chapter 7
A cross between individuals with genotypes a+a b+b × aa bb produces the following progeny: a+a b+b 83 a+a bb 21 aa b+b 19 aa bb 77 (a) Does the evidence indicate that the a and b loci are linked? (b) What is the map distance between a and b? (c) Are the genes in the parent with genotype a+a b+b in coupling configuration or in repulsion? How do you know?
(a) Does the evidence indicate that the a and b loci are linked? The ratio of a+ to a is 104/96, and the ratio of b+ to b is 102/98, both close to 1:1 ratios. The four phenotypic classes are not present in 1:1:1:1 ratios (no need for chi-square test), so they are linked. (b) What is the map distance between a and b? The recombinants are the two phenotypic classes with the fewest progeny: RF = (21 + 19)/200 = 40/200 = 0.2 = 20%; the two genes are 20 m.u. apart. (c) Are the genes in the parent with genotype a+a b+b in coupling configuration or in repulsion? How do you know? They are in coupling configuration because the nonrecombinants are a+b+ and ab.
In tomatoes, tall (D) is dominant over dwarf (d), and smooth fruit (P) is dominant over pubescent (p) fruit, which is covered with fine hairs. A farmer has two tall and smooth tomato plants, which we will call plant A and plant B. The farmer crosses plants A and B with the same dwarf and pubescent plant and obtains the following numbers of progeny: Progeny of Plant A Plant B Dd Pp 122 2 Dd pp 6 82 dd Pp 4 82 dd pp 124 4 (a) What are the genotypes of plant A and plant B? (b) Are the loci that determine height of the plant and pubescence linked? If so, what is the map distance between them? c) Explain why different proportions of progeny are produced when plant A and plant B are crossed with the same dwarf pubescent plant.
(a) What are the genotypes of plant A and plant B? The genotypes of both plants are DdPp. (b) Are the loci that determine height of the plant and pubescence linked? If so, what is the map distance between them? Yes. From the cross of plant A, the map distance is 10/256 = 3.9% or 3.9 m.u. The cross of plant B gives 6/170 = 3.5% or 3.5 m.u. If we pool the data from the two crosses, we get 16/426 = 3.8% or 3.8 m.u. (c) Explain why different proportions of progeny are produced when plant A and plant B are crossed with the same dwarf pubescent plant. The two plants have different coupling configurations. In plant A, the dominant alleles D and P are coupled; one chromosome is D P and the other is d p. In plant B, they are in repulsion; its chromosomes have D p and d P.
*14. In silkmoths (Bombyx mori) red eyes (re) and white-banded wing (wb) are encoded by two mutant alleles that are recessive to those that produce wild-type traits (re+ and wb+); these two genes are on the same chromosome. A moth homozygous for red eyes and white-banded wings is crossed with a moth homozygous for the wild-type traits. The F1 have normal eyes and normal wings. The F1 are crossed with moths that have red eyes and white-banded wings in a testcross. The progeny of this testcross are: wild-type eyes, wild-type wings 418 red eyes, wild-type wings 19 wild-type eyes, white-banded wings 16 red eyes, white-banded wings 426 (a) What phenotypic proportions would be expected if the genes for red eyes and white-banded wings were located on different chromosomes? (b) What is the genetic distance between the genes for red eyes and white-banded wings?
(a) What phenotypic proportions would be expected if the genes for red eyes and white-banded wings were located on different chromosomes? ¼ wild-type eyes, wild-type wings ¼ red eyes, wild-type wings ¼ wild-type eyes, white-banded wings ¼ red eyes, white-banded wings (b) What is the genetic distance between the genes for red eyes and white-banded wings? The F1 heterozygote inherited a chromosome with alleles for red eyes and white-banded wings (re wb) from one parent and a chromosome with alleles for wild-type eyes and wild-type wings (re+ wb+) from the other parent. These are therefore the phenotypes of the nonrecombinant progeny, present in the highest numbers. The recombinants are the 19 with red eyes, wild-type wings and 16 with wild-type eyes, white-banded wings. RF = recombinants/total progeny × 100% = (19 + 16)/879 × 100% = 4.0% The distance between the genes is 4 map units.
In the snail Cepaea nemoralis, an autosomal allele causing a banded shell (BB) is recessive to the allele for unbanded shell (BO). Genes at a different locus determine the background color of the shell; here, yellow (CY) is recessive to brown (CBw). A banded, yellow snail is crossed with a homozygous brown, unbanded snail. The F1 are then crossed with banded, yellow snails (a testcross). (a) What will be the results of the testcross if the loci that control banding and color are linked with no crossing over? b) What will be the results of the testcross if the loci assort independently? c) What will be the results of the testcross if the loci are linked and 20 map units apart?
