Polymeric and pleiotropic action, gene interaction problems
Lessons 54–55 · 2 lessons · A. Zikiryayev, A. To‘xtayev, I. Azimov, N. Sonin. Biology. Basics of Cytology and Genetics, Grade 9, revised and expanded 5th edition. “Yangiyul Poligraph Service”, Tashkent, 2019
55
Practical work 2: solving problems on interaction of non-allelic genes
Textbook: p. 149
GoalSolve problems on non-allelic gene interaction: identify the kind of action from the ratio.
New words
identifying from the ratio · nisbatdan aniqlashrecessive epistasis · retsessiv epistazquantitative trait · miqdoriy belgisum of the classes · sinflar yig‘indisi
Explanation
In a problem look at the F2 ratio: if its parts add up to 16 it is a dihybrid (two genes). 9:3:3:1 — ordinary independent inheritance or the complementary chicken comb; 9:7 — complementary action; 13:3 or 12:3:1 — dominant epistasis; 9:3:4 — recessive epistasis; 15:1 — non-cumulative polymery; 1:4:6:4:1 — cumulative polymery. Then write the parents’ genotypes, find the gametes, build the Punnett square and group the phenotype classes (e.g. in 9:3:4 the 3+1 form one class because of “ee”). To find numbers of offspring, divide the total by 16 and multiply by the class share. Check the result: the classes must add up to the total.
Plant height: base 10 cm, each dominant allele adds 5 cm. aabb = 10 cm; AaBb = 10 + 2 × 5 = 20 cm; AABB = 10 + 4 × 5 = 30 cm. In F2 of AABB × aabb 6/16 of the plants are 20 cm.
Class activity
“Ratio detective”. Each pair secretly writes an F2 ratio (13:3, 9:7, 15:1, 9:3:4) and the others identify the kind of action and the parents’ genotypes.
Practice
1
Which kind of action gives a 9:3:4 ratio in F2?
Recessive epistasis.
2
What is the F2 ratio in non-cumulative polymery?
15 : 1.
3
BbEe × BbEe gave 480 puppies. About how many are yellow (ee)?
120
4
Why does a 9:7 ratio show a changed form of 9:3:3:1?
Three classes (3+3+1 = 7) give the same phenotype because the trait appears only when both dominant genes are present.