Mendelian genetics: Punnett and exercises

In Year 3 of high school, in the human-sciences track as in all the licei that have natural sciences in Year 3, Mendelian genetics arrives in class from January and the test usually falls between February and March. It is a longer window than usual compared with the other posts of this period: I am publishing the guide now because it stays online until the test, and because the topic returns every year, already in Year 2 with Mendel and then again in Year 3 with blood groups and sex-linked inheritance. If you arrive in January with the basics already in order, you are a month ahead.

The thing to study before anything else is cell division. If meiosis is not clear, gametes and Mendel’s ratios seem like a recipe to learn by heart: I talked about it in mitosis and meiosis: differences and exercises. The point needed here is just one, namely that each gamete receives only one allele for each gene, and that pairs of different genes separate independently (in the case of genes on different chromosomes).

This guide is made only for exercises. In genetics tests 90% of the marks lie in knowing how to set up a problem, not in reciting the laws.

The minimum vocabulary

A few words are needed, but they have to be used precisely.

Gene: a stretch of DNA that determines a trait (seed colour, blood group).
Allele: each version of a gene. Every diploid individual has two per gene, one from the mother and one from the father.
Dominant / recessive: the dominant allele shows even when only one copy is present, the recessive only if there are two. Convention: capital letter for the dominant (A), lower case for the recessive (a).
Homozygous / heterozygous: the same two alleles (AA, aa) or different ones (Aa).
Genotype / phenotype: the genotype is the combination of alleles (Aa), the phenotype is the trait you see (yellow seed).

The commonest mistake is saying “the phenotype is Aa”. A dominant phenotype, for example yellow seed, can hide two genotypes: AA or Aa. A recessive phenotype, by contrast, has only one possible genotype: aa. This detail is the key to half an exercise.

The procedure in 4 steps

With this sequence I solve every problem, without skipping steps even when it “seems obvious”.

Write the genotypes of the parents. If the text says “heterozygous”, it is Aa. If it says “recessive”, it is aa. If it says only “dominant phenotype” and gives you nothing else, the genotype can be AA or Aa: you work it out from the offspring.
Write the gametes that each parent can produce. An Aa parent produces two types of gamete, A and a, each with probability 1/2. An AA parent produces only A.
Build the Punnett square: the gametes of one parent along the rows, of the other along the columns, each box is a possible offspring.
Read the ratios, first genotypic and then phenotypic, and answer the question asked.

Every box of the square has the same probability, and that is why you can count.

Monohybrid cross

You follow a single trait. The classic case is the cross between two heterozygotes, $A a \times A a$ . The gametes are A and a for both, and the square gives four boxes: AA, Aa, Aa, aa.

Genotypic ratio:

1 AA : 2 A a : 1 aa

Phenotypic ratio (A dominant over a):

3 dominant : 1 recessive

The 3:1 is Mendel’s historic result. But be careful how you read it: it does not mean that out of four children exactly three will be dominant. It means that each child has a probability of 3/4 of showing the dominant trait.

The test cross

If you have an individual with a dominant phenotype and want to know whether it is AA or Aa, you cross it with a recessive homozygote aa. If it is AA, all the offspring are Aa and show the dominant trait. If it is Aa, half the offspring are Aa and half are aa, so the recessive shows: ratio 1:1. A single recessive offspring is enough to conclude that the dominant parent was heterozygous.

Dihybrid cross

Here you follow two traits on independent genes. Crossing two double heterozygotes, $A a B b \times A a B b$ , each parent produces four types of gamete (AB, Ab, aB, ab), each with probability 1/4. The square has 16 boxes and the phenotypic ratio is

9 : 3 : 3 : 1

that is, 9 with both traits dominant, 3 with the first dominant and the second recessive, 3 with the opposite, 1 with both recessive.

The 16-box square is slow and makes you go wrong when copying. In a test the product rule is better: the two genes are independent, so you calculate the probability of each trait separately and multiply. For $A a B b \times A a B b$ :

probability of A dominant: 3/4; of recessive aa: 1/4;
probability of B dominant: 3/4; of recessive bb: 1/4.

