The Punnett Square Approach For a Monohybrid Cross

The Punnett Square Approach for a Monohybrid Cross

When fertilization occurs between two true-breeding parents that differ in only one characteristic, the process is called a monohybrid cross, and the resulting offspring are monohybrids. Mendel performed seven monohybrid crosses involving contrasting traits for each characteristic. On the basis of his results in F₁ and F₂ generations, Mendel postulated that each parent in the monohybrid cross contributed one of two paired unit factors to each offspring, and every possible combination of unit factors was equally likely.

To demonstrate a monohybrid cross, consider the case of true-breeding pea plants with yellow versus green pea seeds. The dominant seed color is yellow; therefore, the parental genotypes were YY for the plants with yellow seeds and yy for the plants with green seeds, respectively.

A Punnett square, devised by the British geneticist Reginald Punnett, can be drawn that applies the rules of probability to predict the possible outcomes of a genetic cross or mating and their expected frequencies. To prepare a Punnett square, all possible combinations of the parental alleles are listed along the top (for one parent) and side (for the other parent) of a grid, representing their meiotic segregation into haploid gametes.

Then the combinations of egg and sperm are made in the boxes in the table to show which alleles are combining. Each box then represents the diploid genotype of a zygote, or fertilized egg, that could result from this mating. Because each possibility is equally likely, genotypic ratios can be determined from a Punnett square. If the pattern of inheritance (dominant or recessive) is known, the phenotypic ratios can be inferred as well. For a monohybrid cross of two true-breeding parents, each parent contributes one type of allele. In this case, only one genotype is possible. All offspring are Yy and have yellow seeds (see the figure below).

$This illustration shows a monohybrid cross. In the P generation, one parent has a dominant yellow phenotype and the genotype YY, and the other parent has the recessive green phenotype and the genotype yy. Each parent produces one kind of gamete, resulting in an F_{1} generation with a dominant yellow phenotype and the genotype Yy. Self-pollination of the F_{1} generation results in an F_{2} generation with a 3 to 1 ratio of yellow to green peas. One out of three of the yellow pea plants has a dominant genotype of YY, and 2 out of 3 have the heterozygous phenotype Yy. The homozygous recessive plant has the green phenotype and the genotype yy.$

In the P generation, pea plants that are true-breeding for the dominant yellow phenotype are crossed with plants with the recessive green phenotype. This cross produces F₁ heterozygotes with a yellow phenotype. Punnett square analysis can be used to predict the genotypes of the F₂ generation.

A self-cross of one of the Yy heterozygous offspring can be represented in a 2 × 2 Punnett square because each parent can donate one of two different alleles. Therefore, the offspring can potentially have one of four allele combinations: YY, Yy, yY, or yy (see the figure above). Notice that there are two ways to obtain the Yy genotype: a Y from the egg and a y from the sperm, or a y from the egg and a Y from the sperm.

Both of these possibilities must be counted. Recall that Mendel’s pea-plant characteristics behaved in the same way in reciprocal crosses. Therefore, the two possible heterozygous combinations produce offspring that are genotypically and phenotypically identical despite their dominant and recessive alleles deriving from different parents. They are grouped together. Because fertilization is a random event, we expect each combination to be equally likely and for the offspring to exhibit a ratio of YY:Yy:yy genotypes of 1:2:1 (see the figure above).

Furthermore, because the YY and Yy offspring have yellow seeds and are phenotypically identical, applying the sum rule of probability, we expect the offspring to exhibit a phenotypic ratio of 3 yellow:1 green. Indeed, working with large sample sizes, Mendel observed approximately this ratio in every F₂ generation resulting from crosses for individual traits.

Mendel validated these results by performing an F₃ cross in which he self-crossed the dominant- and recessive-expressing F₂ plants. When he self-crossed the plants expressing green seeds, all of the offspring had green seeds, confirming that all green seeds had homozygous genotypes of yy.

When he self-crossed the F₂ plants expressing yellow seeds, he found that one-third of the plants bred true, and two-thirds of the plants segregated at a 3:1 ratio of yellow:green seeds. In this case, the true-breeding plants had homozygous (YY) genotypes, whereas the segregating plants corresponded to the heterozygous (Yy) genotype. When these plants self-fertilized, the outcome was just like the F₁ self-fertilizing cross.

