ppose a farmer is in the business of growing carnations for a local flower shop.
ID: 192615 • Letter: P
Question
ppose a farmer is in the business of growing carnations for a local flower shop. He typically grows red hite, and pink flowers. In his carnations, flower color is determined by a single locus and two alleles. The flower color phenotype exhibits incomplete dominance, so all heterozygous flowers are pink. Homozygous flowers are either red or white The farmer would like the plants to produce enough of each color flower to meet customer demand which is conveniently the standard mendelian ratio of 1:2:1 of white:pink:red flowers. He hypothesizes this ratio can be obtained by letting the flowers self-cross. The farmer plants 1000 total seed of 250 white flowers 250 red flowers, and 500 pink flowers and lets the plants self-fertilize and perpetuate over three growing seasons. He predicts the ratio will remain the same each generation Analyze the averaged collected data from the farmer's field trials and select the conclusion that best fits the farmer's results Traits Red White Pink The farmer will meet demand because the flowers will reliably produce the 1:2:1 ratio of flower colors needed The ratio does not differ significantly from 1:2:1 O Phenotypic 14 14 1/2 Ratio The farmer will not meet demand because the flowers will not reliably produce the 1:2:1 ratio of flower colors needed. The ratio differs significantly from 1:2:1 The farmer will be able to meet demand because the observed and expected values of the pink flowers are equal, therefore the flowers should produce the necessary ratio through self-perpetuation Observed 190 310 500 O Expected 250 250 500 Results of Chi-square analysis X = 28.8 | df= 2 The farmer will not be able to meet demand because a O phenotype expressing incomplete dominance never produces the same color ratio.Explanation / Answer
Hi
The chi square related statistical problems need an hypothesis before the problems can be tackled. We need a null hypothesis and we either accept of reject the null hypothesis.
here, the color of the flowers do not deviate the given ratio of 1:2:1, This is the null hypothesis.
Let us either accept or reject it.
The chi square value is given = 28.8
Df = 2
use the chi square table ( available on net or in any stats book). Look for the row df= 2.( on the left column). Now go across row and find the value under the column p = 0.05. This is the accepted statistical cut off value. We find that the value there is 5.991. This value is smaller than our chi square value of 28.8. This means that there is less than 5% of chance that these variations in color of flowers has occured by chance alone.
Hence we reject the null hypothesis and click on the option 2.
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