Bio 110
005s Gene Linkage, Recombination, Mapping of Genes
Beavers
1. GENETIC LINKAGE, RECOMBINATION AND MAPPING OF GENES
a. QUESTIONS TO CONSIDER:
i. If genes on different chromosomes assort independently, how do genes on
the same chromosome behave?
ii. Why do genes on different chromosomes assort independently?
2. GENETIC LINKAGE
a. (a) 1928 4-generation pedigree following known X-linked red-green
colorblindness and hemophilia A.
i. Neither trait manifested without the other suggests:
1. No _________________________ ______________________
2. The genes are ________________________________
3. Gene for colorblindness is in close proximity to the gene for hemophilia A.
b. (b) 3 generation pedigree following known X-linked red-green colorblindness and
hemophilia B.
i. Are genes linked?
ii. What is the cause of variation in the offspring?
3. GENES THAT TRAVEL TOGETHER
a. The disorders present in the first pedigree are demonstrating genetic linkage.
b. Genetic linkage: the proximity of two or more markers on a chromosome; the
closer together the markers are, the lower the probability that they will be
separated by recombination. Genes are linked when the frequency of parentaltype progeny exceeds that of recombinant progeny.
c. Recombination frequency: P > R indicates linkage
d. Recombination frequency: P = R indicates independent assortment
e. Recombination: in general is the occurrence in progeny of new gene combinations
not seen in previous generations. This can occur in two way:
1. _______________ _______________________ of genes on _______
_________________ chromosomes.
2. ____________________ of genes on the same chromosome during gamete
formation as the result of _____________________ ____________.
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Bio 110
005s Gene Linkage, Recombination, Mapping of Genes
Beavers
f. Syntenic: relationship of two or more loci found to be located on the same
chromosome.
g. Two important themes:
1. The farther apart two genes are on the same chromosome, the greater the
probability of separation through ____________________________.
2. How often genes on the same chromosome separate through recombination
can indicate their relative locations on that chromosome. This allows for
____________ _____________________.
h. Why would you care?
4. DETECTING LINKAGE IN A DIHYBRID
a. Recall fly nomenclature:
i. Gene names are in italics (white eye = w)
ii. Many genes named for mutant phenotype.
iii. If mutant allele is dominant, allele name begins with a capital letter
(mutation: "Deformed" is abbreviated Dfd, note capital D).
iv. If mutant allele is recessive, allele name is completely lowercase ("wingless"
is abbreviated wg).
v. If allele is wild-type (normal), name of gene includes a superscript ( +)
(wild-type red eye flies: w+).
vi. See addition gene nomenclature guidelines in appendix of textbook.
b. The slash symbol: / used to SEPARATE genes found on chromosomes of a pair.
i. Male: X w y / Y represents the genes on his X chromosome and NO
CORRESPONDING GENES on his Y chromosome. What is his phenotype?
ii. Female: Xw y+ / Xw y+. What is her phenotype?
iii. These are:
________________
c. Additional notation for X and Y chromosomes – gene name is superscripted on X.
d. NOTE: no corresponding gene on the ____ chromosome.
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Bio 110
005s Gene Linkage, Recombination, Mapping of Genes
Beavers
e. WHY DO MALES IN THE F1 OF THIS CROSS REFLECT MOTHERS
PHENOTYPE?
Important: Use F2 males to determine linkage patterns in X-linked recessive
crosses. It's similar to a testcross – why?
f. DO FEMALES IN THE F1 REFLECT MOTHER’S PHENOTYPE? WHY?
i. Females are dihybrids with two alleles for each gene they reflect a
combination of their parents and the dominance relations of the alleles
determine phenotype.
g. The crossing of the F1 will answer our question about assortment of genes on the
same chromosome.
i. If independent as Mendel predicted, should result in a 1:1:1:1 ratio of
phenotype.
ii. See chromosome picture: w y+ /w+ y X w y+ / Y
iii. ½ should be of the original parental types and ½ should be of the
recombinant type.
iv. This can be easily counted in males which demonstrate the X received from
the mother.
h. OUR ANSWER IS SEEN IN THE ratio of F2 male offspring.
i. Largest number of offspring represent the parental phenotype.
j. Approximately 1% make up the recombinant phenotype.
k. Genes are LINKED.
l. When genes are linked, parental phenotypic combinations outnumber recombinant
types.
