Genetics and Gene Technology

Lessons

Senior 3 · Biology

Genetics and Gene Technology

5 lessons

Lesson 2 of 5

Monohybrid crosses and complete dominance

25 min

Three to one, again and again

Mendel did not stop at height. He followed seven characteristics of the pea, one at a time. Each time he crossed two pure-breeding forms, then let the offspring pollinate themselves, and counted the second generation. Four of his counts are below.

Characteristic	Dominant form: number	Recessive form: number	Ratio
Stem length	tall: 787	dwarf: 277	2.84 : 1
Seed shape	round: 5,474	wrinkled: 1,850	2.96 : 1
Seed colour	yellow: 6,022	green: 2,001	3.01 : 1
Pod shape	inflated: 882	constricted: 299	2.95 : 1

Every ratio is close to 3 : 1. A pattern this regular, in four quite different characteristics, needs one explanation. In this lesson you will build it with a Punnett square, and then use the same tool to answer a farmer's question: does this animal carry a hidden allele?

Definition

Monohybrid cross: a cross in which the inheritance of only one characteristic (one gene) is followed. Example: tall × dwarf.

P generation: the parents of the cross. F1 (first filial generation): their offspring. F2 (second filial generation): the offspring of F1 crossed with F1, or self-pollinated.

Punnett square: a grid that shows every way the gametes of two parents can combine at fertilisation, and so the expected genotypes of the offspring.

Complete dominance: the dominant allele completely masks the recessive allele, so the heterozygote looks exactly like the homozygous dominant.

Setting out a genetic cross

National examination answers are marked line by line, so set out every cross in the same six steps:

Parental phenotypes.
Parental genotypes, with a key to the letters.
Gametes: each different type once, in a circle.
Fertilisation: a Punnett square.
Offspring genotypes, with their ratio.
Offspring phenotypes, with their ratio.

Worked example

Worked example 1: the P cross. A pure-breeding tall pea plant is crossed with a pure-breeding dwarf plant. Tall (T) is dominant to dwarf (t). Show the F1.

Step	Tall parent		Dwarf parent
Phenotypes	tall	×	dwarf
Genotypes	TT	×	tt
Gametes	all T		all t

Every box is Tt. F1: all Tt, all tall. The F1 carry the dwarf allele, but T masks it.

Worked example

Worked example 2: the F2. Two F1 plants (Tt) are crossed. Predict the F2, and compare with Mendel's count of 1,064 plants.

Each Tt parent makes two kinds of gamete, T and t, in equal numbers (segregation).

Genotypes: 1 TT : 2 Tt : 1 tt.
Phenotypes: TT and Tt are both tall, so 3 tall : 1 dwarf. The chance that any one F2 plant is dwarf is $\frac{1}{4}$ .
Expected in 1,064 plants: $\frac{3}{4} \times 1064 = 798$ tall and $\frac{1}{4} \times 1064 = 266$ dwarf. Mendel counted 787 and 277: close, but not exact, because which gametes meet is a matter of chance.

Warning

A ratio is a probability, not a promise. "3 : 1" means each offspring has a $\frac{3}{4}$ chance of the dominant phenotype. A family of four children need not have exactly three of one kind and one of the other. Ratios come close to the prediction only when the numbers are large, as in Mendel's 1,064 plants. Each fertilisation is a new chance, unaffected by the earlier ones.

The test cross: finding a hidden allele

A tall pea plant might be TT or Tt; you cannot tell by looking. To find out, cross it with a homozygous recessive (tt) plant. This is a test cross. The tt parent gives only t gametes, so it hides nothing: every offspring's phenotype shows which allele came from the plant being tested.

If the tall plant is	Its gametes	Offspring
TT	all T	all Tt: all tall
Tt	T and t	Tt and tt: 1 tall : 1 dwarf

A single recessive offspring proves that the tested parent is heterozygous.

Worked example

Worked example 3: a test cross on a farm. Polled (hornless) cattle are easier to handle. In cattle the polled allele H is dominant to the horned allele h. A farmer has a polled bull and wants to know whether it carries h.

(a) What cows should the bull be mated with? (b) Of ten calves, six are polled and four are horned. What is the bull's genotype? (c) If all ten calves had been polled, could the farmer be certain that the bull is HH?

Solution

Horned cows (hh), because they give only h gametes.
A horned calf is hh: it received h from its mother and h from the bull. So the bull is Hh. Six to four is close to the expected 1 : 1.
Not certain, but nearly. If the bull were Hh, each calf would have a $\frac{1}{2}$ chance of being polled, so the chance of ten polled calves would be $\left(\frac{1}{2}\right)^{10} = \frac{1}{1024}$ , about 0.001. The farmer can be very confident the bull is HH.

