Lecture--Chapter 7 Flashcards
Each organism contains how many genes?
hundreds to thousands; organisms have a small number of chromosomes compared to the estimated numbers of genes
Chromosomes must contain how many different genes?
hundreds to thousands; transmission of genes on a chromosome appears inconsistent with Mendel’s law of independent assortment
____ observed traits that did not assort independently:
Bateson and Punnett
Traits that did not assort independently:
- Sweet peas: flower colour and pollen shape
- Expected F2 generation phenotype ratios, 9:3:3:1
- Observed ratios (rounded), 16:1:1:5
These genes appear to be coupled (linked)
allelic combination of parent(s):
parental
novel combination of alleles; recombinant
nonparental
Groups of genes that tend to be transmitted as a unit are called:
a linkage group; genes that are physically close together on the chromosome
Sweet pea ___ and ___ genes are on the same chromosome:
flower colour; pollen shape
they are linked.
Chromosomes are also ___
linkage groups; a group of genes that are linked together
The number of linkage groups in a species is:
the number of chromosomes in the genome of the species
(for example, in humans there are 22 autosomal linkage groups, there is an X chromosome linkage group, and there is a Y chromosome linkage group in males)
occurs around prophase I of meiosis:
recombination (crossing over)
homologous chromosomes form:
chiasmata
Crossing over during meiosis can:
separate linked genes in diploid eukaryotic species; nonparental (recombinant) phenotype combinations are produced
The frequency of recombinant phenotypes is proportional to:
distance between the genes
A very small frequency of recombinant offspring indicates that:
two loci lie very close to each other on the chromosome
Genes that are far apart on the same chromosome may independently assort from each other due to:
high frequency of cross-over events; they will appear unlinked
Who provided the first direct evidence of linkage?
T.H. Morgan
Studies of several traits that followed an X-linked pattern of inheritance:
- body colour
- eye colour
- wing length
offspring with nonparental phenotypes are the product of a crossover:
Morgan’s Hypothesis
can be used to determine whether the hypothesis is supported:
the Chi Square Test
the genes sort independently:
null hypothesis; provides expected results which can be calculated; genes are analysed in pairs for the test; the predicted ratio is 1:1:1:1.
Calculate the expected values of each of the 4 phenotypes:
- Expected ratio is 1:1:1:1
- 2,205 total offspring
- Expected number of each phenotype (combining males and females) = 1/4 x 2,205 = 551 (your E value)
- X^2 = ((O1 - E1)^2)/(E1) + …
= ((O1 - 551)^2)/(551) + …
= 2109.8
Interpret the calculated chi square value:
- Determine the degrees of freedom to use the chi square table
a. the law of independent assortment states that two different genes randomly assort their alleles during gamete formation. Thus, we have two expected classes–nonrecombinant and recombinant (n=2)
b. df = 1 - For a probability of 0.01 with a df=1, the chi square value is 6.635, which is < 2109.8.
The implications of Morgan’s data:
- The combinations of traits found in the parental generation had the highest proportions; all three genes are located on the X chromosome; therefore, they tend to be transmitted together as a unit
- Flies that had recombined parental traits were also obtained; genes could become unlinked.
- Provided direct evidence that crossing over results in recombination.
- One corn strain had a mutant chromosome 9 with 2 observable abnormalities.
