Genes a, b and c are recessive. Females heterozygous at these three loci are cro
ID: 224018 • Letter: G
Question
Genes a, b and c are recessive. Females heterozygous at these three loci are crossed to phenotypically wild-type males. The progeny are phenotypically shown: Daughters: all + + + Sons: + + + 23 a c b 26 + c + 45 a + b 54 + c b 427 a + + 424 a c + 1 + + b 0 Total 1000 1A. Are the three genes in this cross X-linked or Autosomal? Explain your answer. 1B. Show the order and give the map distances of the genes. 1C. How much interference is there? What does this mean? 1D. Using linkage notation, indicate the genotypes of both parents. In females, are the genes in the cis or trans configuration?Explanation / Answer
Answer:
1A). All the three genes are autosomal
a++/+cb (female) x +++/++ (male)---Parents
If they are X-linked, it would not be possible to produce only one type progeny in F1 becaues of heterozygosuity in female. So all the genes are autosomal.
Explanation:
Shorthand
Number of progeny
+ + +
23
a c b
26
+ c +
45
a + b
54
+ c b
427
a + +
424
a c +
1
+ + b
0
Total 1000
Parental combinations are more so, the parental combinations are a + + & + c b
1. If single cross over (SCO) occurs between a & + and + c
Normal order= a --------+ & + ----- c
After cross over= a ----- c & +------+
a ----- c recombinants are 26+1= 27
+------+ recombinants are 23+0=23
Total recombinants = 1000
RF = (50/1000.)*100 =5%
2. If single cross over (SCO) occurs between + & + and c & b
Normal order= + --------- + & c --------- b
After cross over= + --------- b & c ------ +
+ --------- b recombinants are 54+0= 54
c ------ + recombinants are 45+1=46
Total recombinants = 100
RF = (100/1000)*100 = 10%
3. If single cross over (SCO) occurs between + & b and a&+
Normal order= + --------- b & a------+
After cross over= + --------- + & a---------b
+ --------- + recombinants are 23+45 = 68
a --------- b recombinants are 26+54= 80
Total recombinants = 148
RF = (148/1000)*100 = 14.8%
% RF = Map unit distance
The order of gene is -----
a --5 m.u.-----c-------10 m.u.------b
RF between a & c = 5% = 0.005
RF between w & c = 10 = 0.01
Expected double cross overs = 0.005 * 0.01 = 0.0005
The correct parental genotype of gametes are + c b & a + +
Observed double cross overs are + + b & a c + = 0+ 1
= 1/1000 = 0.001
Observed double cross progeny are more than the expected double cross over progeny.
Coefficent of coincidence (CC) = Obeserved dco / Expected dco
CC = 0.001/0.00005 = 0.05
Interference = 1-CC = 1-0.05 = 0.95
Shorthand
Number of progeny
+ + +
23
a c b
26
+ c +
45
a + b
54
+ c b
427
a + +
424
a c +
1
+ + b
0
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