Chapter Two:
Question #23: Why are the 5’ and 3’ ends of DNA named thus?
Think of the 5’ and 3’ carbons of the deoxyribose in DNA. The DNA chain is created by
linking together nucleotides via a phosphodiester bond. This phosphodiester bond is present
between the phosphate and the 5’ carbon of deoxyribose and between the phosphate and 3’
carbon of deoxyribose. Because of this, the DNA chain has a 5’3’ polarity
Question #25: How many bases are there in 2kb?
2kb= 2000 base pairs= 4000 bases
Chapter Nine:
Question #17: Streptococcus pneumoniae cells of genotype str s mtl - are transformed by
donor DNA in 2 separate experiments. The following table shows the results of this
transformation:
Percentage of Cells Transformed Into:
Experiment
str rmtl -
str smtl +
str rmtl +
str s mtl -  str rmtl – (single trans.)
str smtl + (single trans.)
str rmtl + (double trans.)
1
4.3
0.40
0.17
2
2.8
0.85
0.0066
▪What can you say about each of the DNAs used to transform the cells with respect to whether
the genes in the donor DNA are linked or not?
On page 283 of your text:
“If a+b+ donor DNA is used to transform a-b- recipient cells, then if a and b are closely linked, the
proportion of a+b+ double transformants (aka cotransformants) should EXCEED the
PRODUCT of the proportion of single a+ and b+ transformants.
As a reminder:
Double transformants (cotransformants): a- b-  a+ b+
Single transformants: a- b-  a+ b- or a- b+
SO, if genes are closely linked then there should be more double transformants seen in the
results than the PRODUCT of the single transformants…
(single transormants)(single transformants) < (double transformants)  genes are linked and
transformed as a single unit
(single transormants)(single transformants) ≥ (double transformants) genes are not linked and
are transformed as separate units
Now, Look back at table of Results:
Percentage of Cells Transformed Into:
Experiment
str rmtl -
str smtl +
str rmtl +
1
2
4.3
2.8
0.40
0.85
0.17
0.0066
NOW what can you say about the donor DNA used in the transformation experiments with
respect to whether the genes are linked or not?
Experiment 1? (4.3%)(0.40%) = (0.043)(0.0040) = 0.000172  0.017%
0.017% < 0.17% (observed double transformants)
Linked!
Experiment 2? (2.8%)(0.85%) = (0.028)(0.0085) = 0.000238  0.0238%
0.0238% > 0.0066% (observed double transformants)
Not linked! Separate units!
What are the genotypes of the donor DNAs used in Experiments 1 and 2?
Experiment 1? Has to be able to produce double transformants, so should be str rmtl +
Experiment 2? Transformed as separate unit, so has to be str rmtl – and str smtl +
What does all that look like?!