Lecture 4 Flashcards
Mendel discovered that alleles at a single locus segregate equally because
they are on paired chromosomes,
different loci (and their alleles) independently assort because they are
—- on different chromosomes
ratio when we cross a dihybrid F1 to a tester strain
1:1:1:1
F1 tester cross
pr+/pr vg+/vg x pr/pr vg/vg
gamets produced by left
pr+ vg+ 1339
pr vg 1197
pr+ vg 151
pr vg+ 154
total : 2841
what is expected ratio?
what does it show?
ignore right because only can produce pr/vg
expected: (1:1:1:1)
pr+ vg+ 710
pr vg 710
pr+ vg 710
pr vg+ 710 .
2840
shows: Linked b/c sample size is big enough and very large size difference from expectations
still some recombinance even though linked
why is there still some recombinace even through linked? (when F1 tester cross
pr+/pr vg+/vg x pr/pr vg/vg)
because there is physical crossing over between homologous chromosomes during meiosis 1
When genes are in close proximity on the same chromosome they ___
do not independently assort.
a bias ratio will result favoring the genotype(s) of the parents.
Genes on different chromosomes typically have a
semicolon A/a ; C/c
Unknown linkage is often separated by a
middle dot A/a * D/d
Crossing over “could” occur during ___
but ____
the two-chromosome (meiosis II) stage or the 4 chromatid (meiosis I) stage of meiosis.
but we know that crossing over occurs at relatively early in the 4 chromatid stage (meiosis I).
Even though any single crossing-over event is only between one pair of chromosomes, ____ can occur
multiple crossing over events
crossing over between sister chromatids
can occur but would have no effects because sister chromatids are identical
McClintock and Creighton
crossing over - first discovered over a centry ago
Crossing over appears to occur as the strands of DNA become broken during meiosis and this breakage facilitates the crossing over event
Crossing over frequency is variable by
species, gender, and age.
E.g. male Drosophila have very little or no crossing over,
male humans may cross over less than females.
In recent years (since 2000), there has been a great deal of new data published on recombination as it applies to the human genome:
1) Generally, the rate of human recombination is ~1.6 cross overs per chromosome
2) Some chromosomes display a much higher/lower rate of recombination
3) There is a ten fold difference in rate across Eukaryotes
4) Crossing over tends to occur in certain “hot spots”. These seem to correlate with G/C rich regions, methylation, and are out side of genes and the “control regions” of genes
F1 tester cross
pr+/pr vg+/vg x pr/pr vg/vg
observed:
pr+ vg+ 1339
pr vg 1197
pr+ vg 151
pr vg+ 154
total : 2841
by calculating the cumulative frequency (____) we can ____ because _____
(# recomb /# total)x 100
(151+143)/2841= (0.107)x100=10.7%
we can “map” alleles on chromosomes
frequency of recombination suggests how close these loci are on the same chromosome
Sturtevant and Morgan
(Sturtevant, undergrad student of Morgan) first to deduce this logic, that father apart the genes (loci) are on the chromosome the more frequently they will cross over
noted that the distances implied by the crossing over were relative, not absolute distances
percent of recombinants is synonymous with
“map units” (m.u.) a relative distance measure.
or centimorgans, cM (for T.H. Morgan)
crosses (A-B & A-C) was performed that resulted in the map distances: A and B 5 units, A and C 3 units
what are possibilities for a linkage map?
How know which is correct?
B and C could be 8 units apart or 2
to know which is correct - preform another cross with B + C
(to have total of three, two point crosses)
(OR tri-hybrid (three point) tester cross)
Three Point Cross
v+/v+ * cv/cv * ct/ct cross(x) v/v * cv+/cv+ * ct+/ct+
gametes produced:
F1 trihybrid:
Tester cross:
regardless of the number of loci (genes) we end up performing a cross of ___
Three Point (loci) Cross - Tri-Hybrid
gametes produced: v+ * cv * ct and v * cv+ * ct+
F1 trihybrid: v+/v * cv/cv+ * ct/ct+
Tester cross: v + /v * cv/cv+ * ct/ct+ cross(x) v/v * cv/cv * ct/ct
end up performing a cross of: a heterozygote (dominant) with a homozygous recessive
ignore the gametes on the male side (v/v * cv/cv * ct/ct) during tri-hybrid cross because
the male side (the homozygote) because he can only produce one gamete type
: v cv ct,
(and his gametes do not recombine)
in tri-hybrid results
why we do not see v cv ct as one of the highest proportion (parental)?
