Predicting progeny of recessive mutations using recombination

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I was asked this question on a test and got it wrong, but I'd like to know how to do it. The answers are shown in the blanks below:

You are studying two recessive mutations in the fruit fly D. melanogaster. The red_ mutation causes flies to have red bristles (wild-type flies have black bristles) and the shiny- mutation causes flies to have shiny eyes (wild-type flies have eyes that are pebbly). You mate females from a true-breeding strain with red bristles and pebbly (wild-type) eyes to males from a true-breeding strain with black (wild-type) bristles and shiny eyes. F1 females are then mated to males that have red bristles and shiny eyes to produce F2 progeny.

If you analyzed 1000 male progeny in the F2 generation, how many flies of each possible phenotypic class would you expect, if:

The two traits are determined by two autosomal genes that are 15 cM apart:

``Red bristles, pebbly eyes:_____ 425_______ Red bristles, shiny eyes: ______75____ Black bristles, pebbly eyes:____ 75_____ Black bristles, shiny eyes:____ 425______``

My attempt:

So I know the 15cm=recombinants/parentals * 100 but I don't see how that formula can help.

First, the recombination equation is:

``cM = recombinants/(recombinants AND parentals) * 100``

Let's assign the following genotypes for clarity's sake:

``RR = black bristles Rr = black bristles rr = red bristles SS = pebbly eyes Ss = pebbly eyes ss = shiny eyes F1 cross: rrSS x RRss = RrSs (all progeny) F2 cross: RrSs (F1) x rrss = ?``

You would have the following gametes:

``Parent 1, RrSs = RS, Rs, rS, rs Parent 2, rrss = rs, rs, rs, rs``

Draw your punnet square and you get the following genotypes of the offspring.

Let's translate those to phenotypes.

``RrSs: black pebbly (parental) Rrss: black shiny (recombinant) rrSs: red pebbly (recombinant) rrss: red shiny (parental)``

Now let's look at recombination frequency.

``15cM = (recombinants/(recombinants+parentals)*100 recombinants/(parentals+recombinants) = 15/100``

This tells you for every 15 recombinants, you have 85 parentals. If you scale this up to 1000, that's 150 recombinants for every 850 parentals.

Your 150 recombinants comes from 75 black/shiny and 75 red/pebbly.

Your 850 parentals ceoms from 425 black/pebbly and 425 red/shiny.

Anyways that was my attempt which appears to be directly opposite to what your answer key says. Did you copy it down correctly?

This tutorial will teach you how to predict the segregation of alleles in parents that are heterozygous for different characters. Initially you will work with a tool referred to as a Punnett square, but later you will see how deriving probabilities can help you make the same predictions in a much easier manner. By the end of this tutorial you should have a working understanding of:

• The application of Punnett squares for monohybrid and dihybrid crosses
• Why test crosses are used to determine some genotypes
• Basic probablity theory
• When to apply the Rule of Multiplication and the Rule of Addition

61 Laws of Inheritance

By the end of this section, you will be able to do the following:

• Explain Mendel’s law of segregation and independent assortment in terms of genetics and the events of meiosis
• Use the forked-line method and the probability rules 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 postulates, some of which are sometimes called “laws,” that describe the basis of dominant and recessive inheritance in diploid organisms. As you have learned, more complex extensions of Mendelism exist that do not exhibit the same F2 phenotypic ratios (3:1). Nevertheless, these laws summarize the basics of classical genetics.

Pairs of Unit Factors, or Genes

Mendel proposed first that paired unit factors of heredity were transmitted faithfully from generation to generation by the dissociation and reassociation of paired factors during gametogenesis and fertilization, respectively. After he crossed peas with contrasting traits and found that the recessive trait resurfaced in the F2 generation, Mendel deduced that hereditary factors must be inherited as discrete units. This finding contradicted the belief at that time that parental traits were blended in the offspring.

Alleles Can Be Dominant or Recessive

Mendel’s law of dominance states that in a heterozygote, one trait will conceal the presence of another trait for the same characteristic. Rather than both alleles contributing to a phenotype, the dominant allele will be expressed exclusively. The recessive allele will remain “latent” but will be transmitted to offspring by the same manner in which the dominant allele is transmitted. The recessive trait will only be expressed by offspring that have two copies of this allele ((Figure)), and these offspring will breed true when self-crossed.

Since Mendel’s experiments with pea plants, researchers have found that the law of dominance does not always hold true. Instead, several different patterns of inheritance have been found to exist.

Equal Segregation of Alleles

Observing that true-breeding pea plants with contrasting traits gave rise to F1 generations that all expressed the dominant trait and F2 generations that expressed the dominant and recessive traits in a 3:1 ratio, Mendel proposed the law of segregation . This law states that paired unit factors (genes) must segregate equally into gametes such that offspring have an equal likelihood of inheriting either factor. For the F2 generation of a monohybrid cross, the following three possible combinations of genotypes could result: homozygous dominant, heterozygous, or homozygous recessive. Because heterozygotes could arise from two different pathways (receiving one dominant and one recessive allele from either parent), and because heterozygotes and homozygous dominant individuals are phenotypically identical, the law supports Mendel’s observed 3:1 phenotypic ratio. 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, in which the homologous chromosomes with their different versions of each gene are segregated into daughter nuclei. The role of the meiotic segregation of chromosomes in sexual reproduction was not understood by the scientific community during Mendel’s lifetime.

