5. A female of the genotype a b c / + + + (i.e. heterozygous for three genes on the same chromosome) produces 100 meiotic tetrads (100 cells undergoing meiosis, where recombination can occur). Of these 100 tetrads, 68 show no crossover events. Of the remaining 32, 20 show a crossover between a and b, 10 show a crossover between b and c, and 2 show a double crossover between a and b and between b and c. Of the 400 gametes produced, how many of the 8 different genotypes will be produced? Hint: each tetrad showing no crossover will produce two a b c gametes and two + + + gametes. Assuming the gene order a - b - c and given the above mentioned recombination frequencies, what are the map distances between the loci?

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To determine the number of different genotypes produced from the given information, we need to consider the different combinations of crossovers between the genes a, b, and c.

Let's start by analyzing the tetrads that show no crossover events. Since there are 68 of them, each producing two a b c gametes and two + + + gametes, we have a total of 68 x 2 = 136 a b c gametes and 136 + + + gametes.

Next, let's consider the tetrads that show a crossover between the genes a and b. We are told that 20 tetrads exhibit this crossover event. This means that 20 x 2 = 40 of the gametes will carry the a b combination, while the remaining 20 x 2 = 40 will carry the + b combination. Remember that these 40 gametes with the + b combination already account for some of the gametes mentioned earlier.

Similarly, for the crossover between the genes b and c, we have 10 tetrads producing 20 gametes with the b c combination and another 10 tetrads producing 20 gametes with the b + combination.

Lastly, we have 2 tetrads showing a double crossover between a and b and between b and c. For each of these double crossovers, we will have 2 gametes carrying the a + c combination.

Now, let's summarize the number of gametes with each genotype:

- a b c: 136 (from the no crossover events) + 40 (from the a b crossover events) = 176
- a + c: 2 (from the double crossover events)
- + b +: 136 (from the no crossover events) + 40 (from the a b crossover events)
- + + c: 136 (from the no crossover events) + 20 (from the b c crossover events)
- a + +: 40 (from the a b crossover events)
- + b c: 20 (from the b c crossover events)
- + + +: 136 (from the no crossover events)

Therefore, there will be 176 a b c gametes, 2 a + c gametes, 176 + b + gametes, 156 + + c gametes, 40 a + + gametes, 20 + b c gametes, and 136 + + + gametes.

Now, let's calculate the map distances between the loci using the given recombination frequencies:

The recombination frequency between a and b is: (20 + 2) / 100 = 0.22 or 22%
The recombination frequency between b and c is: (10 + 2) / 100 = 0.12 or 12%

To convert these recombination frequencies to map distances, we can use the formula:

Map distance (%) = (Recombination frequency) x 100

Therefore, the map distance between a and b is 0.22 x 100 = 22 cM (centimorgans) or map units.
And the map distance between b and c is 0.12 x 100 = 12 cM.

So, the map distances between the loci a and b and between b and c are 22 cM and 12 cM, respectively.