I.
Mendelian inheritance has its physical basis in the
behavior of chromosomes during production of gametes
(i.e., sex cells)

Based upon observations, biologists developed the chromosome
theory of inheritance…
1. Mendelian factors or genes are located on
chromosomes
2. It is the chromosomes that segregate and independently
assort
II.
Morgan traced a gene to a specific chromosome

Thomas Hunt Morgan, Columbia University, performed
experiments in the early 1900’s which provided convincing
evidence that Mendel’s inheritable factors are located on
chromosomes.

Morgan selected the fruit fly, Drosophilia melanogaster, as the
experimental organism because:
1. They are easily cultured in the laboratory;
2. They are prolific breeders;
3. They have a short generation time;
4. They have only 4 pairs of chromosomes which are
easily seen with a microscope. (see overhead)

Sex-linked genes: Genes located on sex chromosomes. the
term is commonly applied only to genes on the X chromosome.

Morgan and his colleagues used genetic symbols that are now
“convention”
 A gene’s symbol is based on the first mutant, nonwild type discovered; if recessive, the first letter is
lowercase (e.g., w =white eye allele in
Drosophilia); if dominant, the first letter is
capitalized (e.g., Cy =curly allele in Drosophilia
that causes abnormal, curled wings.)
 the wild type trait is designated by a superscript +
(e.g., Cy+ = allele for normal, straight wings.)


Wild type: Normal or most frequently observed phenotype
Mutant phenotypes: Phenotypes that are alternatives to the wild
type and which are due to mutations in the wild-type gene.

Discovery of Sex-linked genes
 After a year of breeding Drosophilia to find variant
phenotypes, Morgan discovered a single male fly
with white eyes instead of the wild-type red.
Morgan mated this white-eyed male with a redeyed female. (see overhead)
 Morgan deduced that eye color is linked to sex and
that the gene for eye color is located only on the X
chromosome. Premises for this conclusion were:
1. If eye color is located only on the X chromosome,
then females (XX) carry 2 copies of the gene,
while males (XY) have only 1 copy.
2. Since the mutant allele is recessive, a white-eyed
female must have that allele on both X
chromosomes which was impossible for F2
females in Morgan’s experiment.
3. A white-eyed male has no wild-type allele to mask
the recessive mutant allele, so a single copy of the
mutant allele confers white eyes.
III.
Linked genes tend to be inherited together because they are located
on the same chromosome

Genes located on the same chromosome tend to be linked in
inheritance and do not assort independently because they are on
the same chromosome and move together through meiosis and
fertilization.

A dihybrid cross following 2 linked genes will not produce an
F2 phenotypic ratio of 9:3:3:1. Why? (see overhead)

Morgan performed a dihybrid testcross between:
b = black body
b+ = gray body
b+bvg+vg
x
gray, normal wings
IV.
vg = vestigial wings
vg+ = wild-type wings
bbvgvg
black, vestigial wings

Resulting phenotypes of the progeny did not occur in the
expected 1:1:1:1 ratio for a dihybrid testcross.

A disproportionately large number of flies had the phenotuypes
of the parents: gray with normal wings and black with vestigial
wings.

Morgan proposed that these unusual ratios were due to linkage.
Independent assortment of chromosomes and crossing over cause
genetic recombination

Genetic recombination: the production of offspring with new
combinations of traits different from those combinations found
in the parents; results from the events of meiosis and random
fertilization.
1. Recombination of unlinked genes: independent
assortment of chromosomes
 Mendel discovered that some offspring from dihybrid crosses
have phenotypes unlike either parent. An example is the
following cross between pea plants:
YY, Yy = yellow seeds
yy = green seeds
RR, Rr = round seeds
rr = wrinkled seeds
P generation:
YyRr
yellow round
X
yyrr
green wrinkled
Progeny:
¼ Yy Rr
¼ yyRr
green, round
¼ yyrr
} Parental types (50%)
¼ Yyrr
} Recombinant types (50%)
yellow, wrinkled

Parental Types: Progeny that have the same phenotype as one
or the other of the parents.

