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From what I understand, if the tetrads resulting from a cross exhibit the same number of parental ditypes as there are nonparental ditypes, then the genes under consideration are said to be unlinked.
For example, say I cross two gametes containing $ab^+$ and $a^+b$, respectively, and get 170 tetrads exhibiting parental ditype (all offspring are $ab^+$ or $a^+b$), 165 exhibiting nonparental ditype (offspring are $a^+b^+$ or $ab$), and 30 exhibiting tetratype (offspring one of $a^+b^+$, $a^+b$, $ab^+$ or $ab$).
I'm not sure whether to classify these genes as linked or unlinked. I want to say yes, they are indeed linked, but 170 and 165 seem like close enough numbers to suggest they may be unlinked.
I would think the ratio of parental to nonparental ditype has to be considered close enough to 1, but how do we decide a threshold for that?
I'd use a chi-squared test. You calculate what you expect to see if your hypothesis is true, and you take what you do see, and the chi-squared result tells you how likely it is that your results are different enough from what's expected to indicate that your hypothesis is incorrect.
Teaching Genetic Linkage and Recombination through Mapping with Molecular Markers
Most introductory genetics courses cover genetic linkage, a core concept in the CourseSource genetics learning outcome framework. Although it is a classical genetics topic, genetic linkage remains an important concept to understand in order to grasp modern genetics research approaches including Single Nucleotide Polymorphism (SNP) mapping, Genome Wide Association Studies (GWAS), and gene discovery. Typically, genetic linkage is taught in a very traditional way within our introductory genetics classes. Invariably, we see students struggling with the same aspects of linkage: how to distinguish between parental and recombinant combinations of alleles and how to relate phenotype proportions to meiotic processes and outcomes. We designed a lesson that provides a practical and experimental context to target these common student difficulties in learning about linkage and recombination. This student-centered interactive lesson and associated post-class problem set teaches genetic linkage through mapping a gene by determining co-segregation of a phenotype with microsatellite sequences revealed by gel electrophoresis banding patterns. This lesson includes very interactive class sessions and a follow-up problem set and post-test that allows students to develop a deeper understanding of genetic linkage, and provides instructors with insights about student thinking. When we implemented this lesson, we observed a dramatic increase in student understanding of genetic linkage and how to use molecular markers to map the location of genes.
Genetic Problems Solutions Campbell Ch14
Genetics Problems Campbell 1. A man with hemophilia (a recessive , sex-linked condition has a daughter of normal phenotype. She marries a man who is normal for the trait. What is the probability that a daughter of this mating will be a hemophiliac? A son? If the couple has four sons, what is the probability that all four will be born with hemophilia?
2. Pseudohypertropic muscular dystrophy is a disorder that causes gradual deterioration of the muscles. It is seen only in boys born to apparently normal parents and usually results in death in the early teens. (a) Is pseudohypertrophic muscular dystrophy caused by a dominant or recessive allele? (b) Is its inheritance sex-linked or autosomal? (c) How do you know? Explain why this disorder is always seen in boys and never girls.
3. Red-green color blindness is caused by a sex-linked recessive allele. A color-blind man marries a woman with normal vision whose father was color-blind. (a) What is the probability that they will have a color-blind daughter? (b) What is the probability that their first son will be color-blind? (Note: the two questions are worded a bit differently.)
4. A wild-type fruit fly (heterozygous for gray body color and normal wings was mated with a black fly with vestigial wings. The offspring had the following phenotypic distribution: wild type, 778 black-vestigial, 785 black-normal, 158 gray-vestigial, 162. What is the recombination frequency between these genes for body color and wing type.
5. In another cross, a wild-type fruit fly (heterozygous for gray body color and red eyes) was mated with a black fruit fly with purple eyes. The offspring were as follows: wild-type, 721 black-purple, 751 gray-purple, 49 black-red, 45. (a) What is the recombination frequency between these genes for body color and eye color? (b) Following up on this problem and problem 4, what fruit flies (genotypes and phenotypes) would you mate to determine the sequence of the body color, wing shape, and eye color genes on the chromosomes?
6. A space probe discovers a planet inhabited by creatures who reproduce with the same hereditary patterns as those in humans. Three phenotypic characters are height (T = tall, t = dwarf), hearing appendages (A = antennae, a = no antennae), and nose morphology (S = upturned snout, s = downturned snout). Since the creatures were not “intelligent” Earth scientists were able to do some controlled breeding experiments, using various heterozygotes in testcrosses. For a tall heterozygote with antennae, the offspring were tall-antennae, 46 dwarf-antennae 7 dwarf-no antennae 42 tall-no antennae 5. For a heterozygote with antennae and an upturned snout, the offspring were antennae-upturned snout 47 antennae-downturned snout, 2 no antennae-downturned snout, 48: no antennae-upturned snout 3. Calculate the recombination frequencies for both experiments.