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 cross 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 that 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)).

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?

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 into which an r allele sorted would be equally likely to contain either a Y allele or a y allele. Thus, there are four equally likely gametes that can be formed when the YyRr heterozygote is self-crossed, as follows: YR, Yr, yR, and yr. Arranging these gametes along the top and left of a 4 × 4 Punnett square ((Figure)) 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)). These are the offspring ratios we would expect, assuming we performed the crosses with a large enough sample size.

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. Ignoring seed color and 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. Therefore, 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.

The physical basis for the law of independent assortment also lies in meiosis I, in which the different homologous pairs line up in random orientations. Each gamete can contain any combination of paternal and maternal chromosomes (and therefore the genes on them) because the orientation of tetrads on the metaphase plane is random.

Forked-Line Method

When more than two genes are being considered, the Punnett-square method becomes unwieldy. For instance, 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, and then segregate the alleles in each row on forked lines according to the probabilities for individual monohybrid crosses ((Figure)). We then multiply the values along each forked path to obtain the F2 offspring probabilities. Note that 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.

Probability Method

While the forked-line method is a diagrammatic approach to keeping track of probabilities in a cross, the probability method gives the proportions of offspring expected to exhibit each phenotype (or genotype) without the added visual assistance. Both methods make use of the product rule and consider the alleles for each gene separately. Earlier, we examined the phenotypic proportions for a trihybrid cross using the forked-line method now we will use the probability method to examine the genotypic proportions for a cross with even more genes.

For a trihybrid cross, writing out the forked-line method is tedious, albeit not as tedious as using the Punnett-square method. To fully demonstrate the power of the probability method, however, we can consider specific genetic calculations. For instance, for a tetrahybrid cross between individuals that are heterozygotes for all four genes, and in which all four genes are sorting independently and in a dominant and recessive pattern, what proportion of the offspring will be expected to be homozygous recessive for all four alleles? Rather than writing out every possible genotype, we can use the probability method. We know that for each gene, the fraction of homozygous recessive offspring will be 1/4. Therefore, multiplying this fraction for each of the four genes, (1/4) × (1/4) × (1/4) × (1/4), we determine that 1/256 of the offspring will be quadruply homozygous recessive.

For the same tetrahybrid cross, what is the expected proportion of offspring that have the dominant phenotype at all four loci? We can answer this question using phenotypic proportions, but let’s do it the hard way—using genotypic proportions. The question asks for the proportion of offspring that are 1) homozygous dominant at A or heterozygous at A, and 2) homozygous at B or heterozygous at B, and so on. Noting the “or” and “and” in each circumstance makes clear where to apply the sum and product rules. The probability of a homozygous dominant at A is 1/4 and the probability of a heterozygote at A is 1/2. The probability of the homozygote or the heterozygote is 1/4 + 1/2 = 3/4 using the sum rule. The same probability can be obtained in the same way for each of the other genes, so that the probability of a dominant phenotype at A and B and C and D is, using the product rule, equal to 3/4 × 3/4 × 3/4 × 3/4, or 27/64. If you are ever unsure about how to combine probabilities, returning to the forked-line method should make it clear.

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 ((Figure)). To apply these rules, first you must determine n, the number of heterozygous gene pairs (the number of genes segregating two alleles each). For example, a cross between AaBb and AaBb heterozygotes has an n of 2. In contrast, a cross between AABb and AABb has an n of 1 because A is not heterozygous.

General Rules for Multihybrid Crosses
General Rule Number of Heterozygous Gene Pairs
Number of different F1 gametes 2 n
Number of different F2 genotypes 3 n
Given dominant and recessive inheritance, the number of different F2 phenotypes 2 n

Linked Genes Violate the Law of Independent Assortment

Although all of Mendel’s pea characteristics behaved according to the law of independent assortment, 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. However, each chromosome contains hundreds or thousands of genes, organized linearly on chromosomes like beads on a string. The segregation of alleles into gametes can be influenced by linkage , in which genes that are located physically close to each other on the same chromosome are more likely to be inherited as a pair. However, because of the process of recombination, or “crossover,” it is possible for two genes on the same chromosome to behave independently, or as if they are not linked. To understand this, let’s consider the biological basis of gene linkage and recombination.

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 ((Figure)). This process is called recombination, or crossover, and it is a common genetic process. Because the genes are aligned during recombination, the gene order is not altered. Instead, 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, causing extensive shuffling of alleles.

