An individual homozygous for genes cd is crossed with wild type and F1 crossed back with the double recessive. The appearance of the offspring is as follows + + → 903 cd → 897 +d → 98 c+ → 102 The distance between the genes c and d is

Question

An individual homozygous for genes cd is crossed with wild type and F1 crossed back with the double recessive. The appearance of the offspring is as follows + + → 903 cd → 897 +d → 98 c+ → 102 The distance between the genes c and d is (A) 20 map units (B) 9.8 map units (C) 10.2 map units (D) 10 map units

Tardigrade Admin · Accepted Answer

% ðð ðððððððððð¡ððð = ð·ðð tanðð ððð¡ð¤ððð ð¡ð¤ð ððððð = (ððð¡ðð ðð. ðð ððððððððððð¡ ðâðððð¡ð¦ððð /ððð¡ðð ðð. ðð ððððððð¦ × 100 ) × 100 = 200 × 1002000 = 10 % ðð 10 ðð

Q. An individual homozygous for genes cd is crossed with wild type and F1 crossed back with the double recessive. The appearance of the offspring is as follows + + → 903 cd → 897 +d → 98 c+ → 102 The distance between the genes c and d is

20 map units

9.8 map units

10.2 map units

10 map units

% 𝑜𝑓 𝑟𝑒𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 = 𝐷𝑖𝑠tan𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡𝑤𝑜 𝑔𝑒𝑛𝑒𝑠 = Totalno.ofprogeny×100Totalno.ofrecombinantphenotypes​×100 = 200 × 1002000 = 10 % 𝑜𝑟 10 𝑐𝑀

Q. An individual homozygous for genes cd is crossed with wild type and F₁ crossed back with the double recessive. The appearance of the offspring is as follows
+ + → 903
cd → 897
+d → 98
c+ → 102
The distance between the genes c and d is