Bean plants have different symptoms (phenotypes) when infected with a virus. Som

Question

Bean plants have different symptoms (phenotypes) when infected with a virus. Some show local lesions that do not seriously harm the plant while others show general systemic infection. In other words, there are two contrasting phenotypes observed: local lesions and systemic infection. You do not know if these phenotypes are caused by different alleles of the same gene or alleles of different genes. The following genetic analysis was done:

Parental cross: local lesions X systemic infection (both pure breeding). In the F1 generation, you find 100% of bean plants with the local lesion phenotype.

F1 cross: local lesions X local lesions

HYPOTHETICAL RESULTS 1: In the F2 generation, there were 850 bean plants with the local lesion phenotype and 200 bean plants with the systemic infection phenotype (Ratios to consider: 13:3, 15:1, 9:7 and 3:1).

a) State your hypothesis to explain the F2 generation (must be testable).

b) State your null hypothesis:

c) Do a chi square test to accept or reject your hypothesis.

d) State an alternate hypothesis (if you reject your hypothesis)?

Explanation / Answer

a)According to the data, there were 850 bean plants with local lesions while 200 had systemic infection.

Therefore total F2 plants observed = 200 + 850 = 1050

200/1050 = 0.19048 and 850/1050 = 0.8095

These values can be compared only when we consider the ratio 13:3

3/(13+3) = 3/16 = 0.1875

Similarly 13/16 = 0.8125

b) Null hypothesis states that if the calculated Chi square value is less than the tabulated chi square vale then null hypothesis will be accepted. It wil show that there is no significant difference.

c) The expected values (E) according to a 13:3 F2 segregation if there are 1050 F2 will be:

Local lesion phenotype = 3/16 * 1050 = 196.875

Systemic infection = 853.125

Chi square = Summation of (O-E)^2/ Summation of E = 19.5312/1050 = 0.186

Degree of freedom = 2-1 = 1

Tabulate chi square at probablility 0.05 = 3.84 for degree of freedom = 1.

Since calculated vlue is less than tabulated value therefore null hypothesis accepted. F2 genotypes present in 13; 3 RATIO.

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