When two genes are located in close proximity on the same chromosome, they are considered linked, and their alleles tend to be transmitted through meiosis together. To exemplify this, imagine a dihybrid cross involving flower color and plant height in which the genes are next to each other on the chromosome. If one homologous chromosome has alleles for tall plants and red flowers, and the other chromosome has genes for short plants and yellow flowers, then when the gametes are formed, the tall and red alleles will go together into a gamete and the short and yellow alleles will go into other gametes. These are called the parental genotypes because they have been inherited intact from the parents of the individual producing gametes. But unlike if the genes were on different chromosomes, there will be no gametes with tall and yellow alleles and no gametes with short and red alleles. If you create the Punnett square with these gametes, you will see that the classical Mendelian prediction of a 9:3:3:1 outcome of a dihybrid cross would not apply. As the distance between two genes increases, the probability of one or more crossovers between them increases, and the genes behave more like they are on separate chromosomes. Geneticists have used the proportion of recombinant gametes (the ones not like the parents) as a measure of how far apart genes are on a chromosome. Using this information, they have constructed elaborate maps of genes on chromosomes for well-studied organisms, including humans.

Mendel’s seminal publication makes no mention of linkage, and many researchers have questioned whether he encountered linkage but chose not to publish those crosses out of concern that they would invalidate his independent assortment postulate. The garden pea has seven chromosomes, and some have suggested that his choice of seven characteristics was not a coincidence. However, even if the genes he examined were not located on separate chromosomes, it is possible that he simply did not observe linkage because of the extensive shuffling effects of recombination.

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 analogous to (Figure), 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, respectively. As shown in (Figure), 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.

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’s studies in pea plants implied that the sum of an individual’s phenotype was controlled by genes (or as he called them, unit factors), such that every characteristic was distinctly and completely controlled by a single gene. In fact, single observable characteristics are almost always under the influence of multiple genes (each with two or more alleles) acting in unison. For example, at least eight genes contribute to eye color in humans.

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 some cases, several genes can contribute to aspects of a common phenotype without their gene products ever directly interacting. In the case of organ development, for instance, genes may be expressed sequentially, with each gene adding to the complexity and specificity of the organ. Genes may function in complementary or synergistic fashions, such that two or more genes need to be expressed simultaneously to affect a phenotype. Genes may also oppose each other, with one gene modifying the expression of another.

In epistasis , the interaction between genes is antagonistic, such that one gene masks or interferes with the expression of another. “Epistasis” is a word composed of Greek roots that mean “standing upon.” The alleles that are being masked or silenced are said to be hypostatic to the epistatic alleles that are doing the masking. Often the biochemical basis of epistasis is a gene pathway in which the expression of one gene is dependent on the function of a gene that precedes or follows it in the pathway.

An example of epistasis is pigmentation in mice. The wild-type coat color, agouti (AA), is dominant to solid-colored fur (aa). However, a separate gene (C) is necessary for pigment production. A mouse with a recessive c allele at this locus is unable to produce pigment and is albino regardless of the allele present at locus A ((Figure)). 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)). In this case, 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.

Finally, epistasis can be reciprocal such that either gene, when present in the dominant (or recessive) form, expresses the same phenotype. In the shepherd’s purse plant (Capsella bursa-pastoris), the characteristic of 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. That is, every possible genotype other than aabb results in triangular seeds, and a cross between heterozygotes for both genes (AaBb x AaBb) would yield offspring with a phenotypic ratio of 15 triangular:1 ovoid.

As you work through 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. Recall the phenotypic inheritance pattern for Mendel’s dihybrid cross, which considered two noninteracting genes—9:3:3:1. Similarly, we would expect interacting gene pairs to also exhibit ratios expressed as 16 parts. Note that we are assuming the interacting genes are not linked they are still assorting independently into gametes.

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. In addition, genes are assorted into gametes independently of one another. That is, alleles are generally not more likely to segregate into a gamete with a particular allele of another gene. 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, use the forked line or probability methods to predict offspring genotypes and phenotypes rather than a Punnett square.

Although chromosomes sort independently into gametes during meiosis, Mendel’s law of independent assortment refers to genes, not chromosomes, and a single chromosome may carry more than 1,000 genes. 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. However, 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. Therefore, 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. This is called epistasis.

Visual Connection Questions

(Figure) 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?

(Figure) 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.

Multiple Choice

Assuming no gene linkage, in a dihybrid cross of AABB x aabb with AaBb F1 heterozygotes, what is the ratio of the F1 gametes (AB, aB, Ab, ab) that will give rise to the F2 offspring?

The forked line and probability methods make use of what probability rule?

How many different offspring genotypes are expected in a trihybrid cross between parents heterozygous for all three traits when the traits behave in a dominant and recessive pattern? How many phenotypes?

1. 64 genotypes 16 phenotypes
2. 16 genotypes 64 phenotypes
3. 8 genotypes 27 phenotypes
4. 27 genotypes 8 phenotypes

Labrador retrievers’ fur color is controlled by two alleles, E and B. Any dog with the ee__ genotype develops into a yellow lab, while B_E_ dogs become black labs and bbE_ dogs become chocolate labs. This is an example of _____.

Which of the following situations does not follow the Law of Independent Assortment?

