Question

Name: 6. In tomatoes, one gene pair determines beakless vs. beaked fruits, and another determines purple VS. green stems. Several crosses and progeny are shown below. Cross Beakless Beaked Beakless purple 392 Beaked purple 412 Beaked purple X beakless purple b) Beaked purple X beakless purple c) Beakless green X beaked purple d) Beakless purple X beakless purple green green 0 0 115 343 0 203 207 214 198 201 192 598 63 Which alleles (phenotypes) are dominant? (2 pts) Give a possible genotype for each parent plant in the crosses (a-d) above (1 pt. each; 8 pts total) a) b) с d) 7. In four o'clock plants, many flower colors are observed. In a cross involving two true-breeding strains, one crimson and the other white, all of the Fl generation were rose color. In the F generation, four new phenotypes appeared along with the Pi and Fi parental colors. The following ratio was obtained: 1/16 crimson: 2/16 orange: 1/16 yellow: 2/16 magenta: 4/16 rose: 2/16 pale yellow: 4/16 white Propose an what the genotype of each should be: explanation for the inheritance of these flower colors by filling in the table below with Most probable genotype Generation and color id PI crimson P, white Fi rose F2 crimson F2 orange Fa yellow Fa magenta. Fa rose F: pale yellow F2 white

Answer 1

QUESTION 6

In this case, the allele for beakless (B) is dominant over the allele (b) for beaked and the allele (P) for purple is dominant over the allele (p) for green.

In this case we have 412 beaked purple and 392 beakless purple which is similar to a proportion 50%/50% se we need to look that proportion in our cross In this case, we observe a proportion 74 for green and 34 for purple, so we need to find this proportion in the Punnett Square bbPP x BbPP ВВРp x ВВРp ВР Вp BP Вр BP BP bPbP BbPP BbPp bbPP bbPP BBPP BBPp BBPP BBPp BbPP BbPP | bbPp bbPP BBpp ВВРp BBPP BBpp ВВРp BBPP BBPP BbPP BbPP || bbPp bbPP ВВРp BbPP Bbpp || bbPp bbPP 50% beakless 50% beaked ВВРp вBpp | ВВРp | вBpp 4/16 Green or 14 12/16 Purple or 3/4

In this case, you need to obtain a proportion 1:1:1:1 which means that you have to obtain the same amount of offspring of each phenotype Bbpp x bbPp Beaked purple bbPp bbPp bbPp bbPp Beakless purple BbPp BbPp BbPp BbPp 1/4 Beakless green Bbpp Bbpp 1/4 BP BbPp BbPp BP BbPp BbPp bp bp bbPp bbPp bbpp bbPp Beaked green bbpp bbpp Bbpp Bbpp bbpp bbpp 1/4 | Bbpp Bbpp bbpp bbpp 1/4 Bbpp Bbpp bbpp bbpp

BbPp x BbPp Beakless purple Beaked purple bbPP bbPp BBPP BBPp BbPP BbPp ВР BBPP Вp bP bp BBPp BbPP BbPp BBpp BbPp Bbpp bbPp BBPp BbPp Beaked green BbPP BbPp BbPp bbPp bbPp bbpp BbPp Bbpp bbPp bbpp Beakless green BBpp Bbpp Bbpp