Then the probability of a child with A dominant and bb recessive is

\frac{3}{4} \cdot \frac{1}{4} = \frac{3}{16}

and that of an aabb child is

\frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16}

Check that the method works: the probability of a child with both dominant is $\frac{3}{4} \cdot \frac{3}{4} = \frac{9}{16}$ , which is exactly the 9 of the 9:3:3:1. When there are three or more genes the square simply cannot be drawn, whereas the product rule keeps working.

ABO blood groups

The ABO system is the teachers’ favourite exercise because it puts together three new things: three alleles, codominance, dominance.

The gene has three alleles: $I^{A}$ , $I^{B}$ and $i$ . $I^{A}$ and $I^{B}$ are codominant with each other (both show), and dominant over $i$ . The table of genotypes is this.

Group	Possible genotypes
A	$I^{A} I^{A}$ or $I^{A} i$
B	$I^{B} I^{B}$ or $I^{B} i$
AB	$I^{A} I^{B}$
O	$ii$

Each individual has only two alleles, so each parent passes on one of the two. The classic question is “can two parents of group A and B have a child of group O?”. The answer is yes, but only if both are heterozygous ( $I^{A} i$ and $I^{B} i$ ): the O child must receive $i$ from both. If a parent is $I^{A} I^{A}$ , no child can be O.

Two quick rules that every test uses: an AB parent cannot have children of group O, and two parents of group O always have children of group O.

X-linked inheritance

Some genes, such as those for colour blindness and haemophilia, are on the X chromosome and have no counterpart on the Y. That is why the female (XX) has two alleles, while the male (XY) has only one and is called hemizygous: if that allele is recessive, the trait appears, with no need for a second allele.

Alleles are written as superscripts on the X: $X^{D}$ for normal vision, $X^{d}$ for colour blindness. A female $X^{D} X^{d}$ is a carrier: she sees normally, but can pass on the allele. This explains why these diseases are more frequent in males, and why an affected male always receives the X from his mother, never from his father.

The typical mistake is writing a genotype as " $X^{d}$ " for a male without the Y, or forgetting that the father passes the Y to sons and the X to daughters.

Probability: independence, product and sum

All the genetics of exercises rests on three rules of probability, which I go over more calmly in the post on probability and combinatorics.

Independent events (two different children, two different genes): the probability that both occur is the product.
Mutually exclusive events (the child is AA or Aa): the probability that one of the two occurs is the sum.

Incomplete dominance (post-Mendelian genetics)

Not all traits follow the dominant/recessive model. In incomplete dominance the heterozygote has an intermediate phenotype: in snapdragon, red (RR) crossed with white (rr) gives all pink offspring (Rr). Crossing two pinks gives $1 R R : 2 R r : 1 r r$ , that is a phenotypic ratio of 1 red : 2 pink : 1 white, which coincides with the genotypic one. This is the signal to recognise: if the text speaks of an intermediate phenotype, there is no 3:1 ratio. In codominance (like the AB group) the heterozygote shows both traits, without blending.

The mistakes I see most often

Writing the gametes with two alleles of the same gene (for example “Aa” as a gamete): each gamete carries only one allele per gene.
Reading 3:1 as “out of four children exactly three”: it is a probability, not a guaranteed count.
For the dihybrid, drawing 16 boxes and counting them wrongly instead of multiplying.
In X-linked inheritance, writing two alleles on the X for the male.

Worked exercises

Exercise 1: monohybrid and probability

In pea plants yellow seed (G) is dominant over green (g). Two heterozygous plants are crossed. Calculate: a) the probability that a seed is green; b) the probability that, among two seeds, both are green; c) the probability that, among two seeds, one is yellow and one green.

Step 1. Parents: $G g \times G g$ .

Step 2. Gametes: G and g for both, each with probability 1/2.