1. A blond man and a brunette woman produce three offspring over time, all of who have blond hair.
2. A white cow crossed with a brown bull produces roan cattle.
3. Mating a hog with a sow produces six female piglets.
4. Men are more likely to experience hemophilia than women.

Critical Thinking Questions

Use the probability method to calculate the genotypes and genotypic proportions of a cross between AABBCc and Aabbcc parents.

Considering each gene separately, the cross at A will produce offspring of which half are AA and half are Aa B will produce all Bb C will produce half Cc and half cc. Proportions then are (1/2) × (1) × (1/2), or 1/4 AABbCc continuing for the other possibilities yields 1/4 AABbcc, 1/4 AaBbCc, and 1/4 AaBbcc. The proportions therefore are 1:1:1:1.

Explain epistasis in terms of its Greek-language roots “standing upon.”

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.

In Section 12.3, “Laws of Inheritance,” an example of epistasis was given for the summer squash. Cross white WwYy heterozygotes to prove the phenotypic ratio of 12 white:3 yellow:1 green that was given in the text.

The cross can be represented as a 4 × 4 Punnett square, with the following gametes for each parent: WY, Wy, wY, and wy. For all 12 of the offspring that express a dominant W gene, the offspring will be white. The three offspring that are homozygous recessive for w but express a dominant Y gene will be yellow. The remaining wwyy offspring will be green.

People with trisomy 21 develop Down’s syndrome. What law of Mendelian inheritance is violated in this disease? What is the most likely way this occurs?

In any trisomy disorder, a patient inherits 3 copies of a chromosome instead of the normal pair. This violates the Law of Segregation, and usually occurs when the chromosomes fail to separate during the first round of meiosis.

A heterozygous pea plant produces violet flowers and yellow, round seeds. Describe the expected genotypes of the gametes produced by Mendelian inheritance. If all three genes are found on the same arm of one chromosome should a scientist predict that inheritance patterns will follow Mendelian genetics?

Mendelian inheritance would predict that all three genes are inherited independently. There are therefore 8 different gamete genotype possibilities: VYR, VYr, VyR, Vyr, vYR, vYr, vyR, vyr. If all three genes are found on the same chromosome arm, independent assortment is unlikely to occur because the genes are close together (linked).

Biology Test 6: Genetics and Evolution

A condition in which both alleles of a gene pair in a heterozygote are fully expressed, with neither one being dominant or recessive to the other.

enhance an organism's ability to survive and reproduce in specific environments

changes in traits in populations over time
(page 300)

certain traits increase an individual's success at mating explains the development of traits that improve reproductive success but that may harm the individual

the phenomenon by which allele frequencies in a population change as a result of random events, or chance

results from barriers to successful breeding between population groups in the same area

all 13 species descended from a diverged from just a few ancestral finches

directional: favors the formation of more-extreme traits

Embryology - similar early development (vertebrate embryos with tails)

Vestigial Structure - structures with little to no use (flightless bird wings)

Biogeography - species in nearby geographic regions resemble each other (most Austrailian mammals have pouches)

Molecular Homology - similar DNA and amino acid sequences

microevolution: changes within a single gene pool evolution at the genetic level
example: fish of a single species in a pond vary in length

Homologous Recombination

In 1909, Frans Janssen observed chiasmata—the point at which chromatids are in contact with each other and may exchange segments—prior to the first meiosis division. He suggested that alleles become unlinked and chromosomes physically exchange segments. As chromosomes condensed and paired with their homologs, they appeared to interact at distinct points. Janssen suggested that these points corresponded to regions in which chromosome segments exchanged. We now know that the pairing and interaction between homologous chromosomes, or synapsis, does more than simply organize the homologs for migration to separate daughter cells. When synapsed, homologous chromosomes undergo reciprocal physical exchanges at their arms in homologous recombination , or more simply, “crossing over.”

To better understand the type of experimental results that researchers were obtaining at this time, consider a heterozygous individual that inherited dominant maternal alleles for two genes on the same chromosome (such as AB) and two recessive paternal alleles for those same genes (such as ab). If the genes are linked, one would expect this individual to produce gametes that are either AB or ab with a 1:1 ratio. If the genes are unlinked, the individual should produce AB, Ab, aB, and ab gametes with equal frequencies, according to the Mendelian concept of independent assortment. Because they correspond to new allele combinations, the genotypes Ab and aB are nonparental types that result from homologous recombination during meiosis. Parental types are progeny that exhibit the same allelic combination as their parents. Morgan and his colleagues, however, found that when they test crossed such heterozygous individuals to a homozygous recessive parent (AaBb × aabb), both parental and nonparental cases occurred. For example, 950 offspring might be recovered that were either AaBb or aabb, but 50 offspring would also result that were either Aabb or aaBb. These results suggested that linkage occurred most often, but a significant minority of offspring were the products of recombination.

