Genetics Midterm 2 PP

Given the parental cross AAbbdd × aaBBDD, determine:

1. P generation gametes

2. F1 genotype and testcross

3. All 8 possible F2 phenotypic classes

• P gametes: Abd and aBD

• F1 testcross: AaBbDd × aabbdd

• F2 classes: aBD, Abd, aBd, AbD, ABd, abD, ABD, abd

What is the standard three-point testcross setup?

F1 heterozygous at 3 loci × triple-homozygous recessive tester.

Why do F2 phenotypes reveal the F1 gametes in a testcross?

The recessive tester contributes only abd; F1 alleles determine each phenotype.

What offspring ratio is expected for three unlinked loci?

1:1:1:1:1:1:1:1 across 8 classes.

What suggests linkage in a three-point testcross?

A significant departure from the expected equal eight-class ratio.

How are parental and DCO classes identified?

Parental: 2 most frequent. DCO: 2 least frequent.

How many SCO and DCO classes occur in a three-point testcross?

Four SCO classes (two per region) and two DCO classes.

What happens in an SCO versus a DCO?

SCO: crossover in one interval. DCO: crossovers in both intervals.

Parental: Abc/aBC; DCO: ABC/abc. Which gene is middle?

A; only the A allele switches relative to the corresponding parental class.

Parental: ABC/abc; DCO: AbC/aBc. Which gene is middle?

B; the middle allele differs between corresponding parental and DCO classes.

Why can parental–DCO comparison reveal gene order?

A DCO exchanges the middle locus relative to the flanking loci.

Parental: Abd/aBD. DCO: ABD/abd. What is gene order?

B–A–D (or D–A–B). A is in the middle.

Order B–A–D; parental bAd/BaD. Which SCO classes occur in B–A?

BAd and baD (written ABd and abD in A–B–D order).

Order B–A–D; parental bAd/BaD. Which SCO classes occur in A–D?

bAD and Bad (written AbD and aBd in A–B–D order).

What is the recombination-frequency formula?

RF = recombinant progeny ÷ total progeny × 100%.

400 progeny, 110 recombinants. What is the map distance?

110/400 × 100 = 27.5 m.u.

How do you calculate a three-point map interval?

(Both SCO classes in that interval + both DCO classes) ÷ total × 100.

Why must DCO progeny be counted in both intervals?

Each DCO contains one crossover in each of the two regions.

1000 offspring: SCO I = 86+94, DCO = 10+10. Distance I?

(86+94+10+10)/1000 × 100 = 20 m.u.

1000 offspring: SCO II = 97+93, DCO = 10+10. Distance II?

(97+93+10+10)/1000 × 100 = 21 m.u.

Map B–A–D has 20 and 21 m.u. intervals. Distance B–D?

41 m.u.; add both adjacent map intervals.

If an interval is 8 m.u., what percentage are recombinant?

8% total recombinant progeny; about 4% in each reciprocal recombinant class.

What limits three-point map accuracy?

Sample size and gene distance: too close gives few COs; too far allows hidden multiple COs.

Why can distant genes have underestimated recombination?

Multiple crossovers can restore parental allele combinations and go undetected.

What mapping-distance limits does Norris emphasize?

~10 m.u. gives higher confidence; recombination-based distances reach a 50% ceiling.

Why are organisms with many offspring useful for gene mapping?

Large samples help detect rare crossover classes and improve distance estimates.

How can chi-square help analyze a three-point testcross?

Compare observed class counts with the 1:1:1:1:1:1:1:1 expectation for unlinked loci.

Given 8 offspring counts, what is the mapping workflow?

Find P and DCO → middle gene → SCO regions → interval RFs → draw map.