Harriet Creighton and Barbara McClintock
located near the knobbed end:
kernel colour; C = coloured, c = colourless
Endosperm texture:
- Wx = starchy endosperm
2. wx = waxy endosperm
Testcrosses of the linked genes on chromosome 9 and the abnormal chromosome; inspected:
the chromosome structure of the nonparental recombinants with microscopes
could correlate the recombinant offspring with observable physical exchanges between homologous chromosomes
testcross results
an undergraduate in T. H. Morgan’s lab:
Alfred Sturtevant
recombination frequencies can be used to determine the map order of the genes
Alfred Sturtevant
Use mapping to determine the order of genes on a chromosome
Alfred Sturtevant
he created the first linkage map in 1913
Alfred Sturtevant
the percentage of recombinant offspring correlates with:
the distance between the two genes
Genes far apart —>
high recombination frequency (nonparentals are common)
Genes close together —>
low recombination frequency (nonparentals are rare)
Map Units: map distances are based on the frequency of recombination events:
- Frequency of recombination = (# nonparentals / total # offspring) x 100
- 1% frequency = 1 map unit (mu)
- Also called 1 centiMorgan (cM)
determine the recombination frequencies between 5 loci on the X chromosome
Alfred Sturtevant’s Experiment
Alfred Sturtevant’s Experiment: the genes: body colour
yellow (y); wt (y+) is gray
Alfred Sturtevant’s Experiment: the genes: eye colour
- white (w); wt (w+) is red
2. vermillion (v); wt (v+) is red
Alfred Sturtevant’s Experiment: the genes: wing type
- miniature (m); wt (m+) is normal
2. rudimentary (r); wt (r+) is normal
In Alfred Sturtevant’s Experiment, 2 traits were analysed at a time, using dihybrid crosses:
- female heterozygous
2. male hemizygous recessive
Alfred Sturtevant’s Experiment: the interpretation: some dihybrid crosses had low % of nonparental (recombinant) offspring
- only 3% recombinant offspring in crosses of v and m alleles
- these two genes are likely very close together (3 mu)
Alfred Sturtevant’s Experiment: the interpretation: other dihybrid crosses had a higher % of nonparental offspring:
- 33.7% recombinant offspring in crosses of w and m alleles; these two genes are farther apart than y and w (33.7 mu)
- 29.7% recombinant offspring in crosses of w and v; these two genes are closer together than w and m (29.7 mu)
- Therefore, v is closer to m than it is to w, and the gene order is w - v - m.
Alfred Sturtevant’s Experiment: the construction:
Sturtevant proposed the genetic map starting with y
Genetic Mapping Discrepancies: the calculated distance between genes and the frequency of recombination:
do not always agree; for example for y - r genes, the actual map distance is 57.6, but the recombination frequency is 37.5%.
Increasing distance means:
multiple crossover events may occur (but not be detected)
Uses of genetic mapping:
- Determine linkage between genes.
- Provide an understanding of the complexity and genetic organisation of a particular species.
- Help understand evolutionary relationships.
- Diagnose, treat inherited human diseases.
- Predict likelihood of transmitting an inherited disease.
- Provide information for selective breeding in agriculture.
Mapping using ____ can yield information about map distance and gene order:
trihybrid crosses
What are the steps for trihybrid crosses:
- Produce heterozygotes.
- Perform testcross.
- Collect F2 generation data.
- Analyse the data.
Produce heterozygotes:
- Cross two true-breeding strains that differ in three alleles
- F1 progeny are heterozygous in all three genes
Perform testcross:
- F1 heterozygote females crossed with homozygous recessive males.
- Crossover events may occur during meiosis in F1 offspring.
Collect F2 generation data:
usually requires lots of data
Analyse the data:
gene order, map distances, map construction
genetic mapping is possible to correct for:
double crossovers
The discrepancy between the ____ and the ____ is due to the inability to detect double crossovers between 2 gene loci:
map distances; frequency of recombination
Correction for double crossovers can be done by:
multiplying the number of flies (or other variable) where double crossover occurred by 2
for example: for body colour and wing shape (17.8 mu initially)
-map distance = {[179 + 2(2+1)] / 1005} x 100
= 18.4 mu
causes the number of observed double recombinants to be lower than expected:
interference
Use ___ to calculate each crossover event:
product rule
one crossover event reduces the frequency of a second crossover in the same area of a chromosome
positive interference
Calculating interference:
Interference is expressed as: 1 - C, where C = (observed # dbl crossover events) / (expected # dbl crossover events)
crossing over can occur during mitosis:
mitotic recombination
Mitosis does not involve the ___ of chromosomes to form bivalents
homologous pairing
Crossing over in mitosis:
rarely occurs
If mitotic recombination occurs during an early stage of embryonic development:
it may be observable in the somatic cells
daughter cells with recombinant chromosomes:
continue to divide
continuation of division in daughter cells can result in:
a patch of tissue with characteristics that are different from those of the rest of the organism
example of mitotic recombination:
twin spots
Fly strains carry X-linked genes that:
affect body colour and bristle morphology
body colour:
yellow vs. gray
bristle length:
short vs. long