Because the male gametes do not recombine (we can limit our examination to those from the female.)
in tri-hybrid results
What can we conclude about the two largest categories?
they are non- recombining gametes
how to determine the recombination distance for two loci of tir-hybrid tester?
1) place the three pairs of possible loci at the top
2) use the pairs listed and categorize if that row is a recombinant (R) (if those two are seen in one of the non-recombining)
3) sum the total of those rows
4) take this sub-total and divide by the total
what is the recombination distance for v and cv ?
v and ct?
ct and cv?
are four rows of recombinants
45+40+89+94 = 268/1448 = 18.5 (%)
recombination distance for these two loci (v and cv) is 18.5 units
v and ct = (191/1448) = 13.2
ct and cv = (93/1448) = 6.4
mapping results of tri-hybrid tester cross
v and cv = 18.5%
v and ct = 13.2%
ct and cv = 6.4%
data suggests that ___
the tester cross (not the P1) is now accurately written as _____
v and cv are the furthest apart (@ 18.5m.u.)
ct must be in between these, 6.4 from cv and 13.2 from v.
v+ ct cv/v ct+ cv+ cross(x) v ct cv/v ct cv
[slashes signify the homologs of the parents. The dots are gone because they symbolized unknown, now know]
How do we account for the difference of 13.2 and 6.4 = 19.6 NOT 18.5 ?
The rarest two classes of recombinants (the double recombinants) are not included correctly using this method.
If they get included and we recalculate the total distance we get 19.6.
Why not use recalculate distance: Because other method gives a better estimate of map distance.
For three loci on a chromosome there are _____ different orders possible
In a trihybrid tester cross there are ____ genotypes possible
only three (# that could be in middle)
eight (2 OJ, 4 single cross, 2 double cross)
for the two rarest two categories
The loci that is common to both ___ is ___
ct (one not present in non-R category OR in both of two loci involved in those categories, v and ct, cv and ct)
is the one in the middle.
the rarest is the (only) one with a double cross over event
Manhattan plot
plot from a SNP GWAS
Y axis: strength of the correlation (between SNP and trait)
x axis: the particular chromosome and region of that chromosome
(ex. here very strong correlation with one SNP site in particular, NOD2)
(line shows how string the significance need to be to investagate)
difficulties with SNP GWAS
Traits can have many loci that act weakly on the trait are much more challenging to map.
the number of protein coding genes in the human genome by chromosome number
Y chromosome and the M (mitochondrial DNA) lowest
Y has ~ 55 protein coding genes
there is a rough correlation between the physical size of the chromosome and the number of protein coding genes
if the results of a dihybrid to tester cross were much closer (than obv. linkage, ex. image) we would need to preform ___ test for linkage
chi-square
is there linkage?
1) calculate expected
2) calculate chi-square value
3) Sum them
4) Calc degrees of freedom (# of categories - 1)
5) if (sum) exceeds a P of 0.05 (here, 7.815), reject null hypothesis (of non-linkage) thus, deduce that two loci are linked.
if chi square value (sum from formula) meets or exceeds the value in the chart (ex. 7.815)- deviates significantly (thus linked)
what do results of linkage test tell us?
1) AB and ab are in the majority, and are linked by cis (NOT trans)
2) recombinants are (37+41)/200= 39% or 39m.u. apart
cis vs trans conformation
cis - when both dominant alleles (typically wild types) are linked on the same chromosome
trans - when the dominant allele (usually wild type) is linked to the recessive or mutant allele on the chromosome.
