1) all the possible genotypes are also identified by color in the image below :AABBAAbB or AABbaABB or AaBBaAbB or aABb or AabB or AaBbAAbb
aAbb or AabbaaBB
aabB or aaBbaabbThere are 9 different genotypes in total. These can be achived by mixing the alleles of both genes of each parent (just like it's represented in the square).
2) there are 4 phenotypes
-one with an allele A and allele B that would dominate the other. (AABB, aAbB, for example)
- one with all alleles recessive- aabb
-one with the recessive alleles in the gene A but heterozygotic/homozigotic dominant for gene B ( for example: aaBB)
-one with the recessive alleles in the gene B but heterozygotic/homozigotic dominant for gene A ( for example: Aabb)
3) <span>probability of having offspring with the aabb genotype: 1/16
Just by looking at the punnet square, you can see that only one has the aabb genotype, in all the 16 possibilities.
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Answer:
1.31 cM
Explanation:
Total offspring = 2205
Since two genes are involved, F1 progeny should have four types of combination. Out of them two are 17 and 12 which definitely means they are in lesser number. Since recombinants are always less than parental progeny in linkage, the given two types are recombinants.
Recombination frequency = (Number of recombinants / Total progeny) * 100
= [ ( 17 + 12 ) / 2205 ] * 100
= ( 29 / 2205 ) * 100
= 1.31 %
Map distance = Recombination frequency
Hence, distance between two genes = 1.31 cM
Answer:
There are not metric measurment