Worked example

Worked example 4: a recessive condition in humans. Albinism is an inherited condition in which the body makes little or no melanin, the pigment of skin, hair and eyes. People with albinism need protection from the sun and often have low vision; they have the same abilities and rights as anyone else. The allele for normal pigmentation, A, is dominant to the allele for albinism, a. Two parents with normal pigmentation have a child with albinism.

(a) Give the parents' genotypes. (b) What is the chance that their next child has albinism? (c) What fraction of their children with normal pigmentation are expected to carry a?

Solution

The child is aa, so it received a from each parent. Each parent shows normal pigmentation, so each also has A. Both are Aa (carriers).
Aa × Aa gives 1 AA : 2 Aa : 1 aa, so the chance is $\frac{1}{4}$ for every child, whatever happened before.
Among the children with normal pigmentation (AA, Aa, Aa), two of every three are carriers: $\frac{2}{3}$ .

Monohybrids and the Punnett Square Guinea Pigs, from the Amoeba Sisters. Watch how the gametes of each parent go along the top and side of the square, and how the boxes are read as probabilities.

Tip

Start from the recessive. In any genetics problem, find an organism that shows the recessive phenotype first. Its genotype is certain (two recessive alleles), and it tells you that each of its parents gave it one recessive allele.

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Practice: monohybrid crosses

Define (a) monohybrid cross, (b) test cross.
In tomatoes, red fruit (R) is dominant to yellow fruit (r). A homozygous red plant is crossed with a yellow plant. Give the parents' genotypes, their gametes, and the F1 genotype and phenotype.
Two F1 plants from question 2 are crossed. Draw a Punnett square and give the F2 phenotype ratio. If 480 F2 plants are grown, how many are expected to have yellow fruit?
A polled cow and a horned bull have a horned calf (H, polled, is dominant to h, horned). Give the cow's genotype and explain how you know.
In fruit flies, normal wings (V) are dominant to short, crumpled vestigial wings (v). Two normal-winged flies produce 302 normal-winged and 98 vestigial-winged offspring. Give the parents' genotypes and explain.
Explain why a test cross uses a homozygous recessive organism and not a homozygous dominant one.
A round-seeded pea plant is crossed with a wrinkled-seeded plant (round, R, is dominant). The offspring are 51 round and 49 wrinkled. Give the genotype of the round parent, with a reason.
Two parents, both carriers of albinism (Aa), already have three children with normal pigmentation. A relative says their fourth child is "sure" to have albinism. Is this right? Explain.

Show answersHide answers

(a) A cross that follows the inheritance of one characteristic, controlled by one gene. (b) Crossing an organism showing the dominant phenotype with a homozygous recessive organism, to find out whether it is homozygous or heterozygous.
RR × rr; gametes R and r; F1 all Rr, all red.
Rr × Rr gives 1 RR : 2 Rr : 1 rr, so 3 red : 1 yellow. Yellow: $\frac{1}{4} \times 480 = 120$ .
Hh. The horned calf is hh and must have received h from its mother; she is polled, so she also has H.
Both Vv. Vestigial offspring (vv) need v from each parent, and $302 : 98$ is about 3 : 1, the ratio from Vv × Vv.
A homozygous recessive gives only recessive gametes, so it cannot mask anything: each offspring's phenotype shows the allele from the tested parent. A homozygous dominant parent would make every offspring show the dominant phenotype, hiding the answer.
Rr. The ratio is about 1 : 1, as in Rr × rr. If the parent were RR, all the offspring would be round.
No. Each fertilisation is independent, so the chance is still $\frac{1}{4}$ for the fourth child, the same as for each earlier child.

Key points

Set out every cross: phenotypes, genotypes, gametes, Punnett square, offspring genotypes and phenotypes, with ratios.
Homozygous × homozygous recessive: all offspring show the dominant phenotype. Heterozygote × heterozygote: 3 : 1. Heterozygote × homozygous recessive: 1 : 1.
A test cross with a homozygous recessive reveals whether a dominant-looking organism is homozygous or heterozygous.
Ratios are probabilities; they fit best when numbers are large.

Sources and credits

Mendel's counts are from Experiments in Plant Hybridization by Gregor Mendel (1866), in the English translation hosted by Electronic Scholarly Publishing. Mendel's 1866 paper is in the public domain.
Painting: Mendel's Experiments: Tall and dwarf pea plants, Daniel J. Fairbanks, via Wikimedia Commons, licensed under CC BY 4.0.
Video: Monohybrids and the Punnett Square Guinea Pigs, Amoeba Sisters, embedded from YouTube.

Written by Ganzaa: the text, the Punnett squares, the worked examples and the practice questions.

1Three to one, again and again

2Setting out a genetic cross

3The test cross: finding a hidden allele

Three to one, again and again

Setting out a genetic cross

The test cross: finding a hidden allele