Show how the alleles for one trait are allocated or

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show how the alleles for one trait are allocated (or sorted) into each gamete and then offspring independently of the allocation of the second trait - Law of Independent Assortment . The cells inside the table show the resulting combination of alleles in the offspring. The punnett table for one trait - our yellow green and normal green leaf trait - is as follows because an F1 parent is always heterozygous for traits:
9 Punnett table 1 for the single trait cross of heterozygous parents ygrYGR x ygrYGR. This cross produces the F2 second generation. Alleles from parent 1 = ygr YGR Alleles from parent 2 ygr YGR Because YGR is dominant over ygr so those F2 offspring with the heterozygous alleles will have green leaves so only the 1 in four plants with the double recessive alleles ( ygr ygr) will have yellow leaves (or light green). Ratio of leaf color in the F2 offspring is 1:3 yellow to green. In our exercise we are looking at two traits - leaf color and anthocyanin presence in the hypocotyl - in a dihybrid cross of heterozygous parents, so each offspring gets two alleles for each trait one from each parent and this reestablishes the diploid state of the offspring. So our punnet diagramme for two traits is as follows: Punnett table 2 for dihybrid cross ygrYGR anlANL x ygrYGR anlANL Alleles from parent 1 = ygr anl ygr ANL YGR anl YGR ANL Alleles from parent 2 ygr anl ygr ANL YGR anl YGR ANL So in this double trait cross you get 9 phenotypes with YGR and ANL, 3 with YGR and anl, 3 with ANL and ygr and still only 1 double recessive for both traits ygr and anl. ygr ygr YGR ygr ygr YGR YGR YGR ygr anl ygr anl ygr ANL ygr anl YGR anl ygr anl YGR ANL ygr anl ygr anl ygr ANL ygr ANL ygr ANL YGR anl ygr ANL YGR ANL ygr ANL ygr anl YGR anl ygr ANL YGR anl YGR anl YGR anl YGR ANL YGR anl ygr anl YGR ANL ygr ANL YGR ANL YGR anl YGR ANL YGR ANL YGR ANL
10 A ratio of 9:3:3:1 Of course there are more genotypes than phenotypes. To test the Laws of Mendelian genetics we have performed the above dihybrid cross and we need to see if the phenotypes ratios of the offspring are as would be predicted by these laws. On the video for this exercise ( Lab 6 exercise 2 ) you saw an explanation of the above predicted results and a demonstration of how the phenotypes of the Fast plants produced in the above dihybrid cross are “phenotyped” – that is the number of offspring with each combination of phenotype traits were counted. So the phenotyping (which you saw on Lab 6 exercise 2 video) involved counting the number of offspring with each of the following combinations (all the possible combinations): 1. Non-Purple hypocotyls and Yellow Green Leaves 2. Non-Purple hypocotyls and Normal Green leaves 3. Purple hypocotyl and Yellow Green Leaves 4. Purple hypocotyl and Normal Green Leaves So we count the number of plants in each of the above phenotypic class and our results are shown in table 3 below: Table 3 Phenotype expression No of plants Green stem & yellow leaves 66 Green stem & normal yellow green leaves 179

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