I was learning about allele linkage in Biology class and I got confused by the resultant phenotypic ratios.

Say you have a dihybrid cross BbGg x BbGg.

The Punnett square would look like the following (sorry for bad formatting):

     BG...... | Bg.... |  bG.... |  bg

BG   BBGG       BBGg      BbGG      BbGg
---  -----------------------------------
Bg   BBGg       BBgg      BbGg      Bbgg
---  -----------------------------------
bG   BbGG       BbGg      bbGG      bbGg
---  -----------------------------------
bg   BbGg       Bbgg      bbGg      bbgg


This would result in a a 9:3:3:1 phenotypic ratio. (Dom/Dom : Dom/Rec : Rec/Dom : Rec/Rec)

However, if the alleles B/b and G/g were on the same chromosome, the ratios turned out to be around 4:1:1:4 in an example (still Dom/Dom : Dom/Rec : Rec/Dom : Rec/Rec)

The stated reason was that there would only be Dom/Dom and Rec/Rec, and the Dom/Rec and Rec/Dom only showed up due to crossing over. However, this begs the question: Why could Dom/Rec and Rec/Dom not have occurred prior to recombination? My reasoning is that just because the alleles are on the same chromosome does not mean that they must both the autosomal chromosomes for the allele must have either a dominant genotype or recesssive phenotype. That is, before crossover, there should still be a 9:3:3:1 random distribution between phenotypes represented by genotype sets (BbGg/BBGg/BbGG/BBGG):(Bbgg/BBgg):(bbGg/bbGG):(bbgg).

Note: when I say dom/rec for example, I mean that for the first allele the character represented by BB or Bb was expressed, and the "rec" means that for the second allele the character represented by gg was expressed. I am assuming autosomal dominance for simplicity.

If anything is unclear, don't hesitate to comment. I'm pretty tapped out right now. Thanks, hob


closed as unclear what you're asking by canadianer, Corvus, L.B., cagliari2005, WYSIWYG Mar 29 '15 at 7:38

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  • 1
    $\begingroup$ What do you mean by "if the alleles B/b and G/g were on the same chromosome"? Are you saying that the parental genotype is BG/bg? Chromosomes are inherited as a unit. Without crossing over, if the parental genotype is BG/bg, only either BG or bg can be passed on. This means the offspring could be BG/BG, BG/bg or bg/bg. $\endgroup$ – canadianer Mar 27 '15 at 7:43
  • $\begingroup$ No, I mean that allele1 that encodes for B or b and allele2 that encodes for G or g are on the same chromosome. Thanks! $\endgroup$ – bobhob314 Mar 28 '15 at 19:00
  • $\begingroup$ I think you have the wrong terminology. You,re saying that the genes which encode for B or b and G or g are on the same chromosome. This is the situation I described in my first comment. The answer to your question depends on what the parental genotypes are. $\endgroup$ – canadianer Mar 28 '15 at 20:05
  • $\begingroup$ Could you be more specific as to what your question is? $\endgroup$ – L.B. Mar 29 '15 at 0:52

I found the explanation at this site helpful: http://www.ndsu.edu/pubweb/~mcclean/plsc431/linkage/linkage1.htm

I don't have a clear sense of what you mean when you ask about assorted phenotypes occurring before recombination ("crossover"). All of your examples deal with diploid organisms. If they are multicellular then the cells that can undergo meiosis are highly specialized, and eventually become haploid. As the chromosomes are prepared they tend to become compact, and are often not expressed. So it is difficult to assess the phenotype of those haploid cells. The rest of the cells in the organism are diploid and at least some of them can express the alleles that give rise to the phenotypes that you are scoring.

At the moment of fertilization there is no further recombination, as the zygotic cells undergo mitosis. There is no segregation, no independent assortment, no linkage (that you can measure--the genes are still arrayed along their chromosomes). Each mother cell divides to give two perfect copies of itself. Then the daughters divide, etc. Even in the cells that are going to become germ cells, they divide by carbon copy mitosis right up until they enter meiosis.

So there is no mechanism that would lead to genotypes different from the genotype of the zygote (until meiosis).

  • $\begingroup$ But in the genotype of the zygote, it would be possible for it to be bG, for example, right? Therefore after meiosis, bbGG would occur without crossing over, which means that the thing I was taught about BBGG and bbgg being the only possibilities before crossover would be false, right? $\endgroup$ – bobhob314 Mar 28 '15 at 18:59
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    $\begingroup$ As your Punnett square shows, when a bG sperm fertilizes a bG egg then yes, 100% of the viable zygotes will also be bG. In the notation you are using, to indicate both alleles of each gene, you would write it as bbGG as you have. IF the B gene and the G gene are not on the same chromosome (unlinked), THEN you could draw it like this: b/b; G/G. In contrast, if the B gene is linked to the G gene on the same chromosome, then you could draw it like this: bG/bG. The bbGG notation you have been using does not distinguish between these two different situations. $\endgroup$ – mdperry Mar 28 '15 at 19:40

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