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Unformatted text preview: Linkage and Chromosome Mapping (Chapter 7)
Suggested problems: 1-10, 14, 16-19, 22-24, 26, see pp 193-196. Question: Humans have approx. 20,500 genes. How many chromosomes should humans have for the alleles of these 20,500 genes segregate independently of one another? Terminology Linkage. The presence of two or more genes in the
same chromosome, which leads to a tendency for parental traits to be inherited together or travel together. are units of transmission. Genes are units of heredity, whereas chromosomes Genetic Linkage is rule rather than exception. It may be complete, incomplete, or absent (not detectable), depending on the distance between linked genes in a chromosome. Alfred H. Sturtevant discovered linkage and constructed the first genetic map. Example. Genes "a" and "b" are in the same
chromosome: three possible scenarios (below).
a b Linkage complete "a" and "b" are too close to each other.
a < 50 b incomplete "a" and "b" are separated to allow crossing-over between them in up to 100% of meiocytes.
a > 50 b Not detectable "a" and "b" are too far apart to allow crossingover between them in 100% of meiocytes. No Linkage Complete Linkage 4 types of gametes 2 types of gametes Example: two linked characters in tomatoes
4 types of gametes but not equal in frequency Parental Recombinant Vine height (tall vs. dwarf) and fruit shape (round vs. pear) P F1 Tall x Dwarf Tall 3 Tall : 1 Dwarf Round x Pear Round 3 Round : 1 Pear Recombinant F2
Parental ! Both characteristics show monogenic inheritance when one studies them separately. Linkage violates Mendel's Second Law
A dihybrid testcross yields results that are not consistent with the expected 1:1:1:1 ratio. P F1 Tall, Round x Dwarf, Pear (+ + / + +) (dd / pp) Tall, Round (+ d / + p) Testcross + d / + p x dd / pp Cont...
! ! Tall, round Tall, pear Dwarf, round Dwarf, pear ! Observed 81 22 17 79 199 If linkage (4) (1) (1) (4) Expected 49.75 (1) 49.75 (1) 49.75 (1) 49.75 (1) 199 If independent assortment Cont...
Hypothesis. This is a dihybrid testcross ratio of 1:1:1:1 for two genes segregating independently. X2 = 73.85; X2 for 3 df. at 1% is 11.34. There is not even one in thousand chance that the hypothesis is true. The hypothesis must be rejected. How to detect linkage when you see it?
Parental traits tend to be inherited together. Testcrosses produce parental types in excess of >50%. Alleles of two linked genes may be arranged in two different ways in the chromosome.
A A a a B B B B x a a A A b b b b A a a A B b B b Alternative Hypothesis. The vine height and the fruit
shape genes are on the same chromosome; they are not segregating independently. Coupling (cis) x Repulsion (trans) Coupling and repulsion refer only to F1 hybrids.
12 More Terminology
Linkage group. A group of genes inherited together -because they are in the same chromosome. !In any organism,there are as many linkage groups as there are nonhomologous chromosomes. How many linkage groups do you have? Recombination. New combinations of linked genes that result after crossing-over. A B a B a b
Parental (old combination) Recombinant (new combination, nonparental) Cont...
Genetic map. A line diagram showing the (linear) order of two or more linked genes and the distances between them. Map distance. % recombination between any two linked genes x 100. Map unit. 1 % recombination = 1 centi-Morgan (cM). This is approximately one million bp. in duplex DNA. A b Cont...
Cytological map. A line diagram showing the actual physical distances between linked genes.
Genetic and cytological maps have the same gene order but different map distances.
a b c d Some ground rules & facts to remember
!Crossing-over (or recombination) occurs after DNA replication and at pachytene (four-strand stage). !Crossing-over (or recombination) is reciprocal. !% crossing-over and % recombination are fully correlated when map distances are < 10. !% crossing-over exceeds % recombination when map distances are > 10. (i.e., some recombinants are not detected). !Recombination frequency decreases towards the centromere (heterochromatin effect). !Only odd number cross-overs produce detectable recombination. Even number cross-overs undo each other. Genetic map
a Cytological map b c d Undetectable crossing-over or exchange Detectable single crossing-over produces four different gametes (two parentals+two recombinants) Detectable crossing-over or exchange A dihybrid produces two types of gametes if there is no crossing-over between two linked genes but four types gametes if there is crossing-over. Some ground rules & facts to remember
Recombination is detected in heterozygotes only. It also occurs in homozygotes with the same frequency but not detected. Why not? Maximum frequency of recombination between two linked genes is 50% and cannot exceed 50%. All are parental (nonrecombinant) 50
% Recombination 40 30 20 10 10 20 30 40 50 60 70 80
Map Distance 90 100
20 Recombination cannot exceed 50% because:
a) Even if every meiocyte has a crossing-over (i.e. 100% crossing-over) between two linked genes, only 2 of 4 chromatids in a bivalent (tetrad) are involved in x-over. (2/4 = 50%)
A A B B Continued...
b) Double or multiple crossovers decrease the number of detectable recombinants (cancellation effect). !50% recombination between two genes gives the
same result as independent assortment. Then, how does one know whether one is dealing with linkage or independent assortment? meiocytes. This means that when one obtains 50% recombination, all meiocytes (100%) must have had crossing-over (chiasma) between two linked genes. P R x
a b A a b B !1% recombination requires crossing-over in 2% of a b P What percent of the gametes will be recombinants when there is 100% crossing-over between two linked genes? Two-Point Crosses determine linkage, and
distance between two linked genes. Cross parents differing with respect to the alleles of two linked genes, testcross the F1, and analyze the testcross progeny.
