(Solved) Determining parental and F1 genotypes in crosses involving polygenic inheritance

a) First let's go through the matings, assuming pr and vg are on different chromosomes. In the following notation, alleles above the horizontal line are from one homologous chromosome, and alleles below the line are from the other homologous chromosome.

Parents: pr vg x pr+ vg+

pr vg pr+ vg+

F1: pr vg

pr+ vg+

F1 backcross: pr vg male x pr vg female

pr+ vg+ pr vg

Expect in F2: male gametes:

pr vg pr vg+ pr+ vg pr+ vg+

_____________________________ ____________

female gametes pr vg | pr pr vgvg pr pr vg+vg pr+pr vgvg pr+pr vg+vg

This predicts four different phenotypes, purple vestigial, purple long-winged, red-eyed vestigial, and red-eyed long-winged, in equal numbers (each comprising 0.25 of the progeny).

b) The actual results were markedly different. In fact none of the recombinant phenotypes, purple long-winged and red-eyed vestigial, were observed. This indicates that the purple and vestigial genes are linked. Subsequent mapping showed that they are both in the second linkage group (Drosophila has four linkage groups, corresponding to three autosomes and one pair of sex chromosomes). Note that no measurable recombination occurred between the purple and vestigial genes in this backcross; this is a peculiarity of male Drosophila and the heterogametic sex in some other species. Other experiments with heterozygous F1 females do show recombination (see part 1c).

Let's re-examine the predictions of the matings, now that it is clear that the genes are linked. In the notation below, a horizontal line with more than one gene above and below it means that the genes are linked. Again, alleles for one homologous chromosome are above the line, and those for the other chromosome are below it.

Parents: pr vg x pr+ vg+

pr vg pr+ vg+

F1: pr vg

pr+ vg+

F1 backcross: pr vg male x pr vg female

pr+ vg+ pr vg

Expect in F2: male gametes:

pr vg pr+ vg+

_____________________

female gametes pr vg | pr vg pr+ vg+

pr vg pr vg

Thus in the absence of recombination, one obtains equal numbers of purple vestigial and red-eyed long-winged flies in the progeny.

c) In this case, the mating is

F1 backcross: pr vg female x pr vg male

pr+ vg+ pr vg

and recombination does occur (as mentioned in 1.1b, the absence of recombination is peculiar to male Drosophila). Note that the frequency of recombinant types is much less than the 50% predicted for no linkage (see 1.1a). The purple long-winged flies have the genotype

pr vg+

pr vg

and red-eyed vestigial flies have the genotype

pr + vg

pr vg

in both cases resulting from recombination between the purple and vestigial genes. The combined number of recombinants comprises 15.2% of the progeny, and one concludes that the two genes are linked, and are 15.2 map units, or 15.2 centiMorgans apart.

Home Search Q & A Board Gallery Resource Library Blog Dictionary Chat	Quick Links Start a Topic Latest Topics Unanswered Questions Courses Study Tools Browse by Textbook	Social Facebook YouTube Community Stats Who's Online Trending Staff	Help About Us Take the Tour Study Tips Posting Guidelines Terms and Policies Contact Us
Biology Forums - Study Force © 2010-2024 \| Powered by SMF \| SMF © 2015, Simple Machines \| Sitemap Biology-Forums.com is not affiliated with any publisher. Book covers, title and author names appear for reference only.

Determining parental and F1 genotypes in crosses involving polygenic inheritance

Related Topics