Step 3. Square: GG, Gg, Gg, gg. So 3/4 yellow and 1/4 green.

a) The green seed is gg: probability 1/4.

b) Two independent seeds, both green:

\frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16}

c) One yellow and one green can happen in two mutually exclusive ways (first yellow and second green, or first green and second yellow), so you add:

\frac{3}{4} \cdot \frac{1}{4} + \frac{1}{4} \cdot \frac{3}{4} = \frac{3}{16} + \frac{3}{16} = \frac{6}{16} = \frac{3}{8}

Check: the three possibilities (both yellow, both green, one of each colour) must give 1. Both yellow is $9/16$ , so $9/16 + 1/16 + 6/16 = 16/16 = 1$ .

Exercise 2: blood groups

A mother of group A and a father of group B have a child of group O. a) What genotypes do the parents have? b) What is the probability that their next child is of group O, and what the probability that it is of group AB? c) Can a man of group AB be the father of a group O child born to the same mother?

a) The O child is $ii$ , so received $i$ from each parent. The mother is of group A and has an $i$ : genotype $I^{A} i$ . The father is B with an $i$ : $I^{B} i$ .

b) The gametes are $I^{A}$ , $i$ for the mother and $I^{B}$ , $i$ for the father. The square gives four combinations, each with probability 1/4: $I^{A} I^{B}$ (group AB), $I^{A} i$ (group A), $I^{B} i$ (group B), $ii$ (group O). The probability of a group O child is 1/4 and of an AB child is 1/4.

c) An AB man is $I^{A} I^{B}$ : he can pass on only $I^{A}$ or $I^{B}$ , never $i$ . An $ii$ child needs an $i$ from the father, so that man is excluded from paternity. This is the reasoning used to exclude, never to confirm: a compatible group does not prove paternity, an incompatible group rules it out.

Exercise 3: colour blindness

Colour blindness is caused by a recessive allele $X^{d}$ on the X chromosome. A carrier woman ( $X^{D} X^{d}$ ) has children with a man with normal vision ( $X^{D} Y$ ). Calculate the probability that: a) a daughter is colour-blind; b) a son is colour-blind; c) any child (male or female) is a colour-blind male.

Gametes. The mother produces $X^{D}$ and $X^{d}$ , each with probability 1/2. The father produces $X^{D}$ and $Y$ , each with probability 1/2.

Square. The four combinations, each with probability 1/4:

$X^{D} X^{D}$ : healthy female;
$X^{D} X^{d}$ : carrier female, healthy;
$X^{D} Y$ : healthy male;
$X^{d} Y$ : colour-blind male.

a) Among the daughters (the two females) none is colour-blind: the father gave all of them his $X^{D}$ . Probability 0. Half of the daughters are, however, carriers.

b) Among the sons, one in two has $X^{d} Y$ : probability 1/2.

c) If you do not know the sex in advance, the $X^{d} Y$ box is worth 1/4 of all the children. The 1/4 is also obtained as a product: probability of being male (1/2) times probability of receiving $X^{d}$ from the mother (1/2).

If instead the father were colour-blind ( $X^{d} Y$ ) and the mother a carrier, half of the daughters would be colour-blind ( $X^{d} X^{d}$ ): redo the square to check it.

How to train

You need few problems done well, with the four-step procedure written out in full every time. There is the Mendelian genetics quiz, 15 questions with no penalty, always available: use it after trying to solve the exercises on this page, and reread the questions you got wrong looking for which of the four steps you skipped. If you need the general picture of the subject there is the science page, and for those aiming at medicine or biology the post on biology and the filter semester: what to study shows where genetics comes back after high school. For the step from DNA to protein, which usually arrives in Year 4, there is replication, transcription and translation.

If one of these steps does not add up and you cannot work out where you are going wrong, the quickest way is to look together at one of your worked exercises: from the contact page you can ask for a free introductory call and we will see in a few minutes what the sticking point is.

Mendelian genetics: Punnett squares and blood groups