Question

Data 1. Data 2

Rb min y ——— 4. 108

+ + + ———— 34. 257

Rb min + ——— 6. 42

+ + y —————12. 58

Rb + y ————15. 93

+ m + ————3. 89

+ m y ———— 0. 33

Rb + c ————24. 84

There are two sets of data, determine which are the parental recombinant region 1 and 2, the double crossover types and the order that the genes are including their map distance

Calculation of Map Distances A. Group Data Calculations For the following questions, use your group data set (2-4 persons per group): 1. (1 pt) Based on the group data, label the female gamete types in the Type column of the chart as P (parental), RI (recombinant in region 1), R2 (recombinant in region 2), and dco (double crossover) types. 2. (5 pts) Calculate the map distances between all genes. Indicate this on the map diagram below. Gene_- Gene Gene - - mu mu

Answer 1

*There's a mistake in the data, the last row that says "Rb min c" is wrong, that "c" should be a "+"

This whole exercise lacks one important data: the description of the parents that gave those proportions in their nexte generation. I've seen that when that is not specified then it means the parents have the heterozygous genotypes for all the loci, so they have one Rb min y, and the other is + + +.

We have two data sets, so I understand we have to take them as different cases and work them separately.

Data set 1

Let's start by mapping these three caracters. We have to separate all the individuals in two categories: the parental and the recombinant; but we have to do this in the 3 pair of caracters posible, that is: Rb with min, Rb with y, and min with y, so we can know the distance between each of them.

Let's start with the Rb-min pair, so well completely ignore the y alleles. The parentals will be all the data with Rb min state, or + + (only for the locus, ignore the third locus). So we get this:

Parental	Recombinant
56	42

Our total is 98, so the percentage of recombinants is (100/98)*42, which is 42.8. That is the relative distance of Rb and min.

Now well take the pair Rb and y (we'll ignore min). We have the next table:

Parental	Recombinant
56	42

That means the realative distance is (100/98)*42=42.8

And for the last pair min and y (ignore Rb)

Parental	Recombinant
62	36

That means the relative distance between min and y is (100/98)*36=36.7

Now we can complete the map given in the question 2. The Rb locus shows not only a very high distance value (a value of 50 means it may be a different chromosome, non-linked loci), but shows the same distance for both of the other loci, so that means Rb is not linked to neither min nor y, is in another chromosome. Let's map this:

36.7 mu Rb min

Now that we know the order of each locus we can proceed to analyse the recombination. We have only one recombination region (between min and y), because Rb is in another chromosome.

The X lines represent the only recombination region. We ignore Rb, it cannot recombine with the other two loci.

Rb min

So lets label those data using the previous image:

Case	Label
Rb min y	Parental
+ + +	Parental
Rb min +	Recombinant
+ + y	Recombinant
Rb + y	Recombinant
+ min +	Recombinant
+ min y	Parental
Rb + +	Parental

We finished answering for the first data set, now the next.

Data set 2

We do the exact same, let's start by mapping by pairs, first pair Rb - min:

Parental	Recombinant
465	299

Our total is 764, so the relative distance is (100/764)*299=39.1

Now the pair Rb - y:

Parental	Recombinant
547	217

So, the distance is (100/764)*217=28.4

Now the pair min - y:

Parental	Recombinant
482	282

So the distance is (100/764)*282=36.9

Now with these distances let's map:

Now let's answer question two with these reference:

Gene Gene min Gene RB 28.4 mu 36.9 mu

We have to complete the table:

Region 1 Region 2 Rb min Recombination in the region

Case	Label
Rb min y	Parental
+ + +	Parental
Rb min +	Double cross-over
+ + y	Double cross-over
Rb + y	R2
+ min +	R2
+ min y	R1
Rb + +	R1

And we're finished.