Art Connection

This figure shows unlinked and linked gene inheritance patterns. In (a), two genes are located on different chromosomes so independent assortment occurs during meiosis. The offspring have an equal chance of being the parental type (inheriting the same combination of traits as the parents) or a nonparental type (inheriting a different combination of traits than the parents). In (b), two genes are very close together on the same chromosome so that no crossing over occurs between them. Therefore, the genes are always inherited together and all the offspring are the parental type. In (c), two genes are far apart on the chromosome such that crossing over occurs during every meiotic event. The recombination frequency will be the same as if the genes were on separate chromosomes. (d) The actual recombination frequency of fruit fly wing length and body color that Thomas Morgan observed in 1912 was 17 percent. A crossover frequency between 0 percent and 50 percent indicates that the genes are on the same chromosome and crossover sometimes occurs.

In a test cross for two characteristics such as the one here, can the recombinant offspring's predicted frequency be 60 percent? Why or why not?

Chromosome theory of inheritance

Genes are located on chromosomes
We take it for granted today that DNA is the genetic material, and therefore our genes must be located on chromosomes. But like all facts in science, this idea had to be repeatedly tested and found to be true before it could be accepted as fact. The chromosome theory of inheritance, or the idea that genes are located on chromosomes, was proposed based on experiments by Thomas Hunt Morgan using Drosophila melanogaster, or fruit flies. Drosophila are like humans in that an individual with two X chromosomes is female and an individual with one X and one Y chromosome is male (many organisms have other ways of determining gender).
In Drosophila, normal flies have red eyes. Red eye color is dominant. Morgan discovered a recessive mutation (allele) that caused white eyes. When Morgan mated a red eyed female to a white eyed male, all the progeny had red eyes. This result makes perfect sense with a dominant/recessive inheritance pattern, and here is the Punnett square demonstrating that (x^w = recessive white eye mutant allele x^W = dominant red eye wild-type allele):

But Morgan got a surprising result when he made the reciprocal cross, mating white eyed females to red eyed males. Instead of all red eyed progeny, he saw that all the females had red eyes and all the males had white eyes. This result seemed to violate Mendel’s principle of independent assortment, because two different traits (gender and eye color) seemed to be linked. The only way to explain these results was if the gene that caused eye color was located on (linked to) the X chromosome. Here is the Punnett square demonstrating this cross:

These results support the chromosome theory of inheritance because the only way to explain them is if the eye color gene is on the X chromosome. This is sex-linkage, or inheritance of genes that are on the sex chromosomes (X and Y). Sex-linked traits show interesting inheritance patterns in part because females have two copies of each X chromosome, but males only have one. This inheritance pattern means that a male with the recessive allele will always show the recessive trait, because he only has one copy of the allele. In contrast, most genes are located on the autosomes, or non sex chromosomes, where both males and females have two copies of each gene. Recall that all the patterns of inheritance observed by Mendel, including the principle of segregation and the principle of independent assortment are explained by the behavior of chromosomes during meiosis. These principles are part of the chromosome theory of inheritance.
Here is a video explaining these experiments and a little on the implications for humans:

In class, we’ll use phenotypic ratios to determine whether genes are sex-linked and predict offspring phenotypes when genes are sex-linked. We’ll also apply this information to analyze human pedigrees.

Linkage is inheritance of traits in a pattern that violates Mendel’s principle of independent assortment, the idea that alleles for different traits are segregated into gametes independently. Sex-linkage is a special type of linkage, where traits are linked to sex chromosomes. Genetic linkage occurs when the genes controlling two different traits are located near each other on the same chromosome. The basic idea is that if two genes are on the same chromosome, and you inherit the whole chromosome, then you have to inherit those two genes (and whatever alleles they have) together.
However, this is biology so there is a caveat: the phenomenon of crossing over helps to shuffle the alleles for genes located on the same chromosome. A crossover event between the locations of two genes on a chromosome results in genetic recombination, or new combinations of alleles on a chromosome.

Crossing over between genes A and B results in recombinant chromosomes with new allele combinations a, b and A, B, in addition to the original parental combinations A, b and a, B. Image from Wikimedia by user Abbyprovenzano, with CC-BY-SA-3.0 license.

Crossing over occurs during meiotic prophase I, when the homologous chromosomes align and synapse, and results in physically swapping genetic material (DNA) between non-sister chromatids of the paired homologous chromosomes. Because crossing over occurs randomly along the chromosome, the closer two genes are physically located to each other on a chromosome, the less likely that a crossover will occur between them. Conversely, the farther apart two genes are located from each other along the chromosome, the more likely they are to be swapped with the alleles on the homologous chromosome. The image below illustrates this idea:

It may be surprising to realize that two genes on the same chromosome will assort independently (like genes located on separate chromosomes) if they are far enough apart that a crossover almost always occurs between them, producing 50% recombinants (because crossing over involves only two of the 4 chromatids in a synapsed pair of homologous chromosomes, the maximum recombination frequency is 50%).
The video below walks through linkage as a violation of independent assortment and explains how crossing over breaks linkage. Note this video uses an incomplete definition of linkage: linkage occurs when two genes are located close together on the same chromosome and thus tend to be inherited together. It is not sufficient for genes to be on the same chromosome to be linked they also have to be close enough together that crossing over between them is a relatively rare event.

