This question is from a genetics session in my university:
The test cross between a plant with smooth and yellow seeds and a plant with wrinkled and green seeds gives the following results: 140 plants with smooth and green seeds, and 142 plants with wrinkled and yellow seeds. These results are consistent with which case of the following?
- Linked genes
- Epistatic genes
- Genes with intermediate dominance
- Genes with complete dominance and independent assortment
In my answer key the correct choice is linked genes.
What I know is that in the case of linked genes, the outcomes are: either all parental (complete linkage) or the majority of offspring have parental traits and we have few recombinants(partial linkage).
But here it seems like all offspring are recombinants?! How can this be? Is there a mistake in the question, in the answer, or in my information?
Any help is appreciated. Thank you.
There is nothing wrong with the question and it can be explained with not much more than mendelian genetics + linkage.
What do you think are the genotypes of parents?
First hint: If there is more than one phenotype (for the trait) of the offspring at least one of parents must be heterozygote. So the idea of pure lineages goes out of the window here.
Ok, there is at least one heterozygote for each color and smoothness of seeds. What is the genotype of the other parent?
Hint 2: Test cross is a term for cross between recessive homozygote and individual of unknown genome (with dominant phenotype). The test cross reveals genotype of the parent with unknown genome.
Hint 3: 1:1 ratio of phenotypes is typical of cross between a heterozygote and recessive homozygote.
Ok so we have heterozygote and recessive homozygote for both traits. Which traits are dominant and which are recessive?
Hint 4: This question seems to be inspired by original Mendel work. Feel free to read it here to find out which traits are recessive and which are dominant. It is quite a good read, clear in its presentation of evidence and conclusions and I recommend it to everyone.
However, this knowledge is not necessary to answer the question. How are dominant genes organized in parents? Principally we have two options:
(AaBb x aabb) or (Aabb x aaBb)
If we consider organization on chromosomes it expands to basically 3 options:
chr ---A---B--- chr ---a---b--- X chr ---a---b--- chr ---a---b---
chr ---A---b--- chr ---a---B--- X chr ---a---b--- chr ---a---b---
chr ---A---b--- chr ---a---b--- X chr ---a---B--- chr ---a---b---
I leave it to you to figure out which of the tree options leads to result offspring described in the question.
I aggree with you. I think there is a mistake in the question.
Note that the question does not specify whether the parental lineages are pure lineages (fully homozygotes) or not. In all cases, answer 'B' would not explain the observed pattern for the reason you explained. Such pattern would be possible with some complex epistatic interactions if the parental lineages were pure and with complete linkage.
Gene Linkage & Chromosome Maps
Thomas Hunt Morgan studied fruit flies and found that in some crosses, expected outcomes weren't happening. Further experiments confirmed that alleles located on the same chromosome are inherited together.
*Mendel's dihybrid cross AaBb x AaBb would not have yielded a 9:3:3:1 ratio if he had chosen alleles located on the same chromosomes.
A common cross used to demonstrate linkage groups is the cross of a heterozygote wild type vestigial wings/ black body with a recessive mutant.
The cross would look like this
vg+ vg bl+ bl x vg vg bl bl
It may be easier at this point to use the older notation for letters, where the cross would look like AaBb x aabb
There are two possible arrangements for the heterozygote (AaBb) in the above cross.
If the dominant alleles are on different chromosomes (Ab) then it is called TRANS
If the dominant alleles AB are on the same chromosome, it is called a CIS arrangement
Cross produces: 50% wild type / 50 % mutant
If no crossing over has occurred, the outcome will always be 1:1, however this is not what Thomas Hunt Morgan observed.
|Vestigial Wings, Wild||0||17|
|Wild, Black Body||0||17|
Question: How would you explain these results?
Answer: The two offspring that did not look like either parent are called recombinants. They are a result of CROSS-OVER which occurred during meiosis, the alleles switch position.
Using this methodology, the chromosomes of the fruit fly were mapped. Each MAP UNIT represents how far apart the alleles are on the chromosome, the number is based on how often crossing over occurs.
Chromosome 2 on Drosophila Melanogaster (fruit fly)
1. A dumpy winged (dd) fruit fly with long aristae (AA) is crossed with a long winged (Dd) short aristae (aa). Show the cross and the phenotypic proportions.