5. FREQUENCY OF RECOMBINATION IS INDEPENDENT OF THE
ARRANGEMENT OF ALLELES
a. Now compare two experiments:
i. The observed frequencies of the various types of progeny depend on how the
arrangement of alleles in the F1 females originate.
ii. The percentages of recombinant and parental types remain similar due to
location.
b. Two genes are considered LINKED when the number of F2 progeny with parental
genotypes EXCEED the number of F2 progeny with recombinant genotypes.
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Bio 110
005s Gene Linkage, Recombination, Mapping of Genes
Beavers
c. Recombination is not always this LINKED…1 out of 100 that recombine. The
percents can vary…but not beyond 50% which would indicate independent
assortment.
d. Fill in regarding progeny seen in: Linkage P__R Independent assortment P___R
e. Remember that there can be two classes (or more!!) for parental and recombinant.
6. DEFINTIONS
a. PARENTAL TYPES: phenotypes that reflect a previously existing parental
combination of genes that is retained during gamete formation.
b. RECOMBINANT TYPES: phenotypes reflecting a new combination of genes that
occurs during gamete formation.
7. DEFINITIONS
a. PARENTAL CLASSES: combinations of alleles present in the original parental
generation.
b. RECOMBINANT CLASSES: reshuffled combinations of alleles that were not
present in the parental generation.
c. LINKED: describes genes whose alleles are inherited together more often than not;
linked genes are usually located close together on the same chromosome.
8. AUTOSOMAL GENES AND LINKAGE
a. Linked autosomal genes are not inherited according to the 9:3:3:1 Mendelian
ratio.
b. Testcross the F1 double heterozygote to a double recessive: phenotypes of offspring
indicate what kinds of gametes from the mother.
c. Parental classes outnumber recombinant classes – what does this mean?
i. Genes are on the _____________ chromosome.
ii. Genes are __________________.
iii. What should you do if values of parental class comes close to equal values
of recombinants? How do you determine if it is linked or unlinked?
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Bio 110
005s Gene Linkage, Recombination, Mapping of Genes
Beavers
9. The Chi-Square Test and Linkage Analysis
a. The Chi-square test cannot prove a hypothesis, but it can allow rejection of a
hypothesis.
b. Null hypothesis: a statistical hypothesis to be tested and either accepted or rejected
in favor of an alternative.
c. The Chi-square test: In real-world genetic analysis, linkage may not be as tight as
what we have seen. Perhaps genes are linked, but the percentage of recombinant
classes are approaching 50%.
d. How can we have greater assurance of test results?
e. The Chi-square test: has been devised as a quantitative measure of the likelihood that
an experimentally observed deviation from the predictions of a particular hypothesis
could have occurred solely by chance.
f. Helps solve the problem of "chance" in transmission of traits.
g. Chance can produce fluctuations from the mean values expected of unlinked genes
(genes assorting independently).
h. The Chi-square test allows you to answer the question: Are observed data
significantly different from predicted data? The comparison of percent or ratio is not
enough to determine.
i. Absolute numbers obtained are important.
j. They reflect sample size.
k. Larger sample size = greater clarity if hypothesis is correct.
l. Need an accurate NULL hypothesis.
m. If the Chi-square test shows OBSERVED DATA differs SIGNIFICANTLY from
those expected with independent assortment, the researcher can REJECT null
hypothesis of NO LINKAGE.
n. The final result of the calculations is the determination of the "p" value (numerical
probability).
i. p-value: probability these numbers deviated by chance.
o. Probability (p value) HIGH: generally greater than 0.05 (>5%).
i. Tells researcher that there is a greater than 5% probability results they are
seeing happened by chance. This would mean that the deviation is
INSIGNIFICANT because it happened by chance.
p. Probability (p value) LOW: generally less than 0.05 (<5%). No great probability
that the deviation occurred by chance. Reject the null hypothesis.