Simple rules for pedigree analysis
We can’t ask different people to mate and produce lots of offspring so we can test inheritance patterns in humans. Instead we rely on pedigree analysis to infer inheritance patterns. Here is a sample pedigree which explains how to read pedigrees:

The simple rules for pedigree analysis are:

• Autosomal recessive
• affects males and females equally
• both parents must carry allele
• parents may not display trait (carriers)
• typically affects only males
• affected male passes allele to daughters, not to sons
• trait skips a generation

In class, we will practice using these rules to determine the inheritance patterns of traits in different pedigrees.

Powerpoint slides with animated illustration of chromosome movements in mitosis and meiosis, to accompany the Nash case: MollyNashMitosisMeiosisAnimations

Mendel’s Mapped Traits

Homologous recombination is a common genetic process, yet Mendel never observed it. Had he investigated both linked and unlinked genes, it would have been much more difficult for him to create a unified model of his data on the basis of probabilistic calculations. Researchers who have since mapped the seven traits investigated by Mendel onto the seven chromosomes of the pea plant genome have confirmed that all of the genes he examined are either on separate chromosomes or are sufficiently far apart as to be statistically unlinked. Some have suggested that Mendel was enormously lucky to select only unlinked genes, whereas others question whether Mendel discarded any data suggesting linkage. In any case, Mendel consistently observed independent assortment because he examined genes that were effectively unlinked.

63 Chromosomal Theory and Genetic Linkage

By the end of this section, you will be able to do the following:

• Discuss Sutton’s Chromosomal Theory of Inheritance
• Explain the process of homologous recombination, or crossing over
• Describe chromosome creation
• Calculate the distances between three genes on a chromosome using a three-point test cross

Long before scientists visualized chromosomes under a microscope, the father of modern genetics, Gregor Mendel, began studying heredity in 1843. With improved microscopic techniques during the late 1800s, cell biologists could stain and visualize subcellular structures with dyes and observe their actions during cell division and meiosis. With each mitotic division, chromosomes replicated, condensed from an amorphous (no constant shape) nuclear mass into distinct X-shaped bodies (pairs of identical sister chromatids), and migrated to separate cellular poles.

Chromosomal Theory of Inheritance

The speculation that chromosomes might be the key to understanding heredity led several scientists to examine Mendel’s publications and reevaluate his model in terms of chromosome behavior during mitosis and meiosis. In 1902, Theodor Boveri observed that proper sea urchin embryonic development does not occur unless chromosomes are present. That same year, Walter Sutton observed chromosome separation into daughter cells during meiosis ((Figure)). Together, these observations led to the Chromosomal Theory of Inheritance , which identified chromosomes as the genetic material responsible for Mendelian inheritance.

The Chromosomal Theory of Inheritance was consistent with Mendel’s laws, which the following observations supported:

• During meiosis, homologous chromosome pairs migrate as discrete structures that are independent of other chromosome pairs.
• Chromosome sorting from each homologous pair into pre-gametes appears to be random.
• Each parent synthesizes gametes that contain only half their chromosomal complement.
• Even though male and female gametes (sperm and egg) differ in size and morphology, they have the same number of chromosomes, suggesting equal genetic contributions from each parent.
• The gametic chromosomes combine during fertilization to produce offspring with the same chromosome number as their parents.

Despite compelling correlations between chromosome behavior during meiosis and Mendel’s abstract laws, scientists proposed the Chromosomal Theory of Inheritance long before there was any direct evidence that chromosomes carried traits. Critics pointed out that individuals had far more independently segregating traits than they had chromosomes. It was only after several years of carrying out crosses with the fruit fly, Drosophila melanogaster, that Thomas Hunt Morgan provided experimental evidence to support the Chromosomal Theory of Inheritance.

Mendel’s work suggested that traits are inherited independently of each other. Morgan identified a 1:1 correspondence between a segregating trait and the X chromosome, suggesting that random chromosome segregation was the physical basis of Mendel’s model. This also demonstrated that linked genes disrupt Mendel’s predicted outcomes. That each chromosome can carry many linked genes explains how individuals can have many more traits than they have chromosomes. However, researchers in Morgan’s laboratory suggested that alleles positioned on the same chromosome were not always inherited together. During meiosis, linked genes somehow became unlinked.

Homologous Recombination

In 1909, Frans Janssen observed chiasmata—the point at which chromatids are in contact with each other and may exchange segments—prior to the first meiosis division. He suggested that alleles become unlinked and chromosomes physically exchange segments. As chromosomes condensed and paired with their homologs, they appeared to interact at distinct points. Janssen suggested that these points corresponded to regions in which chromosome segments exchanged. We now know that the pairing and interaction between homologous chromosomes, or synapsis, does more than simply organize the homologs for migration to separate daughter cells. When synapsed, homologous chromosomes undergo reciprocal physical exchanges at their arms in homologous recombination , or more simply, “crossing over.”

