1. All of these ratios indicate that two unlinked genes are segregating- they are all
variations on the predicted Mendelian 9:3:3:1 ratio that indicate the genes are being
transmitted independently during meiosis. The genes are interacting in different
ways, however, to influence phenotype. Independent assortment is a consequence
of the genes being on different chromosomes, or so far apart on the same
chromosome that the likelihood of a crossing over occurring between them in any
given meiosis is 100%. Ans: (d) All of the above.
2. Assume the marine trait is sex-linked recessive and purple trait was autosomal
recessive. The initial cross would therefore be:
X+X+ pp x XmYPP
And the progeny would be:
X+YPp x XmX+Pp
This would yield:
¼
X+Xm
¼
XmY
¼
X+X+
¼
X+Y
¾ P_
3/16 black, furry females
¼ pp
1/16 purple, furry females
¾ P_
3/16 black, marine males
¼ pp
1/16 purple, marine males
¾ P_
3/16 black, furry females
¼ pp
1/16 furry, purple females
¾ P_
3/16 furry, black males
¼ pp
1/16 furry, purple males
or 3/8 black, furry females; 1/8 purple, furry females; 3/16 black, furry males; 3/16
marine, furry males; 1/16 purple, furry males; 1/16 purple, marine males. Ans: (e)
all of the above.
3. The situation here is similar to that observed for coat color in mammals- there are
two genes, one governing a variant color set (red and brown, where red (A_) is
dominant) and one governing whether this phenotype can be expressed (where
individuals homozygous recessive (bb) are green, e.g. similar to the albino locus). If
A is the dominant allele at the first locus (A_ = red; aa = brown), and B is the
dominant allele at the second locus (B_ = ability to express red or brown; bb =
green), all of the phenotypes can be explained on the basis of two genes segregating
independently, and cross #9 would represent a progeny ratio resulting from a
dihybrid cross: AaBb x AaBb:
9/16 A_B_ (red): 3/16 aa_B_ (brown): 3/16 A_bb (green): 1/16 aabb (green).
Ans: (a) true.
4. cross #3: AaBB x aabb: ½ AaBb: ½ aaBb;
cross # 10: AaBb x aaBb: 3/8 AaB_:1/8 Aabb:3/8 aaB_: 1/8 aabb
relevant cross: AaBB x aaBb: ½ AaB_: ½ aaB_ = 1R:1B Ans: (a)
5.
f
+
+
p
a
+
f – p interval:
++a: 25
fp+:
20
+++:
3
fpa:
2
total: 50 50/1000 = 0.05 = 5 map units
p – a interval:
+pa: 81
f++:
74
+++:
3
fpa:
2
total: 160 160/1000 = 0.16 = 16 map units
5 + 16 = 21 m.u. Ans: (d)
6. c.o.c = observed double crossovers/expected double crossovers
expected dco = (0.05)(0.16)(1000) = 8
c.o.c = 5/8 = 0.625 Ans: (c)
7. The issue here is… are all these genes linked? We are recovering close
ratios for the segregation patterns of genes loci A and B:
AB: 34 + 204 = 238
aB
AB
ab
Ab
ab: 40 + 210 = 250
Observed:
238
250
250
262
Expected:
250
250
250
250
aB: 41 + 221 = 262
[O-E]
144.0
0.0
0.0
144.0
Ab: 35 + 215 = 250
[O-E] /E
total
1000
1000
2
2
0.58
0.00
0.00
0.58
A chi square analysis indicates that the segregation is consistent with the genes
independently assorting- they are not showing linkage. The B and C loci, however,
appear to be linked:
Bc
BC
bc
bC
BC: 34 + 41 = 75
Observed:
75
75
425
425
bc: 40 + 35 = 75
Expected:
250
250
250
250
bC: 215 + 210 = 425
[O-E] 2
30625.0
30625.0
30625.0
30625.0
[O-E] 2 /E
122.50
122.50
122.50
122.50
Bc: 221 + 204 = 425
150/1000 = 0.15 = 15 m.u. Ans: (c)
1.15
total
1000
1000
490.00
8.
Tetrad Class
1
cd
cd
++
++
41
spore pairs
seg. c
seg. d
2
c+
++
cd
+d
5
1
1
PD
3
cd
++
cd
++
3
2
1
TT
4
c+
cd
++
+d
17
2
2
PD
5
+d
c+
c+
+d
1
1
2
TT
6
c+
+d
cd
++
1
2
2
NPD
7
c+
c+
+d
+d
1
2
2
TT
69
1
1
NPD
For the c gene: ½*(5+3+1+1)/69 = 0.0725 = 7.3 m.u. Ans: (b).
For the d gene: ½*(3+17+1+1)/69 = 0.159 = 16 m.u. If on the same side, then we
would expect 8.7 m.u. between them. If on opposite sides, then we expect 23.3 m.u.
between them –these would be underestimates due to ignoring double crossovers.
Distance between both genes: (1/2*(5+17+1)+3(3))/69 = 29.7 m.u.
9. They are on opposite sides of the centromere, with c closest. Ans: (d).
10. The fibrillin-1 gene is the gene responsible for Marfan Syndrome. As discussed
in class, this gene is autosomal dominant, pleiotropic (has multiple phenotypic
effects), and can show variable expressivity. A quick google search will show it is on
chromosome 15 in humans. Ans: all of the above (e).