2. A fruit fly with short legs (ll) and vestigial wings (ww) is crossed with one that is heterozygous for both traits. Assuming the dominant alleles are on separate chromosomes, show the cross and the expected phenotypic proportions.
3. In fruit flies, red eyes is a dominant allele located on the X chromosome. The recessive condition results in white eyes. The tan body trait is also X-linked and is dominant to yellow bodies. A female who is heterozygous both traits with the dominant alleles located on the same chromosome is crossed with a white eyed, yellow bodied male. Show the cross and the phenotypic proportions (Don't forget these traits are X-linked!)
In pea plants, flower color and pollen shape are located on the same chromosome. A plant with purple flowers and long pollen (AaBb) is crossed with one that is recessive for both traits (aabb).
The results are as follows:
a) Are the chromosomes of the AaBb parent in the cis or trans position? Sketch a punnett square showing the expected offspring.
Independent assortment, linked genes, and recombination
Mendel proposed the Law of Independent Assortment to explain his observations that the outcomes for one gene did not impact the outcomes for another gene. When he did crosses for multiple traits, new combinations occurred in the F2 generation that were not present in the P generation. For example, crossing true-breeding pea plants for yellow round seeds and green wrinkled seeds can result in an F2 generation that includes both yellow wrinkled seeds and green smooth seeds.
If genes are located on different chromosomes, a heterozygote has an equal chance of the dominant or recessive copy ending up in each gamete. This means that in a dihybrid cross there are four combinations that are equally likely to be passed on to the next generation.
To find the expected outcomes for crossing two heterozygotes for both traits, you can make a sixteen box Punnett square. Each parent has four different combinations that can be passed on that will be positioned along the top and side of the square. You can also skip the larger Punnett square and use math to find the combined probability. Multiply the chance of each trait occurring to get the probability that both traits appear together in the offspring.
I prefer this method because it can be scaled up to three or more traits without becoming more difficult. For example, the chances of an individual having the dominant phenotype of three independently assorting traits is 27/64 (3/4 x 3/4 x 3/4). While you could make a 64 box Punnett square, the more boxes there are, the more chances of making a mistake.
Of course, all of this only works if the genes are unlinked. Alleles of genes close to each other on the same chromosome don't have an equal chance of being sorted into different gametes. Through crossing over, linked alleles may end up in different gametes, but at a lower rate than if they were on different chromosomes (or far apart on the same chromosome). The further apart linked genes are located on the chromosome, the more likely they are to be separated through crossing over. When this happens, the chromosome is called a recombinant. Offspring that show a phenotype combination that resulted from crossing over are also called recombinants.
Let's walk through an example of what will happen with linked genes when we cross a parent with both dominant traits to a parent with both recessive traits, using the fictional scenario in the Heredity IV simulation. In this case, the lack of pinna (external ear) and black eyes are dominant to the presence of pinna and white eyes. When our P cross is between black eye/no pinna and white eye/pinna parents (the genes are not sex-linked, so it doesn't matter if males or females have the dominant or recessive traits) the F1 generation results are 100% heterozygous black eye/no pinna. Because the P generation parents are homozygous this is the same result we get when doing a dihybrid cross for unlinked genes (see Heredity I simulation).
When we cross the F1 generation, though, the results will not be the 9:3:3:1 ratio that we expect from independently assorting genes.
If no crossing over occurs during gamete formation in this F1 cross, 75% of the resulting offspring will have both dominant phenotypes and 25% will have both recessive phenotypes. When crossing over occurs, some gametes will have a different combination of alleles on the chromosome and may lead to offspring with one dominant and one recessive phenotype (black eye/pinna and white eyes/no pinna). These offspring are the recombinants. A lower than expected (3/16 + 3/16 = 3/8 or 37.5%) number of recombinant offspring indicates that genes are linked. In the sample data in the picture above, only 2% of the offspring are recombinants. However, in this case, it is difficult to estimate the rate of recombination because crossing over also results in offspring with the dominant phenotype (black eye/no pinna).
In an independently assorting dihybrid cross, a P generation in which one parent has the dominant version and the recessive version of the other and the other parent has the reverse, the F2 generation will have the same 9:3:3:1 expected results. When the two genes are linked, however, the results are different than what occurred when one parent was dominant for both traits and the other parent was recessive for both.
This example uses a P generation cross of black eye/pinna x white eye/no pinna. While the F1 is still 100% dominant phenotypes (black eye/no pinna), each chromosome is what would have been classified as recombinants in the last example.
