Learning Objectives
By the end of this section, you will be able to:
- Explain laws developed by Mendel
- Use the forked-line method to calculate the probability of genotypes and phenotypes from multiple gene crosses
- Explain the effect of linkage and recombination on gamete genotypes
- Explain the phenotypic outcomes of epistatic effects between genes
Mendel generalized the results of his pea-plant experiments into four laws that describe the basis of dominant and recessive inheritance in diploid organisms. More complex extensions of Mendelian genetics exist that do not exhibit the same F2 phenotypic ratios (3:1). Nevertheless, these laws summarize the basics of classical genetics.
Paired Genes
Mendel proposed that paired factors of heredity were transmitted from generation to generation during gamete formation and fertilization,. During his research, Mendel deduced that hereditary factors must be inherited as discrete units, or genes. This contradicted the belief at that time that parental traits were blended in the offspring.
Alleles Can Be Dominant or Recessive
![Photo shows an albino child with his black mother.](https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/198/2016/11/28185201/Figure_12_03_01.jpg)
Figure 1. The child in the photo expresses albinism, a recessive trait.
Mendel’s law of dominance states that in a heterozygote, one trait will conceal the presence of another trait for the same characteristic. The dominant allele will be expressed exclusively. The recessive allele will remain “latent” but can be passed on in subsequent generations. The recessive trait will only be expressed by offspring that have two copies of this allele(Figure 1). With our continued advances in technology, researchers have discovered that this law does not always hold true.
Law of Segregation
Based on his research, Mendel proposed the law of segregation. This law states that paired factors (genes) must segregate equally into gametes such that offspring have an equal likelihood of inheriting either factor. The law supports what Mendel observed in his F1 and F2 data. The equal segregation of alleles is the reason we can apply the Punnett square to accurately predict the offspring of parents with known genotypes. The physical basis of Mendel’s law of segregation is the first division of meiosis, where the homologous chromosomes with their different versions of each gene are segregated into daughter nuclei. Since meiosis was not understood by the scientific community during Mendel’s lifetime, his work cannot be fully appreciated.
Law of Independent Assortment
Mendel’s law of independent assortment states that genes do not influence each other with regard to the sorting of alleles into gametes, and every possible combination of alleles for every gene is equally likely to occur. The independent assortment of genes can be illustrated by the dihybrid cross. A dihybrid is between two true-breeding parents that express different traits for two characteristics. Consider the characteristics of seed color and seed texture for two pea plants. One plant has green, wrinkled seeds (yyrr) and another that has yellow, round seeds (YYRR). Because each parent is homozygous, the law of segregation indicates that the gametes for the green/wrinkled plant all are yr, and the gametes for the yellow/round plant are all YR. Therefore, the F1 generation of offspring all are YyRr (Figure 2).
Art Connection
![This illustration shows a dihybrid cross between pea plants. In the P generation, a plant that has the homozygous dominant phenotype of round, yellow peas is crossed with a plant with the homozygous recessive phenotype of wrinkled, green peas. The resulting F_{1} offspring have a heterozygous genotype and round, yellow peas. Self-pollination of the F_{1} generation results in F_{2} offspring with a phenotypic ratio of 9:3:3:1 for yellow round, green round, yellow wrinkled and green wrinkled peas, respectively.](https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/198/2016/11/28185242/Figure_12_03_02.jpg)
Figure 2. This dihybrid cross of pea plants involves the genes for seed color and texture.
In pea plants, yellow (Y) are dominant to green flowers(y) and round peas (R) are dominant to wrinkled as (r). What are the possible genotypes and phenotypes for a cross between YyRR and yyRr pea plants? How many squares do you need to do a Punnett square analysis of this cross?
For the F2 generation, the law of segregation requires that each gamete receive either an R allele or an r allele along with either a Y allele or a y allele. The law of independent assortment states that a gamete with an r allele would be equally likely to contain either a Y allele or a y allele. There are four equally likely gametes that can be formed when the YyRr heterozygote is self-crossed: YR, Yr, yR, and yr. Arranging these gametes along the top and left of a 4 × 4 Punnett square (Figure 2) gives us 16 equally likely genotypic combinations. From these genotypes, we infer a phenotypic ratio of 9 round/yellow:3 round/green:3 wrinkled/yellow:1 wrinkled/green (Figure 2). These are the offspring ratios we would expect.
