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1. A Drosophila female trihybrid at the Y, Ct and W loci is crossed to a pure­breeding wild­type male. The progeny and their phenotypes are listed below:
Sex
female
Phenotype
y+ ct+ w+
Count
2000
male
y+ ct+ w
y ct w+
y ct+ w+
y+ ct w
y+ ct w+
y ct+ w
y ct w
y+ ct+ w+
773
782
201
209
15
16
3
1
a) On what chromosome are the genes located?
b) Describe the genotype of the mother with respect to the individual chromosomes, give the gene order and map distances between them.
c) Calculate the coefficient of coincidence and interference for this cross.
2.
The above pedigree shows a family which seems to be segregating with an allele at the “B” locus
a) Which individuals in the pedigree had an informative meiosis?
b) Calculate the LOD score for an RF of 0.13631
c) What does your LOD score tell you?
d) While collaborating with another research group in France studying the same condition, they obtain a LOD score of 2.4 at this locus with the same RF, what does this tell you now? e) While sequencing candidate genes in this region you identify a promoter mutation in an enzyme which blocks transcription, what is the most likely basis for the observed autosomal dominance?
f) After publishing a paper on this premature stop codon, your collaborators in France embarrassingly identify a nearby mutation in the coding portion of a gene whose protein product associates as a heterodimer with another gene that has been previously identified as causal in this disorder. What is now the most likely basis for the dominant nature of the disease?
3. The following dihybrid cross was performed for the phenotypic traits: Tall (T > t) and serrated leaves (S > s).
T/T ∙ S/S x t/t ∙ s/s
Producing the expected F1 progeny of T/t ∙ S/s. One F1 dihybrid individual was then test crossed with a tester individual that was pure breeding for the short and smooth leaved phenotypes. The following progeny were observed:
Tall ∙ Serrated
Tall ∙ Smooth
Short ∙ Smooth
Short ∙ Serrated
346
290
320
298
Are the genes for height and leaf type on the same chromosome? Conduct a statistical test of your hypothesis (ie. linked or not linked). 4. A plant species (Foulus smellus) has beautiful flowers with a distinctive (and unfortunate) odor of old shoes. You have isolated five true breeding lines with flowers that smell like vanilla, due to recessive mutations.
You perform a series of crosses of these lines and assess the flower odor of the offspring. 1
2
3
4
5
1
Vanilla
Old Shoe
Old Shoe
Old Shoe
Vanilla
2
Old Shoe
Vanilla
Vanilla
Old Shoe
Old Shoe
3
Old Shoe
Vanilla
Vanilla
Old Shoe
Old Shoe
4
Old Shoe
Old Shoe
Old Shoe
Vanilla
Old Shoe
5
Vanilla
Old Shoe
Old Shoe
Old Shoe
Vanilla
a) How many different genes account for the mutations in these five lines? b) What type of test is this? Could we use this test if the mutations were dominant?
c) What are the genotypes and the pure lines and of the F1 plants?
d) Can we deduce the total number of genes controlling flower odor from this data? Answers
1.
a) The genes must be on the X chromosome because the females are all wild­type. b) The mother must be y+ ct+ w/y ct w+ because these are the most frequent class of progeny. The gene order must be y – w ­ ct because locus out of phase in the double recombinants with respect to the parental type is W. Distances: y­w = 1.75 m.u. ct­w = 20.7 m.u.
c) CoC = obs 2x recomb/exp. 2x recomb 0.002/(0.207 * 0.0175)= 0.56 Interference = 1­CoC = 0.44
2. 1) Every individual in the second generation that had children had an informative meiosis, we don’t know whether any of them are recombinant or parental because we can’t assign a phase for the B2 allele yet
2) Based on question 1, I count 9 parentals and two recombinants, if the RF is 0.13631, the probability of each recombinant is 0.068155 and the parentals are 0.431845
Therefore the LOD score is LOG{(0.431845)9(0.068155)2/(0.25)11}= 1
3) Our LOD score is 1, therefore it is 10x more likely that the disease causing mutation is located within 13.631 centimorgans than assorting independently.
4) LOD scores are cumulative, therefore our new LOD score is 3.4, it is now over 1000x more likely that there is linkage at our given RF, therefore our significance threshold has been reached.
5) Haploinsufficiency
6) Dominant­negative
3. Conduct a chi­squared test with independent assortment as the hypothesis. If independently assorted, we would expect a 1:1:1:1, though we must account for viability by using the allelic proportions that we have observed. Therefore:
T
t
Total
S
s
Total
346
(326.6)
298
(317.4)
644
290
636
(309.4)
320
618
(300.6)
610
1254
Chi­squared =1.15+1.22+1.18+1.25
= 4.80
P < 0.05 at 1df
Therefore we would reject the hypothesis of independent assortment and conclude that the genes are linked.
4. a) Three genes account for these mutations. Lines 1 & 5 and lines 2 & 3 have mutations in the same gene because the crosses do not allow a rescue of the wild­type phenotype. b) A complementation test, it can only be used for recessive mutations.
c) Let A, a1, a5 be alleles of gene 1, B, b2, b3 be alleles of gene 2 and C, c be alleles of gene 3.
Line 1: a1 a1 . BB . CC
Line 2: AA . b2b2 . CC
Line 3: AA . b3b3 . CC
Line 4: AA . BB . cc
Line 5: a5a5 . BB . CC
F1s:
1 X 2: Aa1 . Bb2 . CC
1 X 3: Aa1 . Bb3 . CC
1 x 4: Aa1 . BB. Cc
1 X 5: a1a5 . BB . CC
2 X 3: AA . b2b3 . CC
2 X 4: AA . Bb2 . Cc
2 X 5: Aa5 . Bb2 . CC
3 X 4: AA . Bb3 . Cc
3 X 5: Aa5 . Bb3 . CC
4 X 5: Aa5 . BB . Cc
d) No. This data has allowed us to identify three genes that are involved in determining flower odor, but there may be more genes for which we do not have mutants and therefore cannot be identified.