In this case, the rarest F2 phenotype will be both recessive traits (white eye/no pinna). In the sample data below, only 1% of the offspring have white eyes and the presence of pinna (both recessive traits). Because crossing over can lead to a phenotype that is also possible without crossing over (black eye/no pinna), this cross isn't ideal for estimating the rate of crossing over.
In order to estimate the recombination rate between the genes, we need a cross where there isn't overlap in the phenotypes resulting from crossing over and no crossing over. This can be done with a backcross. In a backcross, F1 heterozygotes are crossed to the parent with recessive phenotypes.
If we perform a backcross with unlinked genes, we should get a 1:1:1:1 ratio (see picture above, using pea seed traits). If the genes are linked there will be an over-representation of the parental phenotypes (dominant for both traits and recessive for both traits).
In this case, unlike the F1 cross examples discussed previously, a crossing over event between the two genes will lead to a recombinant phenotype (black eye/pinna and white eye/no pinna). For this reason, the backcross gives us a way of determining the rate of recombination, which is indicative of the distance between the genes.
The % recombination is equivalent to the map units/centiMorgans (cM) between the genes. This is a relative measure of distance on chromosomes and can be used to map the locations of genes. In the sample data in the picture above, 6.6% of the offspring are recombinants. This data suggests that the eye color and pinna genes are 6.6 cM apart. However, remember that it is best practice to base conclusions on an average of multiple trials.
If genes are located far enough apart on a chromosome, crossing over will occur often enough that the genes assort independently. At 50 cM the recombination rate is 50%, which means the expected 1:1:1:1 ratio for unlinked genes is observed.
For more information on the discovery of linked genes and construction of the first fruit fly chromosome linkage maps, see this article from Nature Education.
The Present and the Future
This chapter discusses how the sex linkage in breast cancer can be accounted for by differences in the hormonal environments of men and women. The incidence in men can be increased probably to a rate similar to that in the female by the administration of estrogens or by feminizing conditions. There is evidence that men with breast cancer metabolize estrogen abnormally. Certain abnormalities in the hormonal milieu, probably acting with other unknown agents, may condition breast tissue to neoplasia. By altering the hormonal milieu from a favorable to an unfavorable one, the growth of breast cancer can be halted and tumor tissue can even be destroyed, but inevitably the cancer adapts itself to this change and sooner or later will grow in the altered environment. Information on the environment in which these responsive tumors grow can now be obtained from blood or urinary hormone assays, and the results of these assays are of some value in predicting the response to endocrine treatment.
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.
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.
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.
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?
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.
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.
(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.
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.
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).
AaBb x AaBb
Lets say each of the genes involves complete dominance and recessiveness. Thus, AA and Aa encode a red phenotype, and aa represents white. Lets also say that BB and Bb are hairy, whereas bb is bald. Lets calculate the genotypes, genotypic ratios, phenotypes, and phenotypic ratios for the above cross:
Thus, our phenotypic ratio is 9:3:3:1 (9 red and hairy, 3 red and bald, 3 white and hairy, and 1 white and bald). Anytime you perform a dihybrid cross, work with two totally heterozygous individuals, and if the alleles exhibit dominance and recessiveness, then you will end up with these same genotypic and phenotypic ratios. NOTE: One thing that you should notice about the dihybrid crosses is that each pair of alleles of a gene behaves independent of the pair of alleles for the other gene. When Mendel performed his dihybrid crosses, he noticed this as well. Therefore, he formulated his Law of Independent Assortment, which states that the members of one pair of factors (alleles) assort (separate) independently of another pair of factors. Thus, knowing all of the above, lets work some dihydrid cross problems.
You know that the possession of claws (WW or Ww) is dominant to lack of claws (ww). You also know that the presence of smelly feet (FF or Ff) is dominant to non-smelly feet (ff). You cross a male who is clawed and has smelly feet with a female who is clawed and has non-smelly feet. All 18 offspring produced have smelly feet, and 14 have claws and 4 are un-clawed. What are the genotypes of the parents?
You have an individual who is totally heterozygous for 2 genes that are not linked (i.e., not on the same chromosome). One gene is for ear size (AA or Aa being big ears whereas aa is for small ears) and the other gene is for buggy eyes (BB and Bb for buggy eyes whereas bb represents normal eyes). If you testcross this individual, what are the resulting genotypes and phenotypes?
