Dr Charu S
Genetics; solved problems and discussions
Following are the solutions and discussion for questions 26, 27 and 30, for the genetics test conducted on 10-3-15
26. Genes ho and dp are 9.0 map units apart, and genes dp and da are 26.3 map units apart. What is the probability of a double crossover in these regions?
- 1.3%
- 2.3%
- 3.4%
- 4.5%
Solution
We know that 1% cross over is equal to 1 MU.
Therefore % of single cross over (SCO) b/w ho and dp is 9%= a frequency of 0.09
SCO b/w dp and da is 26.3 %= frequency of 0.263
Probability of double cross over (DCO)= SCO b/w ho and dp X SCO b/w dp and dw
= 0.09 X 0.263 = 0.0237 =2.3%
27. In maize, two plants that are both heterozygous for the recessive alleles a and b are crossed. What frequency of double-mutant progeny will appear if a and b are 7.2 map units apart and both parents carry a and b in trans?
- 0.036
- 0.0625
- 0.001296
- 0.005184
Solution
When we say that both the parents are trans heterozygous for a and b, the arrangement is like this
Both Parent1 and 2 are a,b+/a+,b.
Each parent will produce the same 4 kinds of gametes. 2 parental types and 2 cross over types. Since a and b are 7.2 map units apart, it means the number of cross overs between a and b will be 7.2%. Therefor cross over gametes will 7.2% and non-crossover gametes will be 92.8%.
Therefore the gametes will be formed as follows
a,b+; parental; 46.4%
a+,b. parental; 46.4%
a+,b+; crossover 3.6%
a,b; crossover 3.6% =0.036 (frequency)
Similarly the other parent will also form the same type of gametes
The frequency of ab gamates from each parent combining to give double mutant progeny (ab/ab) is 0.036 X 0.036=0.001296
30 An experiment was conducted in E. coli to map the following genes (pro, his, bio, met, phe and trp) on a circular map using 3 different Hfr strains.
Strain 1 Order of transfer (early to late): trp met his pro
Strain 2 Order of transfer (early to late): his met trp bio
Strain 3 Order of transfer (early to late): pro phe bio trp
Based on the results what is the most likely map?
c d
Solution
Some bacteria harbour a F or fertility plasmid. It is conjugative plasmid that controls sexual functions of bacteria. Sometimes F plasmid may become integrated in the host chromosome. Such strains which harbour an integrated F plasmid are called High frequency of recombination (Hfr) strains.
In different Hfr strains, the site of integration of the F plasmid may be different, i.e. in some strains integration may be between the bacterial chromosomal genes a and b and in another Hfr strain, integration may be between the genes d and f.
When a Hfr strain (donor) mates with the F- strain (recipient), a nick is initiated at the OriT (Origin of Transfer) sequence of the integrated F plasmid. OriT marks the starting point of conjugative transfer. A part of the F plasmid is followed by the adjacent bacterial gene segments. Obviously bacterial genes closer to the F plasmid are transferred first. Segments of the integrated F plasmid are at the beginning and the end of the DNA being transferred. To transfer the entire plasmid, the entire bacterial chromosome must be transferred first. But the donor and recipient cells separate prior to the complete transfer of bacterial genome and plasmid. The recipient now carries transferred bacterial genes, but remains F- as only a part of the F plasmid is transferred. The donor cell remains HFr.
If we refer to the given data; in strain 1, the first bacterial gene to be transferred is trp, therefore the F plasmid must be integrated just ahead of the trp gene. Following that met is transferred, then his and then pro. Therefore the order of genes on the circular bacterial chromosome is trp, met, his and pro.
If you look at the order of transfer in strain 2, it is clear that trp is between met and bio genes. That trp is adjacent to met is already confirmed from strain 1.
Strain 3 confirms that bio is on the other site of trp and bio is followed by phe and then pro.
The answer is d
No comments:
Post a Comment
Note: Only a member of this blog may post a comment.