Because of independent assortment and dominance, the 9:3:3:1 dihybrid phenotypic ratio can be collapsed into two 3:1 ratios, characteristic of any monohybrid cross that follows a dominant and recessive pattern. Considering only seed texture in the above dihybrid cross, we would expect that three quarters of the F2 generation offspring would be round, and one quarter would be wrinkled. Similarly, isolating only seed color, we would assume that three quarters of the F2 offspring would be yellow and one quarter would be green. The sorting of alleles for texture and color are independent events, so we can apply the product rule. The proportion of round and yellow F2 offspring is expected to be (3/4) × (3/4) = 9/16, and the proportion of wrinkled and green offspring is expected to be (1/4) × (1/4) = 1/16. These proportions are identical to those obtained using a Punnett square. Round, green and wrinkled, yellow offspring can also be calculated using the product rule, as each of these genotypes includes one dominant and one recessive phenotype. Therefore, the proportion of each is calculated as (3/4) × (1/4) = 3/16.
The law of independent assortment also indicates that a cross between yellow, wrinkled (YYrr) and green, round (yyRR) parents would yield the same F1 and F2 offspring as in the YYRR x yyrr cross.
Meiosis I provides the physical basis for the law of independent assortment. During Meiosis I, the different homologous pairs line up in random orientations. Each gamete can contain any combination of paternal and maternal chromosomes due to the randomness of the tetrads in metaphase.
Forked-Line Method
When considering more than two genes, the Punnett-square method is unmanageable. Examining a cross involving four genes would require a 16 × 16 grid containing 256 boxes. It would be extremely cumbersome to manually enter each genotype. For more complex crosses, the forked-line and probability methods are preferred.
To prepare a forked-line diagram for a cross between F1 heterozygotes resulting from a cross between AABBCC and aabbcc parents, we first create rows equal to the number of genes being considered We then segregate the alleles in each row on forked lines according to the probabilities for individual monohybrid crosses (Figure 3). We multiply the values along each forked path to obtain the F2 offspring probabilities. This process is a diagrammatic version of the product rule. The values along each forked pathway can be multiplied because each gene assorts independently. For a trihybrid cross, the F2 phenotypic ratio is 27:9:9:9:3:3:3:1.
![A forked-line diagram is shown for the F_{2} generation of a trihybrid cross of pea plants with the dominant yellow, round, and tall phenotype with pea plants of the recessive green, wrinkled, dwarf phenotype. The top row shows that the color ratio is 3 yellow to 1 green in the F_{2} generation. The second row shows that the probability that plants of either pea color having the round or wrinkled texture is 3 to 1. The third row shows that the probability of plants with either of the above textures having a round or wrinkled phenotype is 3 to 1. The probability of all three phenotypes occurring together is determined by multiplying each individual probability together. The probability ratio is 27 yellow/round/tall: 9 yellow/round/dwarf: 9 yellow/wrinked/tall: 3 yellow/wrinkled/dwarf: 9 green/round/tall: 3 green/round/dwarf: 3 green/wrinkled/tall: 1 green/wrinkled/dwarf.](https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/198/2016/11/28185418/Figure_12_03_03-1024x365.jpg)
Figure 3. The forked-line method can be used to analyze a trihybrid cross. The probability for color in the F2 generation occupies the top row (3 yellow:1 green). The probability for shape occupies the second row (3 round:1 wrinked), and the probability for height occupies the third row (3 tall:1 dwarf). The probability for each possible combination of traits is calculated by multiplying the probability for each individual trait. Thus, the probability of F2 offspring having yellow, round, and tall traits is 3 × 3 × 3, or 27.