(a) What will be the results of the testcross if the loci that control banding and color are linked with no crossing over? With absolute linkage, there will be no recombinant progeny. The F1 inherited banded and yellow alleles (BBCY) together on one chromosome from the banded yellow parent and unbanded and brown alleles (BOCBw) together on the homologous chromosome from the unbanded brown parent. Without recombination, all the F1 gametes will contain only these two allelic combinations, in equal proportions. Therefore, the F2 testcross progeny will be ½ banded, yellow and ½ unbanded, brown. (b) What will be the results of the testcross if the loci assort independently? With independent assortment, the progeny will be: ¼ banded, yellow ¼ banded, brown ¼ unbanded, yellow ¼ unbanded, brown (c) What will be the results of the testcross if the loci are linked and 20 map units apart? The recombination frequency is 20%, so each of the two classes of recombinant progeny must be 10%. The recombinants are banded, brown and unbanded, yellow. The two classes of nonrecombinants are 80% of the progeny, so each must be 40%. The nonrecombinants are banded, yellow and unbanded, brown. In summary: 40% banded, yellow 40% unbanded, brown 10% banded, brown 10% unbanded, yellow
What is the difference between a genetic map and a physical map?
A genetic map gives the order of genes and relative distance between them based on recombination frequencies observed in genetic crosses. A physical map locates genes on the actual chromosome or DNA sequence, and thus represents the physical distance between genes.
What does the interference tell us about the effect of one crossover on another?
A positive interference value results when the actual number of double crossovers observed is less than the number of double crossovers expected from the single crossover frequencies. Thus, positive interference indicates that a crossover inhibits or interferes with the occurrence of a second crossover nearby.Conversely, a negative interference value, where more double crossovers occur than expected, suggests that a crossover event can stimulate additional crossover events in the same region of the chromosome.
Why is the frequency of recombinant gametes always half the frequency of crossing over?
Crossing over occurs at the four-strand stage, when two homologous chromosomes, each consisting of a pair of sister chromatids, are paired. Each crossover involves just two of the four strands and generates two recombinant strands. The remaining two strands that were not involved in the crossover generate two nonrecombinant strands. Therefore, the frequency of recombinant gametes is always half the frequency of crossovers.
List some of the methods for physically mapping genes and explain how they are used to position genes on chromosomes.
Deletion mapping: Recessive mutations are mapped by crossing mutants with strains containing various overlapping deletions that map to the same region as the recessive mutation. If the heterozygote with the mutation on one chromosome and the deletion on the homologous chromosome has a mutant phenotype, then the mutation must be located on the same physical portion of the chromosome that is deleted. If, on the other hand, the heterozygote has a wild-type phenotype (the mutation and the deletion complement), then the mutation lies outside the deleted region of the chromosome. Somatic-cell hybridization: Human and mouse cells are fused. The resulting hybrid cell randomly loses human chromosomes and retains only a few. A panel of hybrids that retain different combinations of human chromosomes is tested for expression of a human gene. A correlation between the expression of the gene and the retention of a unique human chromosome in those cell lines indicates that the human gene must be located on that chromosome. In-situ hybridization: DNA probes that are labeled with either a radioactive or fluorescent tag are hybridized to chromosome spreads. Detection of the labeled hybridized probe by autoradiography or fluorescence imaging reveals which chromosome and where along that chromosome the homologous gene is located. DNA sequencing: Overlapping DNA sequences are joined using computer programs to ultimately form chromosome-length sequence assemblies, or contigs. The locations of genes along the DNA sequence can be determined by searching for matches to known gene or protein amino acid sequences.
Explain how to determine which of three linked loci is the middle locus from the progeny of a three-point testcross.
Double crossovers always result in switching the middle gene with respect to the two nonrecombinant chromosomes. Hence, one can compare the two double crossover phenotypes with the two nonrecombinant phenotypes and see which gene is reversed.
A panel of cell lines was created from mouse-human somatic-cell fusions. Each line was examined for the presence of human chromosomes and for the production of three enzymes. The following results were obtained: On the basis of these results, give the chromosome location of enzyme 1, enzyme 2, and enzyme 3. (see Pic K)
Enzyme 1 is located on chromosome 9. Chromosome 9 is the only chromosome that is present in the cell lines that produce enzyme 1 and absent in the cell lines that do not produce enzyme 1. Enzyme 2 is located on chromosome 4. Chromosome 4 is the only chromosome that is present in cell lines that produce enzyme 2 (C & D) and absent in cell lines that do not produce enzyme 2 (A & B). Enzyme 3 is located on the X chromosome. The X chromosome is the only chromosome present in the three cell lines that produce enzyme 3 and absent in the cell line that does not produce enzyme 3.
