A diploid organism is usually purple due to the following pigment synthesis path

Question

A diploid organism is usually purple due to the following pigment synthesis pathway: A geneticist has isolated two mutant strains, one brown and one orange, each of which is homozygous for a recessive, loss-of-function mutation that inactivates one of the two enzymes in the pathway. When the brown (a/a: B/B) and orange (A/A: b/b) strains were mated, all F_1 progeny were (as expected) purple. These F's_1 were then selfed to produce the next (F_2) generation, which showed phenotypic ratios indicative of recessive epistasis with independent assortment (i.e., 9: 3: 4). If an F_2 individual that is purple is chosen (at random) and mated with an individual F_2 that is orange, what is the probability that only purple progeny will be seen in the F_3 generation (assume each single-pair cross produces hundreds of progeny)? If a purple F_2 is mated with an orange F_2 (as above), what is the probability that ALL THREE phenotypes (purple, orange and brown) will be seen in the F_3 progeny? Pedigree analysis. The pedigree shows the inheritance of a very rare, completely penetrant autosomal recessive disorder. Note that the couple in generation IV are currently expecting their first child (indicated by the "?") What is the probability the child will be affected by the disorder? What is the probability the child will be a carrier (i.e., heterozygous for the causative mutation)? What is the probability the child will be neither affected nor a carrier (i.e., that he/she will be homozygous wild-type)? Gene structure/Splicing problem. "Protein X" consists of a total of 431 amino acids. Your colleague, a biochemist, has purified the protein and determined (via complicated and messy chemical techniques) the sequence of the first 37 amino acids in the protein, which she has reported to you as follows: H_2N- MSNITVDDELNLSREQQGFAEDDFIVIKEERETSLSP... Meanwhile, you have isolated a genomic clone of the gene that codes for protein X, and determined the DNA sequence of the first 227 bases from the 5' end of the gene/transcription unit, as shown below (only the coding non-template strand is indicated). You know that the ORF begins within this sequence (i.e., that it includes the actual start codon), and comparisons with a cDNA clone have indicated that there is a single intron contained within this region (there also others further downstream, but those are not relevant here).

Explanation / Answer

9) Brown X Orange

aaBB x AAbb

F1 : AaBb : Purple

AaBb x AaBb

F2 : 9 purple : 3 orange : 3 brown : 1 (9:3:3:1)

If purple is selected at random from F2 and crossed with orange selected at random from F2, the proability to get all purple progeny is as follows,

Nine purple progeny from F2 has four genotypes : AABB, AABb, AaBB, AaBb

Three orange progeny from F2 has two genotypes :AAbb and Aabb

Possibility 1:

AABB x AAbb

F3 : AABb : all purple : Probability is 1

Possibility 2 :

AABB x Aabb

F3 : AaBb and AABB : all purple : probability is 1

Possibility 3:

AABb x AAbb

F3 : AABb and AAbb : Half purple and half orange : probability is 1/2

Possibility 4 :

AABb x Aabb

F3 : AABb, AAbb, AaBb and Aabb : half purple and half orange : probability is 1/2

Possibility 5 :

AaBB x AAbb

F3 : All purple : probability is 1

Possibility 6 :

AaBB x Aabb

F3 : 3/4 purple and 1/4 orange

Total probability = (1 + 1 + 1/2 + 1/2 + 1 + 3/4 ) x 9/16 (Nine purple from F2) x 3/16 (3 orange from F2 selected)

= 0.45

10) Probability to get all three phenotypes in the F3 generation is Zero

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