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Over the summer you discover a three eyed, fly with legs coming out of its head instead of antennas. To determine how close these alleles (eye trait, legs-on-head trait) are to each other on the same chromosome, you cross this true breeding fly with a normal (2-eyed, antennae) fly. The F1 generation is all wildtype. You then backcross an F1 animal to the original mutant animal which generates; 500 normal flies, 50 2-eyed, head-legged flies, 76 three-eyed, antenna flies and 400 three-eyed, head-legged flies. How far apart are these genes

Sagot :

marianaegarciaperedo

Answer:

These genes are 12.3 MU apart.

Explanation:

Available data:

  • Two diallelic genes coding for eyes number and legs position
  • Normal phenotype: 2-eyed, antennae fly
  • Mutant: three-eyed, fly with legs coming out of its head instead of antennas
  • 1st Cross: a pure breeding wildtype with a pure breeding mutant
  • F1) 100% normal dihybrid
  • 2nd Cross: normal dihybrid from F1 with mutant
  • F2) 500 normal flies, 50 2-eyed, head-legged flies, 76 three-eyed, antenna flies and 400 three-eyed, head-legged flies.

Let us first name the alleles

  • E allele codes for two eyes
  • e allele expresses three eyes
  • L allele codes for legs in the body
  • l allele codes for legs in the head

Normal phenotype → two eyes,antenna in the head → EELL, EeLL, EELl, EeLl

Mutant phenotype → three eyes, legs in the head → eell

1st cross:

Parentals) EELL x eell

F1) 100% EeLl normal individuals

2nd Cross:

Parentals) EeLl  x  eell

F2) 500 normal flies, EeLl

     50 2-eyed, head-legged flies, Eell

     76 three-eyed, antenna flies, eeLl

     400 three-eyed, head-legged flies, eell

When two genes that assort independently are test crossed, they produce a progeny with equal phenotypic frequencies 1:1:1:1.  

However, if these genes are linked, after the test cross, the distribution of phenotypes among the progeny differs from 1:1:1:1. Phenotypes appear in different proportions, so we can assume that these genes are linked.  

In the exposed example, the phenotypic frequency is different from the expected one is genes were not linked.

We can recognize the parental gametes in the descendants because their phenotypes are the most frequent,

Parental

  • 500 EeLl
  • 400 eell

Recombinant

  • 50 Eell
  • 76 eeLl

Now we need to calculate the recombination frequency in order to get the distances between the two genes.

To calculate the recombination frequency, P, we will make use of the following formula:

P = Recombinant number / Total of individuals.

P = 50 + 76 / 50 + 76 + 500 + 400

P = 126 / 1026

P = 0.123

The genetic distance, GD, will result from multiplying that frequency by 100 and expressing it in map units (MU).

The map unit is the distance between the pair of genes for which every 100 meiotic products, one of them results in a recombinant one.  

GD = P x 100

GD = 0.123 x 100

GD = 12.3 MU

Two genes that are very close will have a few recombination events and are strongly bounded. The more separated two genes are, the more chances of recombination there will be. The closer they are, the fewer chances of recombination there will be. Genes that express 50% of recombination frequency or more are not linked genes. As long as they do not exceed this percentage, they are linked.

According to this information, we can assume that genes coding for eyes number and legs position are linked because their recombination frequency is 12.3% (<50%).

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