7. Using the information from problem 6, a further testcross was done using a heterozygote for height and nose morphology. The offspring were tall-upturned nose, 40 dwarf-upturned nose, 9 dwarf-downturned nose, 42 tall-downturned nose, 9. Calculate the recombination frequency from these data then use your answer from problem 6 to determine the correct sequence of the three linked genes.
8. Imagine that a geneticist has identified two disorders that appear to be caused by the same chromosomal defect and are affected by genomic imprinting: blindness and numbness of the limbs. A blind woman (whose mother suffered from numbness) has four children, two of whom, a son and daughter, have inherited the chromosomal defect. If this defect works like Prader-Willi and Angelman syndromes, what disorders do this son and daughter display? What disorders would be seen in their sons and daughters?
9. What pattern of inheritance would lead a geneticist to suspect that an inherited disorder of cell metabolism is due to a defective mitochondrial gene?
10. An aneuploid person is obviously female, but her cells have two Barr bodies. what is the probable complement of sex chromosomes in this individual?
11. Determine the sequence of genes along a chromosome based on the following recombination frequencies: A-B, 8 map units A-C, 28 map units A-D, 25 map units B-C , 20 map units B-D, 33 map units.
12. About 5% of individuals with Downs syndrome are the result of chromosomal translocation. In most of these cases, one copy of chromosome 21 becomes attached to chromosome 14. How does this translocation lead to children with Down syndrome?
13. Assume genes A and B are linked and are 50 map units apart. An individual heterozygous at both loci is crossed with an individual who is homozygous recessive at both loci. (a) What percentage of the offspring will show phenotypes resulting from crossovers? (b) If you did not know genes A and B were linked, how would you interpret the results of this cross?
14. In Drosophila, the gene for white eyes and the gene that produces “hairy” wings have both been mapped to the same chromosome and have a crossover frequency of 1.5%. A geneticist doing some crosses involving these two mutant characteristics noticed that in a particular stock of flies, these two genes assorted independently that is they behaved as though they were on different chromosomes. What explanation can you offer for this observation?
An Introduction to Genetic Analysis. 7th edition.
When RF values are close to 50 percent, the χ 2 test can be used as a critical test for linkage. Assume that we have crossed pure-breeding parents of genotypes A/A · B/B and a/a · b/b, and obtained a dihybrid A/a · B/b, which we have testcrossed to a/a · b/b. A total of 500 progeny are classified as follows (written as gametes from the dihybrid):
From these data the recombinant frequency is 225/500 = percent. This seems like a case of linkage because the RF is less than 50 percent expected from independent assortment. However, it is possible that the two recombinant classes are in the minority merely on the basis of chance therefore, we need to perform a χ 2 test.
The problem then is to find the expectations, E, for each class. For linkage testing, it might be supposed that these expectations are simply given by the 1:1:1:1 ratio of the four backcross classes that we expect when there is independent assortment. For the 500 progeny in our example, then, we would expect 500/4 = in each class. But the 1:1:1:1 ratio is not the appropriate test for linkage, because to get such a ratio two things must be true. There must be independent assortment between the A and B locus, but, in addition, there must be an equal chance for the different genotypes formed at fertilization to reach the age at which they are scored for the test, which usually means that the four genotypes must have equal chance of survival from egg to adult. However, it is often the case that mutations used in linkage tests have some deleterious effect in the homozygous condition so a/a or b/b genotypes have a lower probability of survival than do the wild-type heterozygotes A/a and B/b. We might then be led to reject the hypothesis of independent assortment even when it is correct because the differential survivorship of the genotypes causes deviations from the 1:1:1:1 expected ratio. What we need is a method of calculating the expectations, E, that is insensitive to differences in survivorship.