Rules for Multihybrid Fertilization
Predicting the genotypes and phenotypes of offspring from given crosses is the best way to test your knowledge of Mendelian genetics. Given a multihybrid cross that obeys independent assortment and follows a dominant and recessive pattern, several generalized rules exist. You can use these rules to check your results as you work through genetics calculations (Table 1). To apply these rules, you must determine n, the number of heterozygous gene pairs. A cross between AaBb and AaBb heterozygotes has an n of 2, while a cross between AABb and AABb has an n of 1 because A is not heterozygous.
Table 1. General Rules for Multihybrid Crosses | |
---|---|
General Rule | Number of Heterozygous Gene Pairs |
Number of different F1 gametes | 2n |
Number of different F2 genotypes | 3n |
Given dominant and recessive inheritance, the number of different F2 phenotypes | 2n |
Linked Genes
All of Mendel’s pea characteristics followed the law of independent assortment. But we now know that some allele combinations are not inherited independently of each other. Genes that are located on separate non-homologous chromosomes will always sort independently. Each chromosome contains hundreds or thousands of genes, organized on chromosomes like beads on a string. The segregation of alleles into gametes can be influenced by linkage. Linked genes are located physically close to each other on the same chromosome are more likely to be inherited as a pair. Because of the process of recombination, or “crossover,” it is possible for two genes on the same chromosome to behave independently as if not linked. Let’s consider the biological basis of gene linkage and recombination.
![This illustration shows a pair of homologous chromosomes. One of the pair has the alleles ABC and the other has the alleles abc. During meiosis, crossover occurs between two of the chromosomes and genetic material is exchanged, resulting in one recombinant chromosome that has the alleles ABc and another that has the alleles abC. The other two chromosomes are non-recombinant and have the same arrangement of genes as before meiosis.](https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/198/2016/11/28185525/Figure_12_03_04.jpg)
Figure 4. The process of crossing over, or recombination, occurs when two homologous chromosomes align during meiosis and exchange a segment of genetic material. Here, the alleles for gene C were exchanged. The result is two recombinant and two non-recombinant chromosomes.
Homologous chromosomes possess the same genes in the same linear order. The alleles may differ on homologous chromosome pairs, but the genes to which they correspond do not. In preparation for the first division of meiosis, homologous chromosomes replicate and synapse. Like genes on the homologs align with each other. At this stage, segments of homologous chromosomes exchange linear segments of genetic material(crossing over) (Figure 4). This process is known as recombination and is a common genetic process. Because the genes are aligned during recombination, the gene order is not altered. The result of recombination is that maternal and paternal alleles are combined onto the same chromosome. Across a given chromosome, several recombination events may occur.
Mendel did not mention linkage in his research. Many researchers question whether he encountered linkage, but chose not to publish those crosses out of concern that they would invalidate his data. The garden pea has seven chromosomes and it has been suggested that his choice of seven characteristics was not a coincidence. Even if the genes he examined were not located on separate chromosomes, it is possible that he simply did not observe linkage due to the extensive shuffling effects of recombination.
Scientific Method Connection
Testing the Hypothesis of Independent Assortment
To better appreciate the amount of labor and ingenuity that went into Mendel’s experiments, proceed through one of Mendel’s dihybrid crosses.
Question: What will be the offspring of a dihybrid cross?
Background: Consider that pea plants mature in one growing season, and you have access to a large garden in which you can cultivate thousands of pea plants. There are several true-breeding plants with the following pairs of traits: tall plants with inflated pods and dwarf plants with constricted pods. Before the plants have matured, you remove the pollen-producing organs from the tall/inflated plants in your crosses to prevent self-fertilization. Upon plant maturation, the plants are manually crossed by transferring pollen from the dwarf/constricted plants to the stigmata of the tall/inflated plants.
Hypothesis: Both trait pairs will sort independently according to Mendelian laws. When the true-breeding parents are crossed, all of the F1 offspring are tall and have inflated pods, which indicates that the tall and inflated traits are dominant over the dwarf and constricted traits, respectively. A self-cross of the F1 heterozygotes results in 2,000 F2 progeny.
