Problem Set 4 Three-point
testcross in linkage analysis Three-point testcross in linkage analysis Q1. In Arabidopsis the following mutant alleles have been identified on chromosome 5: c (colorless flowers), I (lumpy leaf) and y (yellow plant). A test-cross between triple recessives and a plant heterozygous for the three genes yielded the following progeny:
a) What were the genotypes of the
parents? A: Since the test-cross is
between triple recessives and a plant heterozygous, the genotypes of the
parents are:
b) What are the genotype and phenotype
of progeny individuals c + l? A: The genotype is: c/c y/+
l/l c) Draw a map showing the gene order
on the chromosome and the map distances between the genes.
The recombinant frequency for genes y and l is the highest. As a result, y and l must be farthest apart. Gene c is between y and l and is closer to y because the recombinant frequency is lower for c and y than c and l.
As the data shown, y and l are farthest apart and have double crossing. To obtain 46.5 map units for genes y and l, double crossing need to be considered and calculated as follows: (121+114+77+73+18+18+22+22)/1000*100%=46.5% Calculate probability Q2. Consider human families with four children. a) What is the probability of the
birth order: Girl-boy-girl-boy? A: Each birth is an independent
event and the probability of getting a girl or a boy is ½. Since
there is only one way to get girl-boy-girl-boy, the probability is: b) What is the probability that two
will be boys and two girls?
As a result, there are six possible birth combinations. Each birth combination probability is: P = ( ½ )4 = 1/16 The total probability is the sum
of each birth combination probability: c) In what proportion of such families
will there be at least two boys? A: For such families to have at least two boys, following are the possibilities:
As a result, there are eleven possible
birth combinations to have at least two boys. Each birth combination probability
is: The total probability is the sum
of each birth combination probability: Gene mapping in bacterial crosses Q3. A cross is made between the Hfr ala+leu+pur+ and F- ala-leu-pur-. Interrupted mating studies showed that ALA entered the bacterium last. Recombinants were selected for ALA+ on medium containing leu and pur and tested for growth in the absence of leu or pur. The following numbers of recombinant classes were found:
a) Why were recombinants selected
for ALA+? Explain briefly. A: The selection of ALA ensures cells that receive fragments containing the last marker ALA would also receive earlier markers leu and pur. b) Draw a map showing the gene order and distances in recombination units. Possible gene order 1: ALA leu pur
Possible gene order 2: ALA pur leu
Now let's summarize the data and compare ALA+ leu- pur+ with ALA+ leu+ pur- :
Looking at the recombinants data, there are 7 recombinants for ALA+ leu- pur+ and 0 recombinants for ALA+ leu+ pur-. This suggests that possible gene order 2 is most likely the correct order. As a result, the gene order is: ALA pur leu The map distance is calculated as
follows: Therefore, the map showing the gene order and distance is drew as follow:
c) Why are ALA+ leu- pur+ recombinants
rare? Explain briefly. d) Why are there no ALA+ leu+ pur-
recombinants? Explain briefly. A: Because there are 4 cross over for the predicted gene order as indicated in part b). Genetic map by phage transduction Q4. A Q2 phage is used to transduce DNA with the markers met+leu+pyr+ into a met-leu-pyr- recipient. Met+ transductants are selected in this experiment and the following classes or phage recovered:
a) What is the co-transductance met/leu
and met/pyr?
b) Draw a map showing the gene order
and co-transductance distances.
Possible gene order 2:
Comparing the transductants, there are 312 co-transductance for the double cross of the non-adjacent genes in gene order 1. However, there are only 2 co-transductance for the double cross of the non-adjacent genes in gene order 2. As a result, it suggests that the gene order 1 is correct: met leu pyr The map showing the gene order and co-transductance distances is:
c) A valid number for pyr leu co-transductance
cannot be calculated in this experiment because of the experimental design.
Why not? Explain briefly.
The other pyr leu co-transductance requires double cross as shown in above. Since met+ transductants are selected in the experiment, this double cross is rare due the fact that the genes are far apart. As a result, the other valid number for pyr leu co-transductance can't be obtained for calculation. d) What are the relative sizes of
a piece of DNA containing these three genes and the piece of DNA included
in the Q2 phage? Explain briefly. e) Why were there only two met+leu-pyr+
transductants? Explain briefly. Identify the longest open reading frame Q5. Identify the longest open reading frame (ORF) in the sequence of based below and list the amino acid sequence coded by this sequence. CGAGATGCCTAAATGAGTTGGCCAGCAGAGCGAGCATGGATGTAATCAG Strand1 A: The open reading frame (ORF) of a sequence starts with AUG/ ATG and ends with a stop codon (UAA/TAA, UAG/TAG, UGA/TGA ). There are six possible open reading frames as listed below (3 from each direction): CGAGATGCCTAAATGAGTTGGCCAGCAGAGCGAGCATGGATGTAATCAG Strand1 GCTCTACGGATTTACTCAACCGGTCGTCTCGCTCGTACCTACATTAGTC Complement 
Looking at the six possible open reading frames, the most likely longest frame is the 4th ORF. The amino acid sequence coded by this nucleotide sequence is: M L A L L A N S F R H L
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