To better understand the type of experimental results that researchers were obtaining at this time, consider a heterozygous individual that inherited dominant maternal alleles for two genes on the same chromosome (such as AB) and two recessive paternal alleles for those same genes (such as ab). If the genes are linked, one would expect this individual to produce gametes that are either AB or ab with a 1:1 ratio. If the genes are unlinked, the individual should produce AB, Ab, aB, and ab gametes with equal frequencies, according to the Mendelian concept of independent assortment. Because they correspond to new allele combinations, the genotypes Ab and aB are nonparental types that result from homologous recombination during meiosis. Parental types are progeny that exhibit the same allelic combination as their parents. Morgan and his colleagues, however, found that when they test crossed such heterozygous individuals to a homozygous recessive parent (AaBb × aabb), both parental and nonparental cases occurred. For example, 950 offspring might be recovered that were either AaBb or aabb, but 50 offspring would also result that were either Aabb or aaBb. These results suggested that linkage occurred most often, but a significant minority of offspring were the products of recombination.

In a test cross for two characteristics such as the one here, can the recombinant offspring’s predicted frequency be 60 percent? Why or why not?

Genetic Maps

Janssen did not have the technology to demonstrate crossing over so it remained an abstract idea that scientists did not widely believe. Scientists thought chiasmata were a variation on synapsis and could not understand how chromosomes could break and rejoin. Yet, the data were clear that linkage did not always occur. Ultimately, it took a young undergraduate student and an “all-nighter” to mathematically elucidate the linkage and recombination problem.

In 1913, Alfred Sturtevant, a student in Morgan’s laboratory, gathered results from researchers in the laboratory, and took them home one night to mull them over. By the next morning, he had created the first “chromosome map,” a linear representation of gene order and relative distance on a chromosome ((Figure)).

Which of the following statements is true?

1. Recombination of the body color and red/cinnabar eye alleles will occur more frequently than recombination of the alleles for wing length and aristae length.
2. Recombination of the body color and aristae length alleles will occur more frequently than recombination of red/brown eye alleles and the aristae length alleles.
3. Recombination of the gray/black body color and long/short aristae alleles will not occur.
4. Recombination of the red/brown eye and long/short aristae alleles will occur more frequently than recombination of the alleles for wing length and body color.

As (Figure) shows, by using recombination frequency to predict genetic distance, we can infer the relative gene order on chromosome 2. The values represent map distances in centimorgans (cM), which correspond to recombination frequencies (in percent). Therefore, the genes for body color and wing size were 65.5 − 48.5 = 17 cM apart, indicating that the maternal and paternal alleles for these genes recombine in 17 percent of offspring, on average.

To construct a chromosome map, Sturtevant assumed that genes were ordered serially on threadlike chromosomes. He also assumed that the incidence of recombination between two homologous chromosomes could occur with equal likelihood anywhere along the chromosome’s length. Operating under these assumptions, Sturtevant postulated that alleles that were far apart on a chromosome were more likely to dissociate during meiosis simply because there was a larger region over which recombination could occur. Conversely, alleles that were close to each other on the chromosome were likely to be inherited together. The average number of crossovers between two alleles—that is, their recombination frequency —correlated with their genetic distance from each other, relative to the locations of other genes on that chromosome. Considering the example cross between AaBb and aabb above, we could calculate the recombination’s frequency as 50/1000 = 0.05. That is, the likelihood of a crossover between genes A/a and B/b was 0.05, or 5 percent. Such a result would indicate that the genes were definitively linked, but that they were far enough apart for crossovers to occasionally occur. Sturtevant divided his genetic map into map units, or centimorgans (cM) , in which a 0,01 recombination frequency corresponds to 1 cM.

By representing alleles in a linear map, Sturtevant suggested that genes can range from linking perfectly (recombination frequency = 0) to unlinking perfectly (recombination frequency = 0.5) when genes are on different chromosomes or genes separate very far apart on the same chromosome. Perfectly unlinked genes correspond to the frequencies Mendel predicted to assort independently in a dihybrid cross. A 0.5 recombination frequency indicates that 50 percent of offspring are recombinants and the other 50 percent are parental types. That is, every type of allele combination is represented with equal frequency. This representation allowed Sturtevant to additively calculate distances between several genes on the same chromosome. However, as the genetic distances approached 0.50, his predictions became less accurate because it was not clear whether the genes were very far apart on the same or on different chromosomes.

In 1931, Barbara McClintock and Harriet Creighton demonstrated the crossover of homologous chromosomes in corn plants. Weeks later, Curt Stern demonstrated microscopically homologous recombination in Drosophila. Stern observed several X-linked phenotypes that were associated with a structurally unusual and dissimilar X chromosome pair in which one X was missing a small terminal segment, and the other X was fused to a piece of the Y chromosome. By crossing flies, observing their offspring, and then visualizing the offspring’s chromosomes, Stern demonstrated that every time the offspring allele combination deviated from either of the parental combinations, there was a corresponding exchange of an X chromosome segment. Using mutant flies with structurally distinct X chromosomes was the key to observing the products of recombination because DNA sequencing and other molecular tools were not yet available. We now know that homologous chromosomes regularly exchange segments in meiosis by reciprocally breaking and rejoining their DNA at precise locations.