Test the hypothesis: Because each trait pair sorts independently, the ratios of tall:dwarf and inflated:constricted are each expected to be 3:1. The tall/dwarf trait pair is called T/t, and the inflated/constricted trait pair is designated I/i. Each member of the F1 generation therefore has a genotype of TtIi. Construct a grid in which you cross two TtIi individuals. Each individual can donate four combinations of two traits: TI, Ti, tI, or ti, meaning that there are 16 possibilities of offspring genotypes. Because the T and I alleles are dominant, any individual having one or two of those alleles will express the tall or inflated phenotypes, respectively, regardless if they also have a t or i allele. Only individuals that are tt or ii will express the dwarf and constricted alleles. As shown in Figure 5, you predict that you will observe the following offspring proportions: tall/inflated:tall/constricted:dwarf/inflated:dwarf/constricted in a 9:3:3:1 ratio. Notice from the grid that when considering the tall/dwarf and inflated/constricted trait pairs in isolation, they are each inherited in 3:1 ratios.
![This figure shows all possible combinations of offspring resulting from a dihybrid cross of pea plants that are heterozygous for the tall/dwarf and inflated/constricted alleles.](https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/198/2016/11/28185611/Figure_12_03_05.jpg)
Figure 5. This figure shows all possible combinations of offspring resulting from a dihybrid cross of pea plants that are heterozygous for the tall/dwarf and inflated/constricted alleles.
Test the hypothesis: You cross the dwarf and tall plants and then self-cross the offspring. For best results, this is repeated with hundreds or even thousands of pea plants. What special precautions should be taken in the crosses and in growing the plants?
Analyze your data: You observe the following plant phenotypes in the F2 generation: 2706 tall/inflated, 930 tall/constricted, 888 dwarf/inflated, and 300 dwarf/constricted. Reduce these findings to a ratio and determine if they are consistent with Mendelian laws.
Form a conclusion: Were the results close to the expected 9:3:3:1 phenotypic ratio? Do the results support the prediction? What might be observed if far fewer plants were used, given that alleles segregate randomly into gametes? Try to imagine growing that many pea plants, and consider the potential for experimental error. For instance, what would happen if it was extremely windy one day?
Epistasis
Mendel implied that the sum of an individual’s phenotype was controlled by genes, such that every characteristic was distinctly and completely controlled by a single gene. We know that this is not always the case.
Link to Learning
Eye color in humans is determined by multiple genes. Use the Eye Color Calculator to predict the eye color of children from parental eye color.
In epistasis, the interaction between genes is antagonistic. One gene masks or interferes with the expression of another. “Epistasis” is a word composed of Greek roots that mean “standing upon.” The alleles being masked or silenced are said to be hypostatic to the epistatic alleles doing the masking.
An example of epistasis is pigmentation in mice. The wild-type coat color, agouti (AA), is dominant to solid-colored fur (aa). A separate gene (C) is necessary for pigment production. A mouse with a recessive c allele is unable to produce pigment and is albino, regardless of the allele present at locus A (Figure 6). Therefore, the genotypes AAcc, Aacc, and aacc all produce the same albino phenotype. A cross between heterozygotes for both genes (AaCc x AaCc) would generate offspring with a phenotypic ratio of 9 agouti:3 solid color:4 albino (Figure 6). In this case, the C gene is epistatic to the A gene.
![A cross between two agouti mice with the heterozygous genotype AaCc is shown. Each mouse produces four different kinds of gametes (AC, aC, Ac, and ac). A 4 × 4 Punnett square is used to determine the genotypic ratio of the offspring. The phenotypic ratio is 9/16 agouti, 3/16 black, and 4/16 white.](https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/198/2016/11/28185647/Figure_12_03_06.jpg)
Figure 6. In mice, the agouti coat color (A) is dominant to a solid coloration, such as black or gray. A gene at a separate locus (C) is responsible for pigment production. The recessive c allele does not produce pigment, and a mouse with the homozygous recessive cc genotype is albino regardless of the allele present at the A locus. Thus, the C gene is epistatic to the A gene.