No matter what the frequencies of a/a or b/b genotypes are among the backcross adults, if there is independent assortment, we expect the frequency of the a · b genotypes to be the product of the frequencies of the a and the b alleles. In our example, the total proportion of a alleles is (135 +)/500, which is indeed the expected 50 percent, but the frequency of alleles is only (135 + b110)/500 = percent. Thus we expect the proportion of a · b genotypes to be 0.50 ×𠁐.49 =𠁐.245 and the number of a · b genotypes in a sample of 500 to be 500 ×𠁐.245 =.5. The same kind of calculation can be performed for each of the other genotypes to give all the expected numbers. The comparison is usually done in a contingency table, as shown in Table 5-3. The expected number for an entry in the contingency table is the product of the proportion observed in its row, the proportion observed in its column, and the total sample size. But the row and column proportions are the row and column totals divided by the grand total, so the actual calculation of the expected number in each entry is simply to multiply the appropriate row total by the appropriate column total and then divide by the grand total. The value of χ 2 is then calculated as follows:
Contingency Table Comparing Observed and Expected Results of a Testcross to Examine Linkage Between Loci A/a and B/b.
The obtained value of χ 2 is converted into a probability by using a χ 2 table (see Table 4-1, page 126). To do so, we need to decide on the degrees of freedom (df) in the test, which, as the name suggests, is the number of independent deviations of observed from expected that have been calculated. We notice that, because of the way that the expectations were calculated in the contingency table from the row and column totals, all deviations are identical in absolute magnitude, 12.5, and that they alternate in sign and so they cancel out when summed in any row or column. Thus, there is really only one independent deviation, so there is only one degree of freedom. Therefore, looking along the one degree of freedom line in Table 4-1, we see that the probability of obtaining a deviation from expectations this large (or larger) by chance alone is 0.025 (2.5 percent). Because this probability is less than 5 percent, the hypothesis of independent assortment must be rejected. Thus, having rejected the hypothesis of no linkage, we are left with the inference that the loci must be linked.
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7.1 Compare recombination and crossover. How are these similar? How are they different?
7.2 Explain why it usually necessary to start with pure-breeding lines when measuring genetic linkage by the methods presented in this chapter.
7.3 If you knew that a locus that affected earlobe shape was tightly linked to a locus that affected susceptibility to cardiovascular disease human, under what circumstances would this information be clinically useful?
7.4 In a previous chapter, we said a 9:3:3:1 phenotypic ratio was expected among the progeny of a dihybrid cross, in absence of gene interaction.
a) What does this ratio assume about the linkage between the two loci in the dihybrid cross?
b) What ratio would be expected if the loci were completely linked? Be sure to consider every possible configuration of alleles in the dihybrids.
7.5 Given a dihybrid with the genotype CcEe:
a) If the alleles are in coupling (cis) configuration, what will be the genotypes of the parental and recombinant progeny from a test cross?
b) If the alleles are in repulsion (trans) configuration, what will be the genotypes of the parental and recombinant progeny from a test cross?
7.6 Imagine the white flowers are recessive to purple flowers, and yellow seeds are recessive to green seeds. If a green-seeded, purple-flowered dihybrid is testcrossed, and half of the progeny have yellow seeds, what can you conclude about linkage between these loci? What do you need to know about the parents of the dihybrid in this case?
7.7 In corn (i.e. maize, a diploid species), imagine that alleles for resistance to a particular pathogen are recessive and are linked to a locus that affects tassel length (short tassels are recessive to long tassels). Design a series of crosses to determine the map distance between these two loci. You can start with any genotypes you want, but be sure to specify the phenotypes of individuals at each stage of the process. Outline the crosses similar to what is shown in Figure 7.8, and specify which progeny will be considered recombinant. You do not need to calculate recombination frequency.
7.8 In a mutant screen in Drosophila, you identified a gene related to memory, as evidenced by the inability of recessive homozygotes to learn to associate a particular scent with the availability of food. Given another line of flies with an autosomal mutation that produces orange eyes, design a series of crosses to determine the map distance between these two loci. Outline the crosses similar to what is shown in Figure 7.8, and specify which progeny will be considered recombinant. You do not need to calculate recombination frequency.
7.9 Image that methionine heterotrophy, chlorosis (loss of chlorophyll), and absence of leaf hairs (trichomes) are each caused by recessive mutations at three different loci in Arabidopsis. Given a triple mutant, and assuming the loci are on the same chromosome, explain how you would determine the order of the loci relative to each other.
7.10 If the progeny of the cross aaBB x AAbb is testcrossed, and the following genotypes are observed among the progeny of the testcross, what is the frequency of recombination between these loci?
7.11 Three loci are linked in the order B-C-A. If the A-B map distance is 1cM, and the B-C map distance is 0.6cM, given the lines AaBbCc and aabbcc, what will be the frequency of Aabb genotypes among their progeny if one of the parents of the dihybrid had the genotypes AABBCC?
