The dihybrid cross Aa Bb x Aa Bb could be written more specifically with regards
ID: 149383 • Letter: T
Question
The dihybrid cross Aa Bb x Aa Bb could be written more specifically with regards to the chromosomes on which one would find the alleles. E.g., we might use the notation Aa||Bb to indicate that the traits are on different chromosomes (each chromosome represented by a "|") and thus should assort independently. However, the traits could also be arranged on a pair of homologous chromosomes as: Ab||aB or AB||ab. All of these in theory could be interpreted as "Aa Bb" (which just indicates that one an individual has those alleles but not how they located on chromosomes). If we use the short-hand of "/" = "||" (i.e., Ab||aB = Ab/aB), answer the following question:
Q2: Consider the cross "Ab/aB" x" Ab/aB" . If the A and B loci are linked and are 28 m.u. apart, what is the expected frequency of observing progeny that present as "ab"? Enter a decimal number to three decimal places (e.g., 0.062). Hint: for this and related questions in this set, you will want to determine the probability of one of the heterozygous individuals to produce each dihybrid gamete type; then, you will need to combine all possibilities for these gametes to mate with the gametes for the other heterozygous individual to generate an offspring with the desired dihybrid phenotype.
Explanation / Answer
Since, the alleles A and B are 28 m.u. apart, therefore, 72% of the gametes of “Ab/aB” x “Ab/aB” will be parental genotype (non-recombinants) and 28% of the gametes will be recombinant. Therefore,
1. Frequency of non-recombinant gametes = 0.72
i.e. Frequency of “Ab” and “aB” = 0.72
Thus, frequency of “Ab” = 0.36
And frequency of “aB” = 0.36
2. Frequency of recombinant gametes = 0.28
i.e. Frequency of “AB” and “ab” = 0.28
Thus, frequency of “AB” = 0.14
And frequency of “ab” = 0.14
Therefore, the probability of progeny with recessive phenotype “ab/ab” will be 0.0196 (0.14 x 0.14)