Now then, after you've completed the problem above, lets ignore the Punnett's square and simply look at the 4 types of offspring from the above cross. What if the actual ratios in your testcross were not 1:1:1:1, but were as follows. What would this represent?
The following is a genetic linkage problem involving 4 genes. You want to determine which of the genes are linked, and which occur on separate chromosomes. You cross two true breeding (i.e., remember that this means that they are homozygous) plants that have the following characteristics:
|PLANT 1||PLANT 2|
|Red flowers||White flowers|
|Spiny seeds||Smooth seeds|
|Long pollen grains||Short pollen grains|
|Late blooming||Early blooming|
Following the above cross, all of the offspring have red flowers, spiny seeds, long pollen grains, and early blooming (meaning, that these traits are dominant). You then testcross the F1 generation, which you should realize by now are totally heterozygous individuals, and obtain the ratios below. What's going on?
|49% red-spiny||25% red-long||25% red-early||25% long-early|
|1% red-smooth||25% red-short||25% red-late||25% long-late|
|1% white-spiny||25% white-long||25% white-early||25% short-early|
|49% white-smooth||25% white-short||25% white-late||25% short-late|
The following is a genetic linkage problem also involving 4 genes. You want to determine which of the genes are linked, which occur on separate chromosomes, and the distances between the linked genes. You cross 2 true breeding (i.e. homozygous) plants that have the following "unusual" characteristics:
|PLANT 1||PLANT 2|
|Red flowers||White flowers|
|Long pollen grains||Short pollen grains|
|Dumb backtalk||Smart backtalk|
|Mean disposition||Nice disposition|
All of the offspring have red flowers, long pollen grains, give smart backtalk, and have a nice disposition (meaning, that these traits are dominant). You then testcross the F1 generation, and obtain the ratios below. How many chromosomes are involved in the linkages, and what are the positions of the linked genes relative to one another?
|45% red-long||25% red-dumb||25% long-dumb||48% red-mean||43% long-mean|
|5% red-short||25% red-smart||25% long smart||2% red-nice||7% long-nice|
|5% white-long||25% white-dumb||25% short-dumb||2% white-mean||7% short-mean|
|45% white-short||25% white-smart||25% short-smart||48% white-nice||43% short-nice|
- The Type A father?
- The Type B father?
- The Type O father?
- The Type AB father?
A brown-eyed, long-winged fly is mated to a red-eyed, long-winged fly. The progeny are: 51 long, red 53 long, brown 18 short, red 16 short, brown Using solely the information provided, what are the genotypes of the parents?
A strange woman has a bizzare condition known as "Cyclops" syndrome, where she has a single eye in the middle of her forehead. The allele for the normal condition (i.e. NO "Cyclops" syndrome) is recessive (cc). Her father is a Cyclops, as well as her mother. Her father's mother was normal. What is the genotype of the strange woman's father?
In calico cats, there is an X-linked gene with 2 alleles that control fur color. BB is a black female B'B' is a yellow female B'B (heterozygous) is a calico female B' is a yellow male and B is a black male. You have recently taken over judge Wapner's job on the People's Court and a woman brings in a black female cat that has given birth to 4 calico female kittens and 2 black male kittens. You must decide which of the defendent's male cats is guilty: the black one or the yellow one.
A common form of red-green color blindness in humans is caused by the presence of an X-linked recessive allele. Given simply that, please answer the following:
- Can two color-blind parents give birth to a normal son or daughter?
- Can two normal parents produce a color-blind daughter?
- Can two normal parents produce a color-blind son?
When studying an inheritance phenomenon, a geneticist discovers a phenotypic ratio of 9:6:1 among offspring of a given mating. Give a simple, plausible explanation of the results. How would you test this hypothesis?
Genetic Variation and Natural Selection: Post-Mendelian Inheritance Factors
Recall from Inheritance that most animals, including you, are diploid, meaning that each of your traits is controlled by the interaction of at least two genes. Mendel's work concentrated on the effect of a single gene from the mother and a single gene from the father to determine the genotype and phenotype of the offspring based on dominant and recessive genes. We now know that often multiple genes are involved in strange and intriguing ways to influence the phenotype of the offspring. The next section focuses on seven such mechanisms.
More than 200 human traits are controlled by a single gene pair, such as dwarfism, cataracts, cystic fibrosis, and albinism.