Epistasis can also occur when a dominant allele masks expression at a separate gene. Fruit color in summer squash is expressed in this way. Homozygous recessive expression of the W gene (ww) coupled with homozygous dominant or heterozygous expression of the Y gene (YY or Yy) generates yellow fruit, and the wwyy genotype produces green fruit. However, if a dominant copy of the W gene is present in the homozygous or heterozygous form, the summer squash will produce white fruit regardless of the Y alleles. A cross between white heterozygotes for both genes (WwYy × WwYy) would produce offspring with a phenotypic ratio of 12 white:3 yellow:1 green.
Epistasis can be reciprocal such that either gene, when present either form, expresses the same phenotype. In the shepherd’s purse plant, seed shape is controlled by two genes in a dominant epistatic relationship. When the genes A and B are both homozygous recessive (aabb), the seeds are ovoid. If the dominant allele for either of these genes is present, the result is triangular seeds. Every possible genotype, other than aabb, results in triangular seeds. A cross between heterozygotes for both genes (AaBb x AaBb) would yield offspring with a phenotypic ratio of 15 triangular:1 ovoid.
When working genetics problems, keep in mind that any single characteristic that results in a phenotypic ratio that totals 16 is typical of a two-gene interaction. Remember the phenotypic inheritance pattern for Mendel’s dihybrid cross, which considered two non-interacting genes—9:3:3:1. Similarly, we would expect interacting gene pairs to also exhibit ratios expressed as 16 parts. We are assuming the interacting genes are not linked but still assorting independently into gametes.
Link to Learning
For an excellent review of Mendel’s experiments and to perform your own crosses and identify patterns of inheritance, visit the Mendel’s Peas web lab.
Section Summary
Mendel postulated that genes (characteristics) are inherited as pairs of alleles (traits) that behave in a dominant and recessive pattern. Alleles segregate into gametes such that each gamete is equally likely to receive either one of the two alleles present in a diploid individual. Genes are assorted into gametes independently of one another. What one gene does has no effect on what another gene does. A dihybrid cross demonstrates independent assortment when the genes in question are on different chromosomes or distant from each other on the same chromosome. For crosses involving more than two genes, the forked line is better able to predict offspring genotypes and phenotypes than a Punnett square.
Although chromosomes sort independently into gametes during meiosis, Mendel’s law of independent assortment refers to genes, not chromosomes. When genes are located in close proximity on the same chromosome, their alleles tend to be inherited together. This results in offspring ratios that violate Mendel’s law of independent assortment. Recombination serves to exchange genetic material on homologous chromosomes such that maternal and paternal alleles may be recombined on the same chromosome. This is why alleles on a given chromosome are not always inherited together. Recombination is a random event occurring anywhere on a chromosome. Genes that are far apart on the same chromosome are likely to still assort independently because of recombination events that occurred in the intervening chromosomal space.
Whether or not they are sorting independently, genes may interact at the level of gene products such that the expression of an allele for one gene masks or modifies the expression of an allele for a different gene in a process known as epistasis.
Additional Self Check Questions
- In pea plants, purple flowers (P) are dominant to white flowers (p) and yellow peas (Y) are dominant to green peas (y). What are the possible genotypes and phenotypes for a cross between PpYY and ppYy pea plants? How many squares do you need to do a Punnett square analysis of this cross?
- Explain epistatis in terms of its Greek-language roots “standing upon.”
Answers
- The possible genotypes are PpYY, PpYy, ppYY, and ppYy. The former two genotypes would result in plants with purple flowers and yellow peas, while the latter two genotypes would result in plants with white flowers with yellow peas, for a 1:1 ratio of each phenotype. You only need a 2 × 2 Punnett square (four squares total) to do this analysis because two of the alleles are homozygous.
- Epistasis describes an antagonistic interaction between genes wherein one gene masks or interferes with the expression of another. The gene that is interfering is referred to as epistatic, as if it is “standing upon” the other (hypostatic) gene to block its expression.