Download Midterm#1 comments#2 Overview- chapter 6 Crossing-over

Midterm#1 comments#2
• So far, ~ 50 % of exams graded, wide range of
results: 5 perfect scores (200 pts)
• Lowest score so far, 25 pts
• Partial credit is given if you get part of the answer
right
• Tests will be returned this Thursday
• Discussion section this week with go over genetic
fundamentals to help you solve problems- you
must understand these fundamentals to pass the
course.
Overview- chapter 6
• In meiosis, recombinant products with new
combinations of parental alleles are generated by:
– independent assortment (segregation) of alleles on
nonhomologous chromosomes.
– crossing-over in premeiotic S between nonsister
homologs.
• In dihybrid meiosis, 50% recombinants indicates
either that genes are on different chromosomes or
that they are far apart on the same chromosome.
• Recombination frequencies can be used to map gene
loci to relative positions; such maps are linear.
• Crossing-over involves formation of DNA
heteroduplex.
Crossing-over
• No loss of genetic material, just formation
of new chromatids
• Parental chromatids are noncrossover
products
• Recombinant chromatids are always
products of crossing-over
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Consequences of crossing-over
• Frequency of recombinant gametes is 0-50%,
depending on frequency of meiocytes with
crossing-over
• Results in deviation from 1:1:1:1 in testcrosses
– parental combination is most frequent
– recombinant combination is rarest
• Allows drawing of linkage maps based on
recombination frequencies (RF)
Linkage maps
• Two genes (loci) on the same chromosome can
exhibit “linkage” = co-inheritance
• The closer two genes are to each other, the more
likely they are to show linkage …why? Because
crossing over occurs by chance and the more DNA
that separates two genes, the greater the likelihood
of a cross over (=recombination event)
• Groups of genes can be mapped using cross overs
to show their relative order on a chromosome or
part of a chromosome ( = linkage group)
• Distance between genes is determined by map
units = recombination frequencies, 1 mu= 1%
recombination between 2 genes
Is recombination perfect for
maps?
• Some cross overs will restore the parental
phase of genes used as markers, as the last
diagram showed
• In an experiment where you follow 3
markers, a double cross over between the 2
outside markers will produce double
recombinants that have the parental allelic
combinations So, recombination underestimates the physical distance (length of DNA)
between the markers, and small intervals are used to sum big distances.
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Recombination frequency (RF)
• Experimentally determined from frequency of
recombinant phenotypes in testcrosses
• Roughly proportional to physical length of DNA
between loci
• Greater physical distance between two loci,
greater chance of recombination by crossing-over
• 1% recombinants = 1 map unit (m.u.)
• 1 m.u. = 1 centiMorgan (cM)
Linkage maps
# observed
140
50
60
150
• RF is (60+50)/400=25%, clearly less than 50%
• Map is given by:
A
B
25 m.u.
Mapping
• RF analysis determines relative gene order
• RF between same two loci may be different in
different strains or sexes
• RF values are roughly additive up to 50%
– multiple crossovers essentially uncouple loci,
mimicking independent assortment
• Maps based on RF can be combined with
molecular and cytological analyses to provide
more precise locations of genes
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What do we need to do trihybrid
mapping?
• This is a linkage analysis or testcross for assigning
relative gene distance (in recombination
frequency) and gene order using 3 markers (loci,
genes) at once, here vestigial, purple, and black
• One parent will be heterozygous for 3 different
genes (construct this genotype by breeding)
• The other parent will be homozygous recessive for
those same genes (find or construct this one too)
• There are 2x2x2 =8 gametic genotypes that are
possible, and we have to be able to infer the
genotype from the phenotype (…that is why we
use the triple recessive in the testcross)
Trihybrid testcross
• Sometimes called three-point testcross
• Determines gene order as well as relative
gene distances
• 8 categories of offspring
– for linked genes, significant departure from
1:1:1:1:1:1:1:1
• Works best with large numbers of offspring,
as in fungi, Drosophila
Analysis of trihybrid testcross data
• Identify pairs of parental and recombinant offspring
– parental (noncrossover); most abundant
– double crossovers; least abundant
– single crossovers; intermediate abundance
• identify on the basis of reciprocal combinations of alleles
• Determine gene order by inspection (the parental
gene order yields double crossovers by switching
middle genes)
• Calculate RF for single crossovers, adding double
crossovers each time
• Draw map
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Genetic maps
• Useful in understanding and experimenting
with the genome of organisms
• Available for many organisms in the
literature and at Web sites
• Maps based on RF are supplemented with
maps based on molecular markers,
segments of chromosomes with different
nucleotide sequences
Tomato genotypes and the CEO
• Your sister’s high end restaurant serves expensive
beaked, yellow tomatoes, and you want to corner
the market in Hawaii, make lots of money, and
retire at 30!
• From Bio 375 you’ve learned that those two
markers should be on different chromosomes
beaked on 1, yellow on 2
• You ask if these are really unlinked genes, so you
plant a garden from seeds you gathered in a
greenhouse experiment of crossing, and now have
parentals and recombinants, but the numbers
aren’t exact and you don’t want to waste your
money…. What to do?
Chi-square test
• Statistical analysis of goodness of fit between
observed data and expected outcome (null
hypothesis)
• Calculates the probability of chance deviations
from expectation if hypothesis is true
• 5% cutoff for rejecting hypothesis
– may therefore reject true hypothesis
– statistical tests never provide certainty, merely
probability
Example of a test?
Start with a cross between A/A.B/B and a/a.b/b, in our case this
could be non-beaked, red x beaked, yellow plants.
Obtain a dihybrid, A/a.B/b (non-beaked, and red), testcross
to a beaked, yellow plant (a/a.b/b)
Get 500 progeny, classified as 142 A.B (non-beaked, red)
133 a.b (beaked, yellow)
113 A.b (non-beaked, yellow)
112 a.B (beaked, red)
Add the classes; parentals = 275 ( 142 +133)
recombinants = 225 ( 113 + 112), so RF= 225/500
=45%, and is <50%, so are these genes linked?
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