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Ch. 6 - Genetic Analysis and Mapping in Bacteria and Bacteriophages

Chapter 6, Problem 9

In a transformation experiment, donor DNA was obtained from a prototroph bacterial strain (a⁺b⁺c⁺) and the recipient was a triple auxotroph (a⁻b⁻c⁻). What general conclusions can you draw about the linkage relationships among the three genes from the following transformant classes that were recovered? a⁺ b⁻ c⁻ 180 a⁻ b⁺ c⁻ 150 a⁺ b⁺ c⁻ 210 a⁻ b⁻ c⁺ 179 a⁺ b⁻ c⁺ 2 a⁻ b⁺ c⁺ 1 a⁺ b⁺ c⁺ 3

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Hi, everyone. Let's take a look at this practice problem together. The D N A from a bacterium with genotype A plus B plus C plus was used to transform a bacteria with genotype ABC. The highest number of transform ints was observed when the recombination occurred between A and C or B and C determine the linked genes. So recall that when transformation occurs, bacteria take up genetic material from the environment. And in our case, recombination occurs and pieces of A plus B plus C plus are broken up and recombined with pieces of A B C. And the recombination rate was the highest between A and C and B and C. So we can also think of this in terms of recombination frequency and it was the highest for these two combinations. So they had a high recombination frequency. Now, during recombination, two genes that are close together would have very few recombination events. So let's take a look at this example. We have genes J and K and they're close together. So they're less likely to be split and broken apart. And therefore they would have a low recombination frequency and it's likely that they're linked Now, in this example, the two genes are further apart. And so it's more likely that they are to be split and recombination events to occur more frequently because of their distance. And so in this instance, they're going to have a high recombination frequency and they are not likely to be linked. And so, if A and C and B and C recombination frequency is the highest, this means that A and C are likely to not be linked and B and C are likely to not be linked. So we can eliminate choices B and C. Now, our only two other possibilities are A and B are linked or none of the above. Well, we can deduce that A and B would have a low recombination frequency because there was not a high number of transform ints for them. So that would suggest that those two are linked. So A A and B are going to be our linked genes. Alright, everyone, I hope you found this helpful and I'll see you for the next practice problem.