How does one test to see if two genes are linked?
One first obtains individuals that are heterozygous for both genes. This may be achieved by crossing an individual homozygous dominant for both genes to one homozygous recessive for both genes, resulting in a heterozygote with genes in coupling configuration. Alternatively, an individual that is homozygous recessive for one gene may be crossed to an individual homozygous recessive for the second gene, resulting in a heterozygote with genes in repulsion. Then, the heterozygote is mated to a homozygous recessive tester and the progeny of each phenotypic class are tallied. If the proportion of recombinant progeny is far less than 50%, the genes are linked. If the results are not so clear-cut, then they may be tested by chi-square, first for equal segregation at each locus, then for independent assortment of the two loci. Significant deviation from results expected for independent assortment indicates linkage of the two genes.
A panel of cell lines was created from mouse-human somatic-cell fusions. Each line was examined for the presence of human chromosomes and for the production of an enzyme. The following results were obtained: (See pic J) On the basis of these results, which chromosome has the gene that codes for the enzyme?
The enzyme is produced only in cell lines B and E. Of all the chromosomes, only chromosome 8 is present in just these two cell lines and absent in all the other cell lines that do not produce the enzyme. Therefore, the gene for the enzyme is most likely on chromosome 8.
Why do calculated recombination frequencies between pairs of loci that are located relatively far apart underestimate the true genetic distances between loci?
The further apart two loci are, the more likely it is to get double crossovers between them. Unless there are marker genes between the loci, such double crossovers will be undetected because double crossovers give the same phenotypes as nonrecombinants. The calculated recombination frequency will underestimate the true crossover frequency because the double crossover progeny are not counted as recombinants.
The locations of six deletions have been mapped to the Drosophila chromosome shown on the following page. Recessive mutations a, b, c, d, e, and f are known to be located in the same region as the deletions, but the order of the mutations on the chromosome is not known. When flies homozygous for the recessive mutations are crossed with flies homozygous for the deletions, the following results are obtained, where "m" represents a mutant phenotype and a plus sign (+) represents the wild type. On the basis of these data, determine the relative order of the seven mutant genes on the chromosome. (see picI)
The mutations are mapped to the intervals indicated on the figure above the table. The location of f is ambiguous; it could be in either location shown above.
What is a lod score and how is it calculated?
The term lod means logarithm of odds. It is used to determine whether genes are linked, usually in the context of pedigree analysis. One first determines the probability of obtaining the observed progeny given a specified degree of linkage. That probability is divided by the probability of obtaining the observed progeny if the genes are not linked and sort independently. The log of the ratio of these probabilities is the lod score. A lod score of 3 or greater, indicating that the specified degree of linkage results in at least a thousand fold greater likelihood of yielding the observed outcome than if the genes are unlinked, indicates linkage.
What effect does the arrangement of linked genes (whether they are in coupling configuration or in repulsion) have on the results of a cross?
The two arrangements have opposite effects on the results of a cross. For genes in coupling configuration, most of the progeny will be either wild type for both genes, or mutant for both genes, with relatively few that are wild type for one gene and mutant for the other. For genes in repulsion, most of the progeny will be mutant for only one gene and wild-type for the other, with relatively few recombinants that are wild-type for both or mutant for both.
What effect does crossing over have on linkage?
generates recombination between genes located on the same chromosome, and thus renders linkage incomplete.
Genes in repulsion
have a wild-type allele of one gene together with the mutant allele of the second gene on the same chromosome, and vice versa on the homologous chromosome.
Genes in coupling configuration
have two wild-type alleles on the same chromosome and the two mutant alleles on the homologous chromosome.
What are two causes of recombination?
loci on different chromosomes that sort independently or by a physical crossing over between two loci on the same chromosome, with breakage and exchange of strands of homologous chromosomes paired in meiotic prophase I.
Incomplete linkage
means that greater than 50% of the gametes produced are nonrecombinant and less than 50% of the gametes are recombinant; the recombination frequency is greater than 0 and less than 50%.
What does the term recombination mean?
meiosis generates gametes with different allelic combinations than the original gametes the organism inherited. If the organism was created by the fusion of an egg bearing AB and a sperm bearing ab, it generates gametes that are Ab and aB.
Complete linkage of two genes
only nonrecombinant gametes will be produced; the recombination frequency is zero.
Independent assortment of two genes
will result in 50% of the gametes being recombinant and 50% being nonrecombinant, as would be observed for genes on two different chromosomes. Independent assortment may also be observed for genes on the same chromosome if they are far enough apart that one or more crossovers occur between them in every meiosis.
