Answer:
- Green/Wide/Long, G-W-L- = 186 individuals
- Yellow/narrow/ long, gg ww L- = 186 individuals
- Green/wide/ short, G-W-ll = 186 individuals
- Yellow/narrow/ short, gg ww ll = 186 individuals
- Green/narrow/ long, G-wwL- = 14 individuals
- Yellow/wide/ long, gg W-L- = 14 individuals
- Green/narrow/short, G-wwll = 14 individuals
- Yellow/wide/ short, ggW- ll = 14 individuals
Explanation:
<u>Available data</u>:
• One gene affecting pod color (green is dominant to yellow)
• One gene affecting pod width (wide is dominant to narrow)
• Both genes are located on chromosome 5 and approximately 7 mu away from each other.
• A third gene, located on chromosome 4, affects pod length (long is dominant to short).
• Cross: A true-breeding wild-type plant (green, wide, long pods) was crossed to a plant with yellow, narrow, short pods.
• The F1offspring were then test-crossed to plants with yellow, narrow, short pods.
• The testcross produced 800 offspring.
<u>Cross 1:
</u>
Parental) GGWWLL x ggwwll
F1) GgWwLl
<u>Cross 2:</u>
Parental) GgWwLl x ggwwll
F2) N=800
To calculate the numbers of the F2 generation, we first need to calculate the frequencies of recombination and parental of the linked genes. We know that they are 7MU apart from each other. <em>The map unit is the distance between a pair of genes for which every 100 meiotic products one results in a recombinant one. </em>
MU = 7 means that there is 7% of recombination for genes that express color and width. This is:
MU = 7 = 7% recombination
We have two possibilities of recombination: green/narrow and yellow/wide. Each of these recombinants has half of the possibilities of occurring, so:
7 map units = 7 % of recombination in total
= % Gw + % gW
= 3.5 % + 3.5 %
Gw = 3.5% = 0.035
gW = 7%/2 = 0.035
For the parentals, we can calculate
100% - 7% = 93% of parental in total
= % of GW + % gw
= 46.5% + 46.5%
GW = 46.5% = 0.465
Gw = 46.5% = 0.465
The length gene is located in a different chromosome so it assorts independently. This is, of the gametes will be long, L, and the other 50% will be short, l.
L = 50% = 0.5
L = 50% = 0.5
Now we need to figure out how to relate these frequencies. All we need to do is to multiply the frequencies of occurrence obtained previously for linked and independent genes, for each possible phenotype. This is:
- Yellow/narrow, gw = 0.465
<u>Phenotypic frequencies:
</u>
- Green/Wide/Long, G-W-L- = GW x L = 0.465 x 0.5 = 0. 2325
- Yellow/narrow/ long, gg ww L- = gw x L = 0.465 x 0.5 = 0. 2325
- Green/wide/ short, G-W-ll = 0.465 x 0.5 = 0. 2325
- Yellow/narrow/ short, gg ww ll = 0.465 x 0.5 = 0. 2325
- Green/narrow/ long, G-wwL- = 0.035 x 0.5 = 0.0175
- Yellow/wide/ long, gg W-L- = 0.035 x 0.5 = 0.0175
- Green/narrow/short, G-wwll = 0.035 x 0.5 = 0.0175
- Yellow/wide/ short, ggW- ll= 0.035 x 0.5 = 0.0175
Finally, as we need to obtain the numbers of the individuals with those phenotypes, we need to multiply each frequency by N, which is the total number of individuals in the F2 (N = 800).
- Green/Wide/Long, G-W-L- = 0. 2325 x 800 = 186 individuals
- Yellow/narrow/ long, gg ww L- = 0. 2325 x 800 = 186 individuals
- Green/wide/ short, G-W-ll = 0. 2325 x 800 = 186 individuals
- Yellow/narrow/ short, gg ww ll = 0. 2325 x 800 = 186 individuals
- Green/narrow/ long, G-wwL- = 0.0175 x 800 = 14 individuals
- Yellow/wide/ long, gg W-L- = 0.0175 x 800 = 14 individuals
- Green/narrow/short, G-wwll = 0.0175 x 800 = 14 individuals
- Yellow/wide/ short, ggW- ll = 0.0175 x 800 = 14 individuals