+ + + + d + p + d d p p "a" and "b" are linked (50 cM)
A a B b AB/ab x ab/ab
a b 1/4 1/4 1/4 1/4 A A a a B b B b P R R P Testcross
P R = 39 = 19.6 cM 199 1:1:1:1 testcross ratio Testcross Progeny 81 + + ; 79 d p 22 + p ; 17 d + Can you conclude if this is really independent assortment or linkage? How? Map Distance (% recomb.) # Recombinants # Recomb. + Parentals Two-point crosses are useful but not adequate; they have disadvantages. They: Underestimate map distance when it is > 10 cM (double crossovers cancel each other). Provide no information about relative positions of two linked genes. Do not allow detection of double crossovers. A vast excess (84% vs. 50%) of parental types among testcross progeny indicates that genes "t" and "d" are linked. Cure: Three-point crosses. Three Point Crosses
Example. George Beadle's corn data (1931).
+ + + gl gl + + va + v va va v v Cont... Testcross (8 gamete types) + + + / gl va v X gl va v / gl va v
(one gamete) ! ! ! F1
gl + + + ! P Test Cross Progeny Phenotypes Normal (WT) Glossy-variable Variable Variable-virescent Glossy-variable-virescent Glossy Glossy-virescent Virescent # of individuals Predicted genotype 235 P + + + 62 S gl va + 40 S + va + 4 dc + va v 270 P gl va v 7 dc gl + + 48 S gl + v 60 S + + v 726 How to analyze data from a threepoint cross? Identify parentals (most frequent classes): 235 and 270. Identify doubles (least frequent classes): 7 and 4. Determine the gene order (what order yields the observed double recombinants?) . P (235) + + + gl + + (7) DC P (270) gl va v + va v (4) DC Cont...
New gene order: v gl va
v gl or va
va gl v + + + I II I + II v-gl gl-va (doubles) v + + + gl va v gl + + + va v + va + gl + 60 62 48 40 4 7 Map distances
(v-gl interval) = (60+62+7+4) = 139 = 18.3% =18.3cM. 726 726 gl-va (interval) = 40+48+7+4 = 99 = 13.6% = 13.6 cM. 726 726
va gl v Cont... Coefficient of coincidence = Obs. freq. Doubles Exp. freq. doubles Observed = 7 + 4 = 1.5 % 726 Expected = 0.136 x 0.183 = 0.025 or 2.5% c.c = 1.5 = 0.6 2.5
0 < c.c < 1.0 Map 13.6 31.9 18.3 Cont...
Interpretation of cc. When there is a crossover in region I, it occurs (coincides with) with a crossover in region II 60% of the time. Coefficient of interference = 1 - c.c. = 1 - 0.6 = 0.4 Interpretation. When there is a crossover in region I, it inhibits (interferes with) a crossover in region II 40% of the time. If observed doubles < expected doubles, there is said to be positive interference. Positive interference is complete (c.c.=0 or c.i.=1.0) A three-point testcross in Drosophila indicates linkage and yields data for mapping Genetic maps of two Drosophila chromosomes Gene Mapping in Humans Mapping using data from pedigrees (difficult) Mapping with molecular markers (easy and getting easier). Utilizes variation in restriction fragment lengths, repeat number in tandem repeats, single nucleotide mutations, etc.
Genetic Map of # I and II Chromosomes in D. melanogaster In situ hybridization (FISH, etc.). Annealing a labeled DNA fragment (gene) to a prophase or metaphase chromosome fixed on a microscope slide. Determines directly in what chromosome and where in the chromosome a gene resides and the physical distance between any two genes in the same chromosome. Whole genome or whole chromosome sequencing. Determines precisely in what chromosome and where in that chromosome a gene resides and its distance from other genes in the same chromosome. No need for crosses. Physical (i.e. deletion) mapping. If a gene product or trait is missing after a chromosome segment is deleted, the gene must reside in the deleted region. In situ hybridization (FISH) localization of two genes (red and green) to two different human chromosomes Gene product present Gene product absent Spectral karyotyping uses chromosome-specific DNA probes for each human chromosome and identifies all of the 24 different chromosomes. Somatic Cell Genetics Human gene mapping by somatic hybridization is based on synteny, that is, the correlation between the presence vs. absence of a human chromosome and the presence vs. absence of a human gene product (typically a protein). Finding correlation between the presence of a human chromosome and a human gene product (protein) is the key to gene mapping by somatic cell hybridization. Making human-mouse somatic cell hybrids In this example, a human protein is present only when a clone has the human chromosome number 4. Therefore, the gene encoding this protein must reside in #4 chromosome. Correlation between the presence of a specific chromosome and a specific gene product (protein): A=chromosome 5; B= chromosome 3; C= chromosome (none) ; and D= chromosome 1 Genetic maps of the X chromosome and #1 chromosome in humans Physical Evidence for crossing-over and recombination (McClintock, 1931) Q. Do non-sister chromatids in a pachytene bivalent actually break and exchange chromatid segments? A. McClintock and Creighton (1931) in maize and Curt Stern in Drosophila (1931) provided unequal evidence that indeed physical breakage and exchange occur. They used cytological markers to show when a marker allele moves from one homolog to the other, a cytological marker (e.g., knob) near the marker also moves. Barbara McClintock (left) and Harriet Creighton (1931) Cytological evidence for physical exchange (crossing-over) between nonsister chromatids Every colorless, waxy maize had a translocated chromosome without a terminal knob indicating that crossing-over involves actual physical breakage and rejoining of nonsister chromatids. Loss of the knob from the translocated chromosome and its appearance in a normal chromosome (colored, starchy) can only be explained by chromatid breakage and rejoining. ...
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This note was uploaded on 04/01/2008 for the course BIOL 2104 taught by Professor Dcimini during the Spring '08 term at Virginia Tech.
- Spring '08
- molecular biology