The Test Cross Distinguishes the Dominant Phenotype

Beyond predicting the offspring of a cross between known homozygous or heterozygous parents, Mendel also developed a way to determine whether an organism that expressed a dominant trait was a heterozygote or a homozygote. Called the test cross, this technique is still used by plant and animal breeders. In a test cross, the dominant-expressing organism is crossed with an organism that is homozygous recessive for the same characteristic.

If the dominant-expressing organism is a homozygote, then all F₁ offspring will be heterozygotes expressing the dominant trait (see the figure below). Alternatively, if the dominant expressing organism is a heterozygote, the F₁ offspring will exhibit a 1:1 ratio of heterozygotes and recessive homozygotes (see the figure below). The test cross further validates Mendel’s postulate that pairs of unit factors segregate equally.

Art Connection

In a test cross, a parent with a dominant phenotype but unknown genotype is crossed with a recessive parent. If the parent with the unknown phenotype is homozygous dominant, all of the resulting offspring will have at least one dominant allele. If the parent with the unknown phenotype is heterozygous, fifty percent of the offspring will inherit a recessive allele from both parents and will have the recessive phenotype.

A test cross can be performed to determine whether an organism expressing a dominant trait is a homozygote or a heterozygote.

In pea plants, round peas (R) are dominant to wrinkled peas (r). You do a test cross between a pea plant with wrinkled peas (genotype rr) and a plant of unknown genotype that has round peas. You end up with three plants, all which have round peas. From this data, can you tell if the round pea parent plant is homozygous dominant or heterozygous? If the round pea parent plant is heterozygous, what is the probability that a random sample of 3 progeny peas will all be round?

Answer

You cannot be sure if the plant is homozygous or heterozygous as the data set is too small: by random chance, all three plants might have acquired only the dominant gene even if the recessive one is present. If the round pea parent is heterozygous, there is a one-eighth probability that a random sample of three progeny peas will all be round.

Many human diseases are genetically inherited. A healthy person in a family in which some members suffer from a recessive genetic disorder may want to know if he or she has the disease-causing gene and what risk exists of passing the disorder on to his or her offspring. Of course, doing a test cross in humans is unethical and impractical. Instead, geneticists use pedigree analysis to study the inheritance pattern of human genetic diseases (see the figure below).

Art Connection

This is a pedigree of a family that carries the recessive disorder alkaptonuria. In the second generation, an unaffected mother and an affected father have three children. One child has the disorder, so the genotype of the mother must be Aa and the genotype of the father is aa. One unaffected child goes on to have two children, one affected and one unaffected. Because her husband was not affected, she and her husband must both be heterozygous. The genotype of their unaffected child is unknown, and is designated A?. In the third generation, the other unaffected child had no offspring, and his genotype is therefore also unknown. The affected third-generation child goes on to have one child with the disorder. Her husband is unaffected and is labeled “3.” The first generation father is affected and is labeled “1.” The first generation mother is unaffected and is labeled “2.” The Art Connection question asks the genotype of the three numbered individuals.

Alkaptonuria is a recessive genetic disorder in which two amino acids, phenylalanine and tyrosine, are not properly metabolized. Affected individuals may have darkened skin and brown urine, and may suffer joint damage and other complications. In this pedigree, individuals with the disorder are indicated in blue and have the genotype aa. Unaffected individuals are indicated in yellow and have the genotype AA or Aa. Note that it is often possible to determine a person’s genotype from the genotype of their offspring. For example, if neither parent has the disorder but their child does, they must be heterozygous. Two individuals on the pedigree have an unaffected phenotype but unknown genotype. Because they do not have the disorder, they must have at least one normal allele, so their genotype gets the “A?” designation.

What are the genotypes of the individuals labeled 1, 2 and 3?

Answer

Individual 1 has the genotype aa. Individual 2 has the genotype Aa. Individual 3 has the genotype Aa.

The Punnett Square Approach For a Monohybrid Cross