Question

In tomatoes, a heterozygous plant with normal fruit and purple stems is crossed with a recessive plant having fasciated fruit and green stems. The following distribution of offspring is observed:
normal fruit, purple stems 38.5% fasciated fruit, green stems 38.5%
normal fruit, green stems 11.5%
fasciated fruit, purple stems 11.5%
What conclusion can be made regarding the loci for fruit shape and stem color? HINT determine the RF
What the loci and RF here mean? How to answer this question?

Answer 1

Answer:

Recombination frequency, P = 0.23.

If the recombination frequency is < 50%, genes are linked. A RF of 23% tells us that these two genes are in the same chromosome, close enough to be linked.

• 1% of recombinations = 1 map unit
• 23% = 23 map units

The genes for stem color and fruit type seem to be in the same chromosome, linked and 23 MP apart.

Explanation:        

In the present example, the genes for stem color and fruit type seem to be linked.

To know if two genes are linked, we must observe the progeny distribution. If individuals, whos genes assort independently, are test crossed, they produce a progeny with equal phenotypic frequencies 1:1:1:1. If we observe a different distribution, that is that phenotypes appear in different proportions, we can assume that genes are linked in the double heterozygote parent.  

In this way, we might verify which are the recombinant gametes produced by the di-hybrid, and we will be able to recognize them by looking at the phenotypes with lower frequencies in the progeny.    

The following distribution of offspring is observed:

• normal fruit, purple stems 38.5%
• fasciated fruit green stems 38.5%
• normal fruit, green stems 11.5%
• fasciated fruit, purple stems 11.5%

38.5% + 38.5% + 11.5% + 11.5% = 100%

N (total number of individuals in the progeny) = 100

• number of individuals with normal fruit, purple stems 38.5
• number of individuals with fasciated fruit green stems 38.5
• number of individuals with normal fruit, green stems 11.5
• number of individuals with fasciated fruit, purple stems 11.5

To calculate the recombination frequency we will make use of the next formula: P = Recombinant number / Total of individuals. The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU).

P = Recombinant number / Total of individuals.

P = 11.5 + 11.5 / 100

P = 0.23  

We need to know that 1% of recombinations = 1 map unit = 1cm. Also, the maximum recombination frequency is always 50%.  This means that if the recombination frequency is < 50%, genes are linked. A RF of 23% tells us that these two genes are in the same chromosome, close enough to be linked. FR 23% = 23 UM.