett square; actual counts vary from it by chance, especially in small numbers of offspring.
Worked example
In tomatoes, purple stem () is completely dominant to green stem (). A heterozygous purple-stemmed plant is crossed with a green-stemmed plant. (a) Give the genotypic and phenotypic ratios of the offspring. (b) What is the probability that a single seedling has a green stem? (c) If the cross produces 48 seedlings, how many are expected to be purple? (d) What is the probability that the first two seedlings are both green?
Step 1 — Write parent genotypes. Heterozygous purple is . Green stem is the recessive phenotype, so that parent must be homozygous recessive, .
Step 2 — Find gametes. The parent makes and gametes; the parent makes only gametes. Put and across the top and and down the side.
Step 3 — Fill the four boxes: , , , .
Step 4 — (a) Count genotypes: two and two , a genotypic ratio of . Count phenotypes: is purple and is green, so the phenotypic ratio is 1 purple : 1 green. Notice both ratios are here, which happens only because no boxes exist.
Step 5 — (b) Two of the four boxes are green, so the probability for any one seedling is , or 50%.
Step 6 — (c) Expected purple count is seedlings. This is an expectation; a real tray might show 21 or 27 purple.
Step 7 — (d) The two fertilizations are independent, so multiply: , a 25% chance both are green.
Step 2 — Find gametes. The parent makes and gametes; the parent makes only gametes. Put and across the top and and down the side.
Step 3 — Fill the four boxes: , , , .
Step 4 — (a) Count genotypes: two and two , a genotypic ratio of . Count phenotypes: is purple and is green, so the phenotypic ratio is 1 purple : 1 green. Notice both ratios are here, which happens only because no boxes exist.
Step 5 — (b) Two of the four boxes are green, so the probability for any one seedling is , or 50%.
Step 6 — (c) Expected purple count is seedlings. This is an expectation; a real tray might show 21 or 27 purple.
Step 7 — (d) The two fertilizations are independent, so multiply: , a 25% chance both are green.
Practice questions
In guinea pigs, black fur () is completely dominant to white fur (). Two black guinea pigs are crossed and produce a white offspring. What are the genotypes of the two black parents?
- and
- and
- and
- and
Answer: and
The white offspring is , so it received one allele from each parent. That rules out any parent, since has no recessive allele to pass on. Both parents show the black phenotype, so neither can be . The only genotype that is black and carries is , so both parents are . As a check, gives a 3 black : 1 white expected ratio, which allows white offspring.
A pea plant with round seeds (round, , is dominant to wrinkled, ) is crossed with a wrinkled-seed plant. All 32 offspring have round seeds. State the most likely genotype of the round parent, explain your reasoning, and explain why a single wrinkled offspring would have changed your answer.
Answer: The round parent is most likely (homozygous dominant). Because the wrinkled parent is and donates only gametes, every offspring's phenotype reveals the allele from the round parent. If that parent were , about half the offspring — roughly 16 of 32 — would be and wrinkled. Seeing zero wrinkled offspring in 32 strongly indicates the round parent donated every time, so it is . A single wrinkled offspring would prove it carries and must be , since offspring require a recessive allele from each parent.
This is a testcross. Crossing with a homozygous recessive individual exposes the hidden allele of the dominant-looking parent, because the recessive parent contributes nothing that can mask it. The reasoning is probabilistic: with only two or three offspring, an parent could by chance produce no wrinkled seeds, which is why a large number of offspring makes the conclusion convincing.
Two parents heterozygous for a recessive trait already have three children who all show the recessive phenotype. What is the probability that their next child shows the recessive phenotype?
Answer:
Each fertilization is an independent event using a fresh pair of gametes, so earlier children do not change the odds. The Punnett square gives one box out of four, so every child has a chance of the recessive phenotype. Thinking the odds must now shift toward the dominant phenotype to 'balance out' the ratio is the most common error here; the ratio describes long-run expectations, not a quota within one family.
FAQ
- What is the difference between a genotypic ratio and a phenotypic ratio?
- A genotypic ratio counts allele combinations separately, so gives . A phenotypic ratio counts only what you can observe, so and are grouped and the same cross gives 3 dominant : 1 recessive. Always label the categories when you write a ratio.
- Why didn't my results match the 3:1 ratio my Punnett square predicted?
- Punnett squares give expected probabilities, not guarantees. Fertilization is a chance event, like coin flipping, so small numbers of offspring often deviate from the prediction. With hundreds or thousands of offspring the observed proportions move much closer to , which is why Mendel counted such large numbers of plants.
- How do I know a parent's genotype if I can only see its phenotype?
- An individual with the recessive phenotype must be homozygous recessive, so its genotype is certain. An individual with the dominant phenotype could be homozygous dominant or heterozygous; write it as and use offspring evidence. If it produces any recessive offspring, it must be heterozygous. A testcross with a homozygous recessive partner is the standard way to find out.
- Do I need a bigger Punnett square when a parent is homozygous?
- No. A one-gene cross always uses a 2-by-2 square, because each parent contributes one allele. A homozygous parent simply has the same allele labeling both of its rows or columns, which makes two of the boxes identical. Larger grids come in later with dihybrid crosses, which track two genes at once.
Learn this with a teacher, not a page
The Crimsora tutor teaches Monohybrid Crosses & Punnett Squares live — explaining on a whiteboard, asking you questions, and adapting to where you get stuck.