Suppose that in a bird species three different genes control striped versus soli

Question

Suppose that in a bird species three different genes control striped versus solid feathers, small versus large beaks, and broad versus narrow talons and each gene has completely dominant and recessive alleles. These three genes are found on the same chromosome. Birds heterozygous for all three genes have the dominant striped-feather, small-beak, narrow-talons phenotype and are mated to birds that are homozygous recessive for all three genes. Analysis of the phenotypes of 300 offspring gives the following results:

A) Which gene is in the middle of the other three? Briefly explain what observation about crossing over allows you to make this determination.

B) Determine the recombination frequencies for the feather and beak genes and between the feathers and talons genes?

Phenot Feathers Beak Small Small Small Small Large Large Large Large Talons Broad Broad Narrow Narrow Broad Broad Narrow Narrow Number of Offsprin Striped Solid Striped Solid Striped Solid Striped Solid 112 26 23 107 21

Explanation / Answer

The given data showa that the total number of progenies is 310 (not the 300 as given in the question, so question is solved by taking total number of progenies as 310)

    As the highest two numbers shows the parental type which is here: sBn(112) and SbN(107);

And the smallest number of classes will show the double cross over : sBN(1) and Sbn(2)

Now we will compare the parental type with the double cross over type class. The recombined gene in double cross over will be in middle.

On comparison we will get to know that the double cross over class sBN is close to the parental type sBn. In this the only difference is found between N/n.

This indicated that the “N” would be in the middle. So the probable sequence of genes is S—N—B

We have got the single cross over class i.e. SNB(26) , SnB(18), sNb(21), snb(23).

Now parental type would be: snB and SNb

In classes of single cross over the reciprocals are SNB and snb; SnB and sNb. On comparing SNB with the parent SNb we found that the recombination lies between N and B.

Similarly on comparing SnB with the parent snB we found the recombination between S and N.

So the recombination frequency between S & N will be= [(no. of progenies in single cross over+ no. of progenies in double cross over)/ total number of progenies]*100.

= [(18+21+2+1)/310]*100

= 13.54

Recombination frequency between N & B will be: [(26+23+2+1)/310]*100 = 16.77

So the recombination frequency between S & B will be: 13.54+ 16.77 = 30.31

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