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Question

An Avian Biologist studying a species of parakeet in Australia noticed that in the wild the parakeets had either red or green tail feathers. After many crosses the scientist obtained a true breeding red individual and crossed it to a second true breeding red individual that she had obtained in a separate series of crosses. Much to her surprise when the F1 progeny were obtained all of the birds were green. She then self crossed the F1 birds with the following results: 45 green birds and 35 red birds. After seeing these results she thought she understood what was going on but continued crossing F1 birds over the next year and obtained these results: 921 green birds and 704 red birds.

What mode(s) of inheritance are demonstrated by the above crosses? How many genes are involved?





What were the genotypes of the original true breeding red stocks and the green progeny?














Why was it important to continue crossing F1 birds over the course of that year instead of relying only on the original results?
15 minutes ago - 3 days left to answer.

Answer 1

There must be 2 pairs of genes (or two pairs of alleles). Here's my reasoning.

Parents: red x red
F1: all green
F2: 9 green: 7 red

1. Looking at the F2 ratio, it reminded me of the 9:3:3:1 ratio from a dihybrid cross when the F1 are heterozygous for both traits (RrBb x RrBb).
2. 9:3:3:1 can be translated into 9:7 if the 3:3:1 are combined into one group (3+3+1=7).
3. That means that the F1 must have been heterozygous for each of two pairs of alleles ... must have been RrBb.
4. Then how are the F2 genotypes sorted out to make the 9:7? In a regular dihybrid cross, the Punnett square has:
--9 boxes with at least one R and one B (RRBB, RRBb, RRbB, RrBB, RrBb, RrbB, rRBB, rRBb, rRbB) These 9 boxes represent the 9 green of the F2. So it must take at least one R and one B to make a green bird.
--3 boxes with at least one R, but no B (RRbb, Rrbb, rRbb) and these must be red
--3 boxes with at least one B, but no R (rrBB, rrBb, rrbB) and these must be red
--1 box with no B, no R (bbrr) and this one must be red
SO ... it must require a minimum of one dominant allele in each pair to produce green birds.
5. So how can we discover the parental genotypes? The F1 must have been RrBb. The parental true-breeding reds cannot have been RRBB x rrbb if the rule is that it requires a minimum of one dominant allele in each pair to produce green birds. Following that rule, RRBB would be green and rrbb would be red ... but the parents were said to both be red.
Another option: One parent was RRbb. That bird is red because it doesn't have a capital B. The other parent was rrBB - also red. These birds would breed true because RRbb x RRbb will produce all RRbb = red. Same for the other parental line.

Parents: RRbb x rrBB
F1: all RrBb
F2: 9 R_B_: 3 R_bb: 3 rrB_: 1 rrbb

Why continue crossing birds? The more offspring counted, the more likely that the ratio of phenotypes is accurate and not just according to chance or coincidence. It's the same idea as repeating an experiment many times to verify results.