Whenever two genes combine and the offspring's phenotype is a compromise between the effects of the two genes, then neither gene has expressed dominance over the other gene. In fact, one gene is incompletely dominant over the other gene that is inactive. Mendel's work with peas never encountered incomplete dominance because the heterozygous genotype always expressed the dominant gene in the phenotype.
The active gene is apparently not strong enough to compensate for the loss of effect by the inactive allele. The classic example of incomplete dominance is the cross between a red snapdragon and a white snapdragon. According to Mendel, the offspring should be either red or white. In reality they are all pink, indicating that neither allele is completely dominant but one is incompletely dominant. Therefore, a weak expression of one allele, the red one, is actually expressed in a heterozygous offspring.
Incomplete dominance becomes evident in the F2 generation as detailed in the following Punnet square analysis of a fictitious trait: B = blue, b = yellow. A cross between a blue (BB) parent and a yellow (bb) parent, BB bb, should give all blue offspring in the F1 generation. Refer to the illustration Parents: BB bb.
Instead, the Bb genotype is green, a blending of the two colors. This indicates that the B gene is not strong enough to dominate the effect of the b allele completely. If two heterozygous individuals are mated, the results are even more intriguing. Refer to the illustration Parents: Bb Bb.
The genotypes are BB (1) Bb (2) bb (1). A 1:2:1 ratio is typical of Mendelian inheritance patterns. However, the phenotypic ratio is: blue (1), green (2), yellow (1). This 1:2:1 ratio is not expected by Mendel's predictions.
Codominance is a type of allele dominance that occurs when both alleles are active and expressed, as sometimes happens in a heterozygous individual. This phenomenon is most commonly displayed in the roan coat color of horses and cows. A roan color is actually the result of a codominance by a gene for red hair and a gene for white hair. In a roan-colored animal, there are equal numbers of both colors evenly dispersed within the coat, creating a ghostly brown color. Another example is Erminette chickens. They are a mottled black and white color because they have roughly equal numbers of black and white feathers. In their case, a gene for white feathers and a gene for black feathers are codominant in the heterozygous individuals.
Although individual humans carry a maximum of two alleles that control the expression of a trait, often there are more than two different alleles in a given population that code for the same trait. When three or more alleles are available for a particular gene, the inheritance of those alleles follows a multiple alleles pattern. The most common case for demonstrating multiple alleles is the inheritance of human blood types: A, B, AB, and O. Combinations of three possible alleles give rise to the four blood types. Because you can only have two alleles for any gene, not everyone carries every type of allele available but within a population, all of the multiple alleles are present. Sexual reproduction within that population keeps the alleles available for future generations and continues the genetic variation.
An expanded Punnet square shows the possible blood type combinations that exist in the human population. Refer to the illustration Human population blood type combinations.
Remember although there are three possible alleles listed on each axis, a human can only possess two of these. However, other individuals in the human population may carry a different set of two alleles, so that all three types are present and active in the human population.
Each chromosome normally carries thousands of genes. When certain neighboring genes on the same chromosome are inherited together, they are considered . Linked genes are an anomaly in Mendel's theory of independent assortment because they tend to remain together as a segment and therefore function as a cluster of genes rather than individual genes. In humans, the autosomal genes for red hair, freckles, and a fair complexion are linked genes.
Sex-Linked, or X-Linked Genes
The linkage in sex-linked traits is completely different from linked genes. A sex-linked trait is one whose allele is located on one of the sex chromosomes, usually the X for autosomal traits. Most sex-linked genes are recessive and are only expressed in homozygous females and males. Because males normally inherit one X chromosome from the mother only, the recessive traits found on the X chromosome are passed from mother to son. It is more difficult to pass an X-linked trait to a daughter because she inherits an X from both parents, so both parents would have to possess and pass on the recessive gene. The recessive gene on the X chromosome is expressed in males because it is the only gene present. In females, it can be hidden by the presence of a dominant gene on the other X chromosome. In humans, sex-linked traits include colorblindness, albinisms, and hemophilia.
Refer to the illustration Colorblindness pedigree. A heterozygous ?carrier? mates with a ?carrier? female (X c = gene for colorblindness). A carrier is an individual who is heterozygous for a condition but does not display the trait. Therefore, a carrier for colorblindness is not colorblind but can pass on the gene for colorblindness.