7.12 Genes for body color (B black dominant to b yellow) and wing shape (C straight dominant to c curved) are located on the same chromosome in flies. If single mutants for each of these traits are crossed (i.e. a yellow fly crossed to a curved-wing fly), and their progeny is testcrossed, the following phenotypic ratios are observed among their progeny.
b) Why are the frequencies of the two smallest classes not exactly the same?
7.13 Given the map distance you calculated between B-C in question 12, if you crossed a double mutant (i.e. yellow body and curved wing) with a wild-type fly, and testcrossed the progeny, what phenotypes in what proportions would you expect to observe among the F2 generation?
7.14 In a three-point cross, individuals AAbbcc and aaBBCC are crossed, and their F1 progeny is testcrossed. Answer the following questions based on these F2 frequency data.
7.6: Genetic Mapping
- Contributed by Todd Nickle and Isabelle Barrette-Ng
- Professors (Biology) at Mount Royal University & University of Calgary
Because the frequency of recombination between two loci (up to 50%) is roughly proportional to the chromosomal distance between them, we can use recombination frequencies to produce genetic maps of all the loci along a chromosome and ultimately in the whole genome. The units of genetic distance are called map units (mu) or centiMorgans (cM), in honor of Thomas Hunt Morgan by his student, Alfred Sturtevant, who developed the concept. Geneticists routinely convert recombination frequencies into cM: the recombination frequency in percent is approximately the same as the map distance in cM. For example, if two loci have a recombination frequency of 25% they are said to be
25cM apart on a chromosome (Figure (PageIndex<9>)). Note: this approximation works well for small distances (RF<30%) but progressively fails at longer distances because the RF reaches a maximum at 50%. Some chromosomes are >100 cM long but loci at the tips only have an RF of 50%. The method for mapping of these long chromosomes is shown below.
Note that the map distance of two loci alone does not tell us anything about the orientation of these loci relative to other features, such as centromeres or telomeres, on the chromosome.
Figure (PageIndex<9>): Two genetic maps consistent with a recombination frequency of 25% between A and B. Note the location of the centromere. (Original-Deyholos-CC:AN)
Map distances are always calculated for one pair of loci at a time. However, by combining the results of multiple pairwise calculations, a genetic map of many loci on a chromosome can be produced (Figure (PageIndex<10>)). A genetic map shows the map distance, in cM, that separates any two loci, and the position of these loci relative to all other mapped loci. The genetic map distance is roughly proportional to the physical distance, i.e. the amount of DNA between two loci. For example, in Arabidopsis, 1.0 cM corresponds to approximately 150,000bp and contains approximately 50 genes. The exact number of DNA bases in a cM depends on the organism, and on the particular position in the chromosome some parts of chromosomes (&ldquocrossover hot spots&rdquo) have higher rates of recombination than others, while other regions have reduced crossing over and often correspond to large regions of heterochromatin.
Figure (PageIndex<10>): Genetic maps for regions of two chromosomes from two species of the moth, Bombyx. The scale at left shows distance in cM, and the position of various loci is indicated on each chromosome. Diagonal lines connecting loci on different chromosomes show the position of corresponding loci in different species. This is referred to as regions of conserved synteny. (NCBI-NIH-PD)
When a novel gene or locus is identified by mutation or polymorphism, its approximate position on a chromosome can be determined by crossing it with previously mapped genes, and then calculating the recombination frequency. If the novel gene and the previously mapped genes show complete or partial linkage, the recombination frequency will indicate the approximate position of the novel gene within the genetic map. This information is useful in isolating (i.e. cloning) the specific fragment of DNA that encodes the novel gene, through a process called map-based cloning.
Genetic maps are also useful to track genes/alleles in breeding crops and animals, in studying evolutionary relationships between species, and in determining the causes and individual susceptibility of some human diseases.
Figure (PageIndex<11>): A double crossover between two loci will produce gametes with parental genotypes. (Original-Deyholos-CC:AN)
Genetic maps are useful for showing the order of loci along a chromosome, but the distances are only an approximation. The correlation between recombination frequency and actual chromosomal distance is more accurate for short distances (low RF values) than long distances. Observed recombination frequencies between two relatively distant markers tend to underestimate the actual number of crossovers that occurred. This is because as the distance between loci increases, so does the possibility of having a second (or more) crossovers occur between the loci. This is a problem for geneticists, because with respect to the loci being studied, these double-crossovers produce gametes with the same genotypes as if no recombination events had occurred (Figure (PageIndex<11>)) &ndash they have parental genotypes. Thus a double crossover will appear to be a parental type and not be counted as a recombinant, despite having two (or more) crossovers. Geneticists will sometimes use specific mathematical formulae to adjust large recombination frequencies to account for the possibility of multiple crossovers and thus get a better estimate of the actual distance between two loci.