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Bio 110
005s Gene Linkage, Recombination, Mapping of Genes
Beavers
10.CHI-SQUARE TEST ON MONOHYBRID AND DIHYBRID RATIOS
a. How close to an expected result is close enough?
b. A statistical test is needed to check such ratios against expectations: X2 = ∑ (OE)2/E for all classes
c. Chi-square test, or X2 test. (O = observed, E = expected) checks actual results to
expected results.
d. Observed results are compared with those predicted by a hypothesis.
e. General scientific convention is that a hypothesis will be rejected as false if there is
a probability of less than 5 percent of observing a deviation from expectations at
least as large as the one actually observed.
11.CRITICAL VALUES OF X2 DISTRIBUTION
12.Degrees of Freedom
a. Degrees of freedom is the number of categories or classes being tested minus 1.
b. Degrees of freedom can also be thought of as opportunities for change. For
example, if five random samples are drawn from a given population, there are four
opportunities for change, or four degrees of freedom.
c. The X2 test relies on sample sizes – larger samples mean greater reliability.
13.CHI-SQUARE TEST SUMMARY
a. Used to evaluate the probability that differences between PREDICTED results and
OBSERVED results are due to random sampling error.
b. “p” values of less than 0.05 allow rejection of the null hypothesis of no linkage and
accept the alternative of LINKAGE BETWEEN THE TWO GENES.
14.RECOMBINATION
a. 1909: Belgian cytologist Janssen describes chiasmata.
b. 1910: Morgan suggest chiasmata are sites of chromosome breakage and exchange.
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Bio 110
005s Gene Linkage, Recombination, Mapping of Genes
Beavers
No physical evidence until Creighton, McClintock (corn) and Stern (Drosophila)
demonstrated reciprocal exchange of parts between female and male chromosomes.
c. They saw in meiosis the physical evidence of CROSSING OVER which produced
RECOMBINATION.
d. Cytogenetics: is a branch of genetics that is concerned with the study of the structure
and function of the cell, especially the chromosomes. A cytogenetic chromosome
result would be visualization of chromosomes under a light microscope.
e. Physical markers: cytogenetically visible abnormalities that make it possible to keep
track of specific chromosome parts from one generation to the next.
f. Genetic markers: genes identifiable through phenotypic variants that can serve as
points of reference in determining whether particular progeny are the result of
recombination.
15.Recombination Frequencies: How far apart are two genes?
a. Chiasmata are  physical crossing over between genes leading to  recombination.
Varying linkage rates INDICATE different distances BETWEEN genes and
opportunities for recombination.
Therefore, genes must exist along a chromosome linearly, and can be mapped by
recombination frequencies: the percentage of recombinant progeny.
1% RF = 1cM = 1 m.u.
b. Recombination frequency: the percentage of recombination progeny; can be used
as an indication of the physical distance separating any two genes on a
chromosome.
16.RECOMBINATION FREQUENCIES BETWEEN TWO GENES NEVER EXCEEDS
50%.
a. It has never been experimentally observed to be greater than 50%.
b. Recombination frequencies near 50% suggest:
i. Two genes - on different chomosomes.
ii. Two genes lie far apart on the SAME chromosome.
iii. Two genes can be identified as syntenic if they can be tied to an
intermediary between them.
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Bio 110
005s Gene Linkage, Recombination, Mapping of Genes
Beavers
17.Two Point Crosses Establish Relative Gene Positions
a. Two point crosses: crosses tracing two genes at a time. This would establish genes
as linear.
b. DATA from breeding experiments demonstrate recombination frequency between
two genes, or the spatial arrangement of genes.
c. Once ORDER OF GENES becomes established, left to right becomes arbitrary.
18.ANALYZING RESULTS OF A THREE-POINT CROSS
19.CORRECTION FOR DOUBLE CROSSOVERS
20.INTERFERENCE and COEFFICIENT OF COINCIDENCE
a. Chromosomal Interference: the phenomenon of crossovers not occurring
independently. Interference = 1 – coefficient of coincidence.
b. Coefficient of coincidence: the ratio between the actual frequency of double
crossovers observed in an experiment and the number of double crossovers expected
on the basis of independent probabilities. coc = freq observed/freq expected
c. Complete interference would be 1: meaning no double crossovers occur in the
progeny in a particular gene region.
21.Genetic Maps and Reality Summary
a. The best way to create an accurate map is summing many small intervals.
b. Recombination rates differ from species to species.
c. Recombination can vary WITHIN a species: DROSOPHILA male has NO
RECOMBINATION DURING MEIOSIS.
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