Review Sturtevant’s process to create a genetic map on the basis of recombination frequencies here.

Mendel’s Mapped Traits

Homologous recombination is a common genetic process, yet Mendel never observed it. Had he investigated both linked and unlinked genes, it would have been much more difficult for him to create a unified model of his data on the basis of probabilistic calculations. Researchers who have since mapped the seven traits that Mendel investigated onto a pea plant genome’s seven chromosomes have confirmed that all the genes he examined are either on separate chromosomes or are sufficiently far apart as to be statistically unlinked. Some have suggested that Mendel was enormously lucky to select only unlinked genes whereas, others question whether Mendel discarded any data suggesting linkage. In any case, Mendel consistently observed independent assortment because he examined genes that were effectively unlinked.

Section Summary

Sutton and Boveri’s Chromosomal Theory of Inheritance states that chromosomes are the vehicles of genetic heredity. Neither Mendelian genetics nor gene linkage is perfectly accurate. Instead, chromosome behavior involves segregation, independent assortment, and occasionally, linkage. Sturtevant devised a method to assess recombination frequency and infer linked genes’ relative positions and distances on a chromosome on the basis of the average number of crossovers in the intervening region between the genes. Sturtevant correctly presumed that genes are arranged in serial order on chromosomes and that recombination between homologs can occur anywhere on a chromosome with equal likelihood. Whereas linkage causes alleles on the same chromosome to be inherited together, homologous recombination biases alleles toward an independent inheritance pattern.

Visual Connection Questions

(Figure) In a test cross for two characteristics such as the one shown here, can the predicted frequency of recombinant offspring be 60 percent? Why or why not?

(Figure) No. The predicted frequency of recombinant offspring ranges from 0% (for linked traits) to 50% (for unlinked traits).

(Figure) Which of the following statements is true?

1. Recombination of the body color and red/cinnabar eye alleles will occur more frequently than recombination of the alleles for wing length and aristae length.
2. Recombination of the body color and aristae length alleles will occur more frequently than recombination of red/brown eye alleles and the aristae length alleles.
3. Recombination of the gray/black body color and long/short aristae alleles will not occur.
4. Recombination of the red/brown eye and long/short aristae alleles will occur more frequently than recombination of the alleles for wing length and body color.

Review Questions

X-linked recessive traits in humans (or in Drosophila) are observed ________.

1. in more males than females
2. in more females than males
3. in males and females equally
4. in different distributions depending on the trait

The first suggestion that chromosomes may physically exchange segments came from the microscopic identification of ________.

Which recombination frequency corresponds to independent assortment and the absence of linkage?

Which recombination frequency corresponds to perfect linkage and violates the law of independent assortment?

Critical Thinking Questions

Explain how the Chromosomal Theory of Inheritance helped to advance our understanding of genetics.

The Chromosomal Theory of Inheritance proposed that genes reside on chromosomes. The understanding that chromosomes are linear arrays of genes explained linkage, and crossing over explained recombination.

Art Connections

Figure In a test cross for two characteristics such as the one shown here, can the predicted frequency of recombinant offspring be 60 percent? Why or why not?

Figure No. The predicted frequency of recombinant offspring ranges from 0% (for linked traits) to 50% (for unlinked traits).

Figure Which of the following statements is true?

1. Recombination of the body color and red/cinnabar eye alleles will occur more frequently than recombination of the alleles for wing length and aristae length.
2. Recombination of the body color and aristae length alleles will occur more frequently than recombination of red/brown eye alleles and the aristae length alleles.
3. Recombination of the gray/black body color and long/short aristae alleles will not occur.
4. Recombination of the red/brown eye and long/short aristae alleles will occur more frequently than recombination of the alleles for wing length and body color.

Free Response

Use the probability method to calculate the genotypes and genotypic proportions of a cross between AABBCc and Aabbcc parents.

Considering each gene separately, the cross at A will produce offspring of which half are AA and half are Aa B will produce all Bb C will produce half Cc and half cc. Proportions then are (1/2) × (1) × (1/2), or 1/4 AABbCc continuing for the other possibilities yields 1/4 AABbcc, 1/4 AaBbCc, and 1/4 AaBbcc. The proportions therefore are 1:1:1:1.

Explain epistatis in terms of its Greek-language roots “standing upon.”

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.

In Section 12.3, “Laws of Inheritance,” an example of epistasis was given for the summer squash. Cross white WwYy heterozygotes to prove the phenotypic ratio of 12 white:3 yellow:1 green that was given in the text.

The cross can be represented as a 4 × 4 Punnett square, with the following gametes for each parent: WY, Wy, wY, and wy. For all 12 of the offspring that express a dominant W gene, the offspring will be white. The three offspring that are homozygous recessive for w but express a dominant Y gene will be yellow. The remaining wwyy offspring will be green.

Watch the video: DNA recombination basic (July 2022).

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