Recombinants: Progeny whose phenotypes differ from either
parent.

The genes for seed shape and seed color assort independently of
one another because they are located on different chromosomes
which randomly align during metaphase of meiosis I.
2. Recombination of Linked Genes: Crossing Over

If genes are totally linked, some phenotypic combinations
should not appear. Sometimes, however, the unexpected
recombinant phenotypes do appear! (see overhead)
b = black body
b+ = gray body
b+bvg+vg
X
gray, normal wings
vg = vestigial wings
vg+ = wild-type wings
bbvgvg
black, vestigial wings

Morgan’s results showed that the two genes were neither
unlinked nor totally linked.

17% of the progeny were recombinants…the linkage must be
incomplete!

V.
Crossing Over during meiosis accounts for the recombination
of linked genes. the exchange of parts between homologous
chromosomes breaks linkages in parental chromosomes and
forms recombinants with new allelic combinations.
Geneticists can use recombination data to map a chromosome’s
genetic loci

Scientists used the recombination frequencies between genes to
map the sequence of linked genes on particular chromosomes.
1. Morgan’s Drosophilia studies showed that some genes
are linked more tightly than others.
--For example, the recombination frequency between the
b and vg loci is 17%
--the recombination frequency is only 9% between b and
cn, a third locus on the same chromosome (cinnabar
eyes)

Sturtevant used recombination frequencies between genes to
assign them a linear position on a chromosome map.

He defined one map unit as 1% recombination frequency, called
a centimorgan, in honor of Morgan.

Using crossover data, a map may be constructed as follows:
1. Establish the relative distance between those genes
farthest apart or with the highest recombination
frequency.
b---------------------------------------------------vg
------------------17--------------------------
2. determine the recombination frequency between the
third gene (cn) and the first (b).
cn---------------b
------9----
3. Consider the two possible placements of the third gene:
a. cn------------b-------------------------------------vg
-----9-------------------17---------------
b. b------------cn------------------------------------vg
-----9-----------------?------------------
-----------------------17-----------------------
4. Determine the recombination frequency between the
third gene (cn) and the second (vg) to eliminate the
incorrect sequence.

The correct sequence is b----cn----vg; the distance between cn
and vg is 17cM – 9 cM = 8 cM. However, the recombination
frequency data between cn and vg was 9.5%.
What causes this discrepancy?
Loci
b and vg
cn and b
cn and vg
Recombination
Frequency
17.0%
9.05
9.5%
Approximate Map
Units
18.5
9.0
9.5
Note that there are actually 18.5 map units between b and vg.
This is higher than that predicted from the recombination
frequency of 17.0%. Because b and vg are relatively far apart,
double crossovers occur between these loci and cancel each other
out, leading us to underestimate the actual map distance.

If genes are so far apart that the recombination frequency is
50%, they are indistinguishable from unlinked genes that assort
independently.

Linked genes that far apart can be mapped if additional
recombination frequencies can be determined between
intermediate genes and each of the distant genes

Maps based on crossover data only give information about the
relative position of linked genes on a chromosome. another
technique, Cytological Mapping, pinpoints the actual location
of genes and the real distance between them.
For practice, try to predict the map of the following testcross results:
Loci (Genotypes) Actual Results
bvg+/bvg
b+vg+/bvg
bvg/bvg
b+vg/bvg
cn and b
b and vg
vg and cn
156
915
894
135
Recombination
Frequency
1. ?
Approximate
Map Units
2. ?
15%
1. ?
3. ?
Recombination Frequency = recombinants
total offspring
x 100%
Answer:
1. R.F.= 291/2100 x 100% = 13.86%
2. 13.86 cM
3.
a. cn----------------------------b-----------------vg
-------------15----------------13.9--
b. cn-------vg-------------------------------------b
--1.1---------------13.9-------------
vg and cn could either be 28.9 cM or 1.1 cM apart,
further cross data is needed to confirm!