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KendraJo KendraJo
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Posts: 2
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9 years ago
You decide to designate the alleles of the four loci as either additive (contributing to fruit segment number and represented by a superscript “+”) or non-additive (not contributing to fruit segment number and represented by a superscript “0”). Using this convention, choose the correct genotype for the two pure lines and the F1, and indicate how many additive alleles each genotype has.
Complete the diagram below by dragging one blue label to each blue target and one pink label to each pink target. Labels may be used once, more than once, or not at all.
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wrote...
Staff Member
Educator
9 years ago
HI Kendra

See if this helps Downwards Arrow
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elena1234,  foofoofluffypoo,  nikkinikki,  teeprice,  pcarawa,  jdean1020,  bio_man,  DJANGOOO,  real_madrid11,  tblack7,  came,  Michelle Harms
Mastering in Nutritional Biology
Tralalalala Slight Smile
wrote...
9 years ago
 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.
wrote...
Staff Member
Educator
9 years ago
Awesome Slight Smile
mmiglio
Mastering in Nutritional Biology
Tralalalala Slight Smile
KendraJo Author
wrote...
9 years ago
Yes thank you for the help!
wrote...
Staff Member
Educator
9 years ago
NP
Mastering in Nutritional Biology
Tralalalala Slight Smile
wrote...
9 years ago
Thank you Padre!
wrote...
9 years ago
Thank you! This really helped!
wrote...
4 years ago Edited: 4 years ago, Kelly Thompson
.
wrote...
4 years ago
Thank you so much!
wrote...
3 years ago
Thank you!
wrote...
3 years ago
This makes sense now Thank You
wrote...
3 years ago
Thank you!
wrote...
3 years ago
Thank you!
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