Two genes of a flower, one controlling blue (B) versus white (b) petals and the other controlling round (R) versus oval (r) stamens, are linked and are 10 map units apart. You cross a homozygous blue oval plant with a homozygous white round plant. The resulting F1 progeny are crossed with homozygous white oval plants, and 1,000 offspring plants are obtained. How many plants of each of the four phenotypes do you expect?

The answer is: 450 each of blue oval and white round (parentals) and 50 each of blue round and white oval (recombinants).

Here's how I solve it: BBrr (homozygous blue oval) X bbRR (homozygous white round) gives four times BbRr (heterozygous blue round). Then BbRr (the resulting F1 progeny) X bbrr (homozygous white oval) gives BbRr, bbRr, Bbrr, and bbrr. I don't get how the numbers 450 and 50 are related to the 10 map units apartness, but BbRr (blue round) and bbrr (white oval) are parentals.

Source: Campbell Biology 2017, page 313.

According to my book, a map unit is a unit of measurement of the distance between genes. One map unit is equivalent to a 1% recombination frequency.

  • 1
    $\begingroup$ What is the definition of map unit according to your book? $\endgroup$ Commented Feb 14, 2018 at 9:45
  • $\begingroup$ @LinuxBlanket, I'm Viktor but I have to have at least 50 reputation to add a comment. $\endgroup$
    – ViktorM
    Commented Feb 14, 2018 at 11:55
  • $\begingroup$ Related: Genetic linkage greater than 50 centimorgans $\endgroup$
    – Remi.b
    Commented Aug 15, 2018 at 16:09

2 Answers 2


I don't get how the numbers 450 and 50 are related to the 10 map units apartness,

Understanding how to use the map unit information is the point of the question.

If the B and R genes were unlinked, a gamete with the B allele would have a 50% change of having the R allele, and a 50% chance of having the r allele. It would just be luck which chromosome went in to the gamete with the B-containing chromosome.

But in this question, the B and R genes are on the same chromosome. Given the breeding scheme, the F1s have one Br chromosome, and one bR chromosome. Unless crossing over happens, the gamete that gets B has to get r as well. Crossing over happens 10% of the time. So instead of the gametes BR, Br, bR, br all being equally likely coming from the F1, their odds are actually 5%, 45%, 45%, 5%.


The two chromosome copies that bear the alleles in the F1 are

B-{10 map units}-r  -  Blue oval
b-{10 map units}-R  -  White round

As you said, one map unit is equivalent to a 1% recombination frequency. Ten map units = 10% recombination frequency. It means that the progeny will have a blue round/white oval phenotipe only 10% of the times, while 90% of the time it'll show the parental phenotype.

Hence the 450/50 (90%/10%) proportions of the F2.


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