2)The kaffir lily, Clivia miniata , is a flowering plant in the family Amaryllid

Question

2)The kaffir lily, Clivia miniata, is a flowering plant in the family Amaryllidaceae native to woodland habitats in South Africa. (It has been domesticated as an ornamental plant and can be found planted throughout Golden Gate Park.) Plants produce clustered flowers in either orange or yellow. A horticulturist traveled to South Africa and collected both orange and yellow varieties; she has repeatedly selfed these plants to produce true-breeding populations of each color type. When the pure-bred orange plant is cross-pollinated with a pure-bred yellow plant, all offspring are orange. When these orange F1 plants are intercrossed, you observe 250 orange flowering offspring and 83 yellow flowering offspring.

a)Propose how this trait is inherited; assign symbols to the genes & alleles; and state their relationship to one another. (2 points)

b)You travel to the same South African population visited by the horticulturist above. Of 300 randomly sampled Clivia plants, 220 produce yellow flowers. Based on this information, your answer in (a), and assuming Hardy-Weinberg equilibrium, predict the number of orange Clivia plants that are homozygous at this color locus. Show your calculations. (4 points)

c)You spend the next several months conducting test crosses on each of the orange-flowering plants from (b) by crossing them to yellow-flowering plants. You observe the following:

                  7 Clivia plants produced all orange offspring
                  73 Clivia plants produced ½ orange offspring and ½ yellow offspring

d)Based on this information and the information from (b), conduct a statistical test to determine whether the wild type population of Clivia miniata tested in (b) is in Hardy-Weinberg equilibrium at this locus. Show all calculations and clearly state your conclusion. (5 points)

Explanation / Answer

A - There are two alleles that are responsible for the flower colour. Orange colour and yellow colour. The orange colour is dominant over yellow colour.

Suppose Y is the allele for orange colour and y for the yellow colour

When the genotype of the plant is YY or Yy, flower colour will be orange and when the genotype os yy then the flower colour will be yellow.

B- Given that out of 300 flowers, 220 found to be yellow colour and rest i.e. 80 will be orange in colour.

According to Hardy-Weinberg law p+q =1

Here p is the allelic frequency of allele Y and q be the allelic frequency of y allele.

Also, the genotypic frequency is given by p2+q2+2pq =1.

In our case q2 = 220/300 = 0.73

q = 0.85

So p = 1-0.85 = 0.15

frequency of homozygous orange flower will be 0.15*0.15 = 0.0225.

so number of homozygous orange flower will be 0.0255 *300 = 7

So number of heterozygous flower will be 80-7 = 73.

C- When we perform the test-cross of homozygous orange flower we will get all the progenies which will have the orange colour.

- When we perform the test-cross of heterozygous orange flower we will get half of the progenies which will have orange colour flower and half of the progeny will have the yellow flowers.

As we can see that the observed plant's ratio is equal to the expected plant ratio. So the wild type population is in the Hardy-Weinberg equilibrium for the colour gene

Observed Expected O-E (O-E)2 (O-E)2/E Homozygous 7 7 0 0 0 Heterozygous 73 73 0 0 0 Total 80 0 0

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