Linkage And Mapping Of Genes

[TECHXHEAD]

I would appreciate help with this problem, if at all possible!

Homework Statement

Snapdragons, homozygous for the recessive allele ls of the leaf form gene, have serrated leaves, the normal allele l+ is dominant and responsible for a smooth leaf phenotype. Individuals homozygous for the recessive allele pW of the petal color locus, have white petals; the normal red color results from the dominant p+ allele. A cross is performed between plants homozygous for serrated leaves, red petals, and plants homozygous for smooth leaves, white petals.

F1 snapdragons of genotype: l s p + / l + p W are testcrosses with snapdragons of the two recessive alleles. The progeny obtained are shown below. Calculate the recombination frequency and map distance between the two genes:

a. Without using the mapping function

AND

b. Using the mapping function.


Phenotype of Progeny/Number

Serrated Leaves - Red Petals/264
Smooth Leaves - White Petals/37
Serrated Leaves - Red Petals/41
Smooth Leaves - White Petals/258

 

[nate808]

I would be happy to assist you with this problem. Based on the information provided, we can determine the following genotypes for the parent plants:

Parent 1: ls/ls pp/pp (serrated leaves, white petals)
Parent 2: l+/l+ p+/p+ (smooth leaves, red petals)

The F1 generation has the genotype l s p + / l + p W, which is a result of the cross between the two parent plants. Now, to determine the recombination frequency and map distance between the two genes, we can use the following formulas:

Recombination frequency = (Number of recombinant progeny / Total number of progeny) x 100
Map distance = Recombination frequency x 100

a. Without using the mapping function:
Based on the given progeny numbers, we have 264 recombinant progeny out of a total of 600 progeny (264+37+41+258). Therefore, the recombination frequency is (264/600) x 100 = 44%. The map distance between the two genes is also 44%.

b. Using the mapping function:
To use the mapping function, we need to first determine the coefficient of coincidence (c) and interference (i).

Coefficient of coincidence (c) = Observed double recombinants / Expected double recombinants
= (37 x 41) / (264 x 258)
= 0.0059

Interference (i) = 1 - c
= 1 - 0.0059
= 0.9941

Now, we can use the formula: Map distance = (-log(1-c)) / 2
= (-log(0.0059)) / 2
= 3.24 cM

Therefore, using the mapping function, the map distance between the two genes is 3.24 cM.

I hope this helps. Let me know if you have any further questions. Best of luck with your problem!

 

[Blueshift5]

I would approach this problem by first understanding the basics of linkage and mapping of genes. Linkage refers to the phenomenon where genes located on the same chromosome tend to be inherited together, while mapping refers to the process of determining the relative positions of genes on a chromosome.

Based on the given information, we can determine that the two genes, leaf form and petal color, are located on the same chromosome as they are linked. The cross between the two homozygous plants (ls/ls pp/p+) produces F1 offspring with the genotype l s p + / l + p W, indicating that the two genes are on the same chromosome.

To calculate the recombination frequency and map distance between the two genes, we can use the formula: recombination frequency = (number of recombinant offspring/total number of offspring) x 100%. Recombinant offspring are those that show a different combination of traits compared to the parents.

a. Without using the mapping function:

The total number of offspring is 264 + 37 + 41 + 258 = 600. The number of recombinant offspring is 37 + 41 = 78. Therefore, the recombination frequency is (78/600) x 100% = 13%.

b. Using the mapping function:

To calculate the map distance, we need to use the mapping function: map distance = (1/2) x recombination frequency. Therefore, the map distance between the two genes is (1/2) x 13% = 6.5 cM (centimorgans).

In conclusion, the two genes, leaf form and petal color, are located on the same chromosome and have a recombination frequency of 13% and a map distance of 6.5 cM. This information can be used for further genetic studies and breeding experiments.