2 organisms, AABBCCDDEE and aabbccddee are mated to produce an F1 that is self fertilized. If the capitol letters are dominant, independantly assorting alleles:
A) how many different genotypes will occur in the F2?
B)what portion of the F2 genotypes will be recessive for all five loci?
check my thinking.
Parent one is homozygous in dominant, and parent two is homozygous in recessive. Therefore F1 cannot be anything but heterozygous.( AaBbCcDdEe), one genotype.
Now in F2, each parent will contribute either dominant or recessive, so there are three types in each gene locus (for instance AA, Aa, or aa. Given this there must be a total of 15 types (3types per locus, 5 loci). What portion is recessive for all five? recessive (ie, aa) in each loci is a 1/4 probability, so ...(1/4)^5 seems to be the probability it occurs at 5 loci. Check my thinking, population genetics is not my area of expertise.
To determine the number of different genotypes that will occur in the F2 generation, we need to consider the independent assortment of alleles at five different loci.
A) Number of Different Genotypes in the F2 Generation:
For each locus, there are two possible alleles due to the presence of both homozygous dominant (AA) and heterozygous (Aa) genotypes. Since there are five independent loci, we can multiply the number of possible genotypes at each locus together to find the total number of different genotypes in the F2 generation.
In this case, both organisms have identical genotypes AABBCCDDEE and aabbccddee, respectively. When they are mated, all the dominant alleles (capital letters) will be expressed in the F1 generation since they are dominant to the recessive alleles (lowercase letters). So, the genotype of the F1 generation will be AaBbCcDdEe.
In the F2 generation, the F1 organisms will self-fertilize. To determine the number of different genotypes, we can use the Punnett square method. Since the F1 generation is heterozygous (AaBbCcDdEe) for all five loci, we need to consider the combinations of these alleles.
For each locus, there are four possible outcomes: homozygous dominant (AA), heterozygous (Aa), heterozygous (Aa), and homozygous recessive (aa). Since there are five loci, we can multiply the number of possible outcomes at each locus together to find the total number of different genotypes in the F2 generation.
Possible outcomes at each locus: 4 (AA, Aa, Aa, aa)
Total number of genotypes in the F2 generation: 4^5 = 1024
Therefore, there will be 1024 different genotypes in the F2 generation.
B) Portion of F2 Genotypes that are Recessive for all Five Loci:
To calculate the portion of F2 genotypes that will be recessive for all five loci, we need to determine the probability of having a recessive genotype at each locus (aa) and multiply those probabilities together.
Since each locus has two possible alleles (A, a), and the probability of having a recessive genotype (aa) is 1/4, the probability of having a recessive phenotype at each locus is 1/4.
Since there are five independent loci, we can multiply the probability of having a recessive phenotype at each locus together to find the portion of F2 genotypes that will be recessive for all five loci:
Probability of recessive genotype at each locus: 1/4
Portion of F2 genotypes that are recessive for all five loci: (1/4)^5 = 1/1024
Therefore, only 1 out of 1024 genotypes in the F2 generation will be recessive for all five loci.