In the F1 generation, one colorblind female and one carrier female each mate with a noncolorblind male. In the F2 generation, individuals 7 and 8 are brothers, one is colorblind, the other is not. The colorblind male inherited the colorblind gene from the heterozygous carrier mother the noncolorblind son inherited the normal X chromosome that did not contain the gene for colorblindness. Likewise, individual 9 could only receive a gene for colorblindness from his mother, number 5. The sister, number 10, is a heterozygous carrier.
Certain genes are activated by hormones secreted by the sex organs, so they display different phenotypes when found in males than they do when found in females, even though their genotypes are identical. The standard sex-influenced study in humans is male pattern baldness, which is influenced by testosterone and can affect both males and females. However, it is dominant in males and recessive in females! In the homozygous state, both males and females become bald in the heterozygous genotype, females show no signs of baldness, but males become bald. Interestingly, sex-influenced genes are generally not found on the sex chromosomes, so both sexes can have the same genotype and different phenotypes!
Polygenic inheritance is the opposite of pleiotropic inheritance, in which a single gene affects several characteristics.
So far we have limited gene expression to those traits expressed by the action of one gene. In most primates, including you, traits are governed by the interaction of multiple genes. The genes of polygenic traits may be interspersed on the same chromosome or on completely different chromosomes. The additive effects of numerous genes on a single phenotype create a continuum of possible outcomes. A typical example is eye color in humans, which under normal genotypes is in varying shades of ebony brown to crystal blue, to Kelly green, and all points in between. Polygenic traits are also most susceptible to environmental influences. For instance, tallness is controlled by polygenes for skeleton height, but their effect may be retarded by malnutrition, injury, and disease.
Natural environmental influences on the expression of genotypes tend to be an advantage for the individual, whereas man-made environmental influences are both positive and negative for the individual. Natural environmental influences include the phenomenon of color change in the Arctic fox from red-brown in the summer months to pure white during the winter season for better camouflage. The genes that produce the red-brown summer pigment are blocked by cold temperatures, causing the hair to grow with no color (therefore, white). Another colorful example is the interaction between the color of the hydrangea flower, which is blue in acidic soils and pink in alkaline soils. A recent study also linked improved diet in infants and adolescents to a taller average height in the United States and Europe with the opposite effect in famine-stricken populations. The genetic complement of an individual is inherited however, the environmental effect on these genes may alter their application and expression.
Given that the trait of interest is either autosomal or sex-linked and follows by either complete dominance or incomplete dominance, a reciprocal cross following two generations will determine the mode of inheritance of the trait.
White-eye mutation in Drosophila melanogaster Edit
Sex linkage was first reported by Doncaster and Raynor in 1906  who studied the inheritance of a colour mutation in a moth, Abraxas grossulariata. Thomas Hunt Morgan later showed that a new white-eye mutation in Drosophila melanogaster was also sex-linked. He found that a white-eyed male crossed with a red-eyed female produced only red-eyed offspring. However, when they crossed a red-eyed male with a white-eyed female, the male offspring had white eyes while the female offspring had red eyes. The reason was that the white eye allele is sex-linked (more specifically, on the X chromosome) and recessive.
The analysis can be more easily shown with Punnett squares:
As shown in Table 1, the male offspring are white-eyed and the female offspring are red-eyed. The female offspring are carrying the mutant white-eye allele X(mut), but do not express it phenotypically because it is recessive. Although the males carry only one mutant allele like the females, the X-chromosome takes precedence over the Y and the recessive phenotype is shown.
As shown in Table 2, all offspring are Red-eyed. The males are free of the mutation. The females however, are carriers. 
My Biology Final
2. Describe the methods Mendel used in his plant-breeding experiments.
He bred pea plants for seven years. One of his methods included tying a little cloth bag around the flowers that way pollen from other plants couldn't fertilize it. He also crossed true-breeding plants to test the particulate hypothesis.
- Explain Mendel's principle of segregation.
- Describe how probability applies to genetics.
- Contrast genotype and phenotype.
- Explain Mendel's principle of independent assortment.
- Describe how alleles interact in intermediate inheritance.
- Describe inheritance patterns involving multiple alleles.
- Explain how polygenic inheritance can result in a wide range of phenotypes.
- Describe how environmental conditions affect phenotype expression.
1. Describe how alleles interact in intermediate inheritance.
In some organisms, neither allele is dominant. This pattern doesn't support the blending theory, because parent phenotypes can reappear in the F2 generation.
