(3 pts) Suppose you have two flowering plants of the same species: one has red-c
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Question
(3 pts) Suppose you have two flowering plants of the same species: one has red-colored flowers, the other has white flowers. Both plants are pure-breeding for this trait of flower color. You perform a cross between these plants and all the F offspring have red flowers. You then mate the F plants and obtain 142 red-flowered plants, 48 purple-flowered plants and 63 white-flowered plants. A. (i pt) What pattern of inheritance best describes how flower color is transmitted in these plants? What phenotypic ratio is expected for this pattern of inheritance. Justify your answer; what phenotypic characteristic(s) of inheritance led you to this conclusion (and it is NOT Incomplete Dominance? Whycan't it be incomplete dominance? This Panern1nhertance is best reprasented by Recessive, epistastSybecauase it appears to tolo e 8 24 phenoty were ina F2generatjon, and the naw phenotgpe won- tau) is in the Fi gereration sis to test this hypothesis. Do you accept or reject the hypothesis o mplete, donin YOU MUST SHOW ALL YOUR WORK, INTERPRETATION AND SUPPORT FOR YOUR CONCLUSION X23(-12-142.3125)Yn2.3125 Phenotype | oiserved expected n ould expec 2.3125 f freed om resuit ih a p vanue greaer ox 253-.9515 thain o0s, there fore the hyporhesis is uExplanation / Answer
Answer:
Based on the given data:
A) What pattern of inheritance best describes how flower color is transmitted in these plants? What phenotypic ratio is expected for this pattern of inheritance. Justify your answer, what phenotypic characteristic of inheritance led you to this conclusion. Why cant it be incomplete dominance.
B) Use chi square analysis to test this hypothesis. Do you accept or reject the hypothesis.
The expected phenotypes are calculated and compared with observed phenotypes
Degree of freedom = number of categories - 1 = 3 - 1 = 2
P-value corresponding to a chi-square value of 0.00834 with 2 degree of freedom is P-value = 0.9959. The result is not significant at p < 0.05. Since the P-value (0.9959) is greater than the significance level (0.05), therefore Null Hypothesis is not rejected. Based on the evidence, there is no significant difference between the observed frequency and expected frequency in epistasis.
Phenotype Observed Frequency Expected Frequency (O-E) (O-E)^2/E Red Flowered plants 142 =253*9/16 = 142.312 -0.3125 0.00069 Purple Flowered plants 48 =253*3/16 =47.437 0.5625 0.00667 White flowered plants 63 =253*4/16 = 63.25 -0.25 0.00099 Total 253 253.0 Chi square (sum) 0.00834Related Questions
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