New Gene Prediction Method Capitalizes On Multiple Genomes
Researchers at Stanford University report a new approach to computationally predicting the locations and structures of protein-coding genes in a genome. Gene finding remains an important problem in biology as scientists are still far from fully mapping the set of human genes.
Furthermore, gene maps for other vertebrates, including important model organisms such as mouse, are much more incomplete than the human annotation. The new technique, known as CONTRAST (CONditionally TRAined Search for Transcripts), works by comparing a genome of interest to the genomes of several related species.
CONTRAST exploits the fact that the functional role protein-coding genes play a specific part within a cell and are therefore subjected to characteristic evolutionary pressures. For example, mutations that alter an important part of a protein's structure are likely to be deleterious and thus selected against. On the other hand, mutations that preserve a protein's amino acid sequence are normally well tolerated. Thus, protein-coding genes can be identified by searching a genome for regions that show evidence such patterns of selection. However, learning to recognize such patterns when more than two species are compared has proved difficult.
Previous systems for gene prediction were able to effectively make use of one additional 'informant' genome. For example, when searching for human genes, taking into account information from the mouse genome led to a substantial increase in accuracy. But, no system was able to leverage additional informant genomes to improve upon state-of-the-art performance using mouse alone, although it was expected that adding informants would make patterns of selection clearer.
CONTRAST solves this problem by learning to recognize the signature of protein-coding gene selection in a fundamentally different way from previous approaches. Instead of constructing a model of sequence evolution, CONTRAST directly 'learns' which features of a genomic alignment are most useful for recognizing genes. This approach leads to overall higher levels of accuracy and is able to extract useful information from several informant sequences.
In a test on the human genome, CONTRAST exactly predicted the full structure of 59% of the genes in the test set, compared with the previous best result of 36%. Its exact exon sensitivity of 93%, compared with a previous best of 84%, translates into many thousands of exons correctly predicted by CONTRAST but missed by previous methods. Importantly, CONTRAST's accuracy using a combination of eleven informant genomes was significantly higher than its accuracy using any single informant. The substantial advance in predictive accuracy represented by CONTRAST will further efforts to complete protein-coding gene maps for human and other organisms.
Further information about existing gene-prediction methods and the advance CONTRAST brings to the field can be found in a minireview by Paul Flicek, which accompanies the article by Batzoglou and colleagues.
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Let's conclude QTL mapping = process of locating genes with effects on quantitative traits using molecular markers QTL mapping strategies = based on measuring the mean difference between lines with contrasting marker alleles QTL mapping = preliminary step in the discovery of useful genes for marker-aided backcrossing So far, only successful with disease resistance and stress tolerance genes having very large effects QTL mapping = basic research activity requiring careful planning of crosses and high-precision phenotyping Kearsey, M.J. and Pooni, H.S. 1996. The genetical analysis of quantitative traits. Chapter 7 Bernardo, R. 2002. Breeding for quantitative traits in plants. Chapters 13 and 14 Recombination
Note that all the diploid progeny fungi from the mating of mutant strains 1 and 2 have the ability to grow on arginine, and this complementation does not require any change in the two chromosomes (Figure 1.6.). The only thing that is happening is that the functional alleles of each gene are providing active enzymes. If genes 1 and 2 are on the same chromosome, at a low frequency, recombinations between the two chromosomes in the diploid can lead to crossovers, resulting in one chromosome with wild-type alleles of each gene and another chromosome with the mutant alleles of each gene (Figure 1.7). This can be observed in fungi by inducing sporulation of the diploid. Each spore is haploid, and the vast majority will carry one of the two parental chromosomes, and hence be defective in either gene 1 or gene 2. But wild type recombinants can be observed at a low frequency these will be prototrophs. The double-mutant recombinants will be auxotrophs, of course, but these can be distinguished from the parental single mutants by the inability of the double mutants to complement either mutant strain 1 or strain 2.
Figure 1.7. Recombination between homologous chromosomes in a diploid
Note that this recombinationis a physical alteration in the chromosomes. The frequency of its occurrence is directly proportional to the distance the genes are apart, which is the basis for mapping genes by their recombination distances. Recombination occurs in a small fraction of the progeny, whereas all the progeny of a complementing diploid have the previously lost function restored.
X-linked recessive traits in humans (or in Drosophila) are observed ________.
- in more males than females
- in more females than males
- in males and females equally
- 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?