2. Describe inheritance patterns involving multiple alleles.
One pattern involving multiple inheritance is codominance. Codominance is when the heterozygote expresses both traits. It shows the traits of both alleles.
3. Explain how polygenic inheritance can result in a wide range of phenotypes.
Polygenic inheritance is when two or more genes affect a single trait. The more genes that affect a single character, the more possible phenotypes.
4. Describe how environmental conditions affect phenotype expression.
Temperature, nutrition, exercise, illnesses, and exposure to sunlight are environmental factor that could affect phenotype expression.
The genes located on the same chromosome are called linked genes. The phenomenon of staying together of all the genes of a chromosome is called linkage. Every organism possesses numerous characters. These characters are controlled by thousands of genes. But they have limited number of chromosomes. Therefore. each chromosome carries many genes on it. So the gene present on the same chromosome is linked to each other. Gene linkage is a physical relationship between genes. Linked genes cannot assort independently during meiosis. So they do not obey Menders law of independent assortment. Gene linkage also minimizes the chances of genetic recombination and variations among offspring.
Work of William Bateson and R.0 Punnet
William Bateson and R.0 Punnet were study flower colour and pollen shape in sweat pea in 1906. Purple (PP) flower colour is dominant over red (pp) flower colour. Similarly. long pollen grain (LL) is dominant over round pollen grain (11). They cross pure line of purple long (PPLL) with pure line of red, round (pp11). They obtained purple long in F1 They then intercrossed the F1. But they obtained strange results in F. They obtained following phenotypes:
These results do not give 9:3:3:1 ratio. Thus they do not follow
Menders law of independent inheritance. The progeny of F2 has large number of parental types (Purple. long and red. round) than recombinant type (Purple. round and red, long). Bateson and Punnett suggested that the appearance of large number of parental types in F, showed that some physical connection is present between these characters. They believed that some attractive forces are present between parental alleles. So they tend to remain together. They named this process as coupling. But they believed some negative forces are produced during recombination. These are called repulsion.
Morgan performed experiments on Drosophila. He studied the affect of linkage on the inheritance of two di Ifere. , t characters. He followed two characters, body colour and wing size ,h 7 drosophila:
(a) Wild type: Wild-type flies have gray bodies and normal ‘sings. (I)) Mutants: Mutant phenotypes for these characters are black body
and vestigial wings. The vestigial wings are much smaller than
The alleles for these traits arc represented by the follow ing symbols: = gray, b = black vg = normal wings. vg vestigial wings. They are not sex linked. Their loci are on autosomes.
Morgan crossed female di-hybrids (h h vg’ vg) with males. These males %%ere mutant with black bodies and vestigial wings (b b vg N g). It is a Mendelian testcross. According to Menders la) of independent assortment. Morgan’s Drosophila testcross should produce four phenotypic classes of offspring. These classes should – be equal in number:
OBSERVED 965 944 206 185
The actual results were verb different. There were disproportionate numbers of wild-type (gray-normal) and double mutant (black-vestigial) flies among the offspring. These two phenotypes corresponded to the phenotypes Of the two parents. Morgan concluded that body color and wing shape are inherited together in a specific combination. The genes for these two characters are located on the same chromosome. The other two phenotypes (gray-vestigial and black-normal) had much less number. These phenotypes were present among the offspring of Morgan’s cross. These new combinations of the two characters were formed by crossing over.
Linkage in corn
The phenomenon of linkage is more obvious in corn. Hutchion performed experiments on corn. lie crosses coloured, normal (full endosperm) seed (CCSS) with colourless and shrunken seeds (ccss). lie obtained coloured normal in F1. Then he test crossed this F1 with recessive (ccss). He obtained four classes of progeny. But these classes were not present in I: I:1:1 as expected by Mendel. It has more parental types (CCSS and ccss) than recombinant type (CCss or ccSS). It shows that both genes are linked.
A chromosome carries its linked genes en bloc in the form of a linkage group. The numbers of linkage groups are equal to the number of homologous pairs of chromosomes. Man has 23 linkage groups. Genes for colour blindness, haemophilia, gout etc form one linkage group on human X — chromosome. Similarly, gene for sickle cell anaemia, leukemia and albinism make another linkage group on human chromosome II. Number of linked groups in Drosophila are 4 and pea has 7.