Having trouble with the last portion of the worksheet (except the critical x 2 v

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Having trouble with the last portion of the worksheet (except the critical x2 value question). Could someone please help me?

Sec Genetics Chi-square Practice Na n incomplete dominance alleles of true-breeding (homozygous) parents blend together. Red flowers crossed with white flowers produce a heterozvaous F1 generation of pink flowers. A well-studied example n humans straight hair from the other parent will give a hair texture that is a blend of the two: wavy hair occurs in the genes for hair. Inheriting a gene for curly hair from one parent and a gene In codominance alleles of true-breeding ( flowers crossed with white flowers produce a heterozygous F1 generation of flowers that are red wit white spots or white with red spots. A codominant genetic trait in humans occurs wi someone inherits one A allele and one B allele they will not have a blood type in between those two but homozygous) parents are both expressed in the offspring. Red th ABO blood types. If will instead have a blood type AB in which both alleles are expressed. The grid below is another example of Chi-square analysis. This is an example to illustrate incomplete dominance in which neither allele is dominant or recessive. in the example below using snapdragon flowers both parents P1 are true-breeding (homozygous) A plant with red flowers is crossed with a plant with white flowers. The heterozygous F1 generation all have pink flowers. If two pink flowered plants are crossed what will be the phenotypes in the F2 generation? Test your hypothesis by performing a Chi-square analysis What is the expected ratio of red : white : pink phenotypes in the F2 generation? You should draw a Punnett square on the back to work out your hypothesis Hypothesis: There will be phenotyre raro in the F ganern Observation of 115 F2 generation plants showed 30 with red flowers, 24 with white flowers, and 61 with pink flowers. Predict the expected number of F2 plants out of 115 that will have red or white or pink flowers. hite Pink Red 2.8 ,75 30 .03 57.S Expected (e) Observed (o) Deviation (d = e-o) d squared (d2) d2 divided by e = d2e Chi-square (X2) value = sum of (d2/e) values-Ed2e) 24 61 .75 3.5 a.as .05 otal number of classes Therefore degree of freedom Therefore critical x2 value for p = 0.05 is The calculated x2 value is greater than or less than the critical value (circle one). Therefore the hypothesis is not rejected or rejected (circle the correct conclusion) (lab manual page 154) and What were the genotypes of the P1 parental generation?

Explanation / Answer

Total number classes= 3

Degree of freedom(df) = Total number classes - 1 = 3-1 = 2

Therefore chi square value for p=0.05 is 5.991

The calculated chi square value (given 1.03) is less than the critical value

Therefore the hypothesis is not rejected ( that shows chance did not produce the results, the results actually followed the expected ratio)

Genotypes of P1 generation are RR (Red Flowers) and rr (White flowers)

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