3. In pea plants, autotetraploid plants were produced by colchicine treatment. A
ID: 206011 • Letter: 3
Question
3. In pea plants, autotetraploid plants were produced by colchicine treatment. Assume that one homozygous autotetraploid pea plant (genotype as "AAAA" express the dominant trait RED flowers. This pea plant is crossed to one autotetraploid pea plant with WHITE flowers that is also homozygous (genotype as “aaaa"). If only ONE dominant allele (A) is sufficient to produce red flowers. Assume that synapsis between chromosome pair is random during meiosis. a. Indicate genotype and phenotype for Fl plants, respectively. (4 points) AaAa: red If FI plants are self-crossed, indicate the phenotypes, corresponding genotypes, and the ratio of the phenotypes for F2 plants, respectively. (For genotype, use dashes where appropriate) (8 points) b. AaAa x AaAa For gametes: 1 AA, 4Aa, and laa 35 A___(red): 1 aaaa (white)Explanation / Answer
a)
Parents: Homozygous dominant pea plant (Red flowers) X Homozygous recessive pea plant (White flowers)
AAAA aaaa
Gametes: AA aa
Progeny (F1 plants): AAaa or AaAa
(Heterozygous dominant pea plant with red flowers)
Since it is given that ONE dominant allele, i.e., A is sufficient to produce red flowers, therefore phenotype of the progeny will be red and the genotype is AaAa.
b) Here synapsis between chromosome pair is random, this means that all possible pairing combinations are equally likely. Now consider the four homologous chromosomes A/a/A/a be 1, 2, 3, and 4, then the probable combinations will be 1-2/3-4, 1-3/2-4, and 1-4/2-3. So, three types of gametes formed are Aa/Aa, AA/aa, and Aa/aa and all the gametes are produced at a frequency of 1/6. Thus, on selfing F1 plants, the frequency of AaAa progeny from selfing must be 1/36. Hence, the expected ratio is 35 A_ _ _ (red):1 aaaa (white)
Gametes: 1 AA, 4 Aa, and 1 aa
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