Question

how do you solve these?

14.26 The questions that follow make use of this тар: + 10 15 20 10 ki 20 + 151 Calculate a. the frequency of jb gametes from a J B/j b genotype, 0. the frequency of AM gametes from an a M/Am genotype. c. the frequency of JBD gametes from a j B dij bo genotype. d. the frequency of JB d gametes from a j B dij bo genotype. e. the frequency ofjb dlj bd genotypes in aj B dij bo jB dj b D mating. the frequency of Ak F gametes from an AKI genotype.

this is all the information available. it says to calculate for each

Answer 1

a. The units between j and b are 20 in the given map.So, the gametes are of 20% recombinant  type and 80% parental type. In j B/j b parent 7 b gametes are half the parental type or 40% of the total gametes.

b. Between A M there are 65 map units so this show independent assortment as the frequency of recombinant type gametes should not exceed 50 percent. In an a M/A m parent, the A M gametes are half of the recombinant type gametes. There will be 25% A M gametes.

c. In a jB d/l bD individual,/B D gametes are produced by double cross over : a crossover in the interval j - b and a crossoverin the interval b - d. As there are 20 map units between j and band 10 map units between b and d, the frequency of doublecrossoversis expected to be 0.20 x 0.10 x 100'2,:2"/u. I B D gametes are half of the double crossovers produced, so will be seen1 percent of the time.

d. In a jB d/j b D individual,lB d gametesare one-half of thegametesproduced by a single crossover in the interval T - b. Sincethere are 20 map units between these genes, 10 percent of thegametes will be / B d. e .
e.By the reasoning in( c ) ,j b d gametes will be seen l percent of the time in a j B d / l h D individual. To obtain a j h d/j b dgenotype, one must obtain such gametes from bothparents.

f. The frequency of this is (0.01 x 0.01) x 100'/" : 0.01'2,. f. Basedon the map distances,one expects 10 percent recombination between aand k and 50 percent recombination between k and /. (Again, notethat even though k and / are more than 50 map units apart, oneobserves only 50 percent recombination between these two genes.In an A K F/a kf individual, an Ak F gamete results from a doublecrossover: one crossover between n and k (10 percent chance) andone' crossover between k and/(50 percent chance).The chance of bothcrossovers occurring simultaneously is 0.10 x 0.50 : 0.05, or 5percent. Since A k F gametes are half of the doublecrossoversproduced, they will be 2.5 percent of thetotal.