1.) In tomato plants, for three traits: solid leaves ( M_ ) are dominant to mott
ID: 54968 • Letter: 1
Question
1.) In tomato plants, for three traits: solid leaves (M_) are dominant to mottled leaves (mm), normal height (D_) is dominant to dwarf height (dd), and smooth skin (P_) is dominant to peach skin (pp). Suppose that you perform the following cross: MmDdPp x mmddpp. You plant 1000 tomato seeds resulting from the cross, and get the following results:
Use this information to complete the following linkage map. A, B, and C are distances, and “gene 1” and “gene 2” are genes. Gene m has already been placed on the linkage map.
A =
B =
C =
gene 1 =
gene 2 =
Choose from the answer choices below:
a) 5 cM
b) 7 cM
c) 12 cM
d) 15.8 cM
e) 17 cM
f) 17.4 cM
g) p
h) d
2.) In chupacabras, brown (B) fur is dominant to black (b), and red eyes (G) are dominant to green (g). A chupacabra that is homozygous for black fur and green eyes mates with one that is homozygous for brown with red eyes. The F1 are then mated in a series of testcrosses, producing the following progeny. Do the genes assort independently? Round your expected values AND your final answer to one decimal place. For example, 321.12345 = 321.1. R
a) Chi-square value =
b) Do the genes assort independently ? (type "yes" or "no")
Phenotype Number Brown, Red eyes 185 Brown, Green eyes 164 Black, Red eyes 162 Black, Green eyes 179 Total 690Explanation / Answer
1. The cross is:
MmDdPp..................................... .......................................................x...............................................mmddpp.
Gametes: ..MDP,mdp,Mdp,MDp,MdP,mDP,mdP,mDp...........................................................mdp
F1: MmDdPp : solid leaves, normal height, smooth skin = 420
mmddpp: mottled, dwarf, peach = 416
Mmddpp: solid, dwarf, peach = 52
MmDdpp: solid, normal, peach = 21
MmddPp: solid, dwarf, smooth = 2
mmDdPp: mottled, normal, smooth = 62
mmddPp: mottled, dwarf, smooth= 23
mmDdpp: mottled, normal, peach = 4
· Largest number of offsprings: parental genotype= (1) solid leaves, normal height, smooth skin = 420 &
(2) mmddpp: mottled, dwarf, peach = 416
· Smallest number of offsprings: double cross-over leading to these genotypes = (1) MmddPp: solid, dwarf, smooth = 2 & (2) mmDdpp: mottled, normal, peach = 4
ORDER of gene: m.d.p (so no need to change the order of genes in the representation of genotype in the F1)
Now, Distance between m & d = ( Single crossing over between m & d + Double crossing over )/ Total =
(52+62) + (2+4) / 1000 = 120/1000 = 0.12 = 12 cM
Now, Distance between d & p = ( Single crossing over between d & p + Double crossing over )/ Total =
((21+23)+(2+4))/1000=50/1000= 0.05 = 5 cM
Distance between m & p = ( Single crossing over between m & p + 2 X Double crossing over )/ Total =
(21+23+52+62)+2X(2+4)/1000 = 170/1000 = 0.17= 17 cM
ANSWER: A = 12 cM
B = 5 cM
C = 17 cM
gene 1 = d
gene 2 = p
2. In chupacabras, brown (B) fur is dominant to black (b), and red eyes (G) are dominant to green (g).
A chupacabra that is homozygous for black fur and green eyes mates with one that is homozygous for brown with red eyes.
Hence, the cross is : bbgg................X......................................BBGG
gametes:........................bg...............................................BG
F1:.............................................BbGg ( brown fur + red eyes)
Now, the F1 is test-crossed.
BbGg....................................................................X.....................................bbgg
gametes: BG,Bg, bG,bg..............................................................bg
F2:...........BbGg( brown fur, red eye)...........Bbgg(brown fur, green eye)..........bbGg( black fur, red eye).............bbgg (black fur, green eye)
Hence all the four phenotypes appear in 1:1:1:1 ratio
Hence out of 690 progenies, each phenotype's expected appearance should be = 690/4 = 172.5
NULL HYPOTHESIS: Gene assortment occurs independently.
PHENOTYPE
Observed number (O)
Expected number(E)
(O-E)2/E
Brown, red eyes
185
172.5
0.905
Brown, green eyes
164
172.5
0.418
Black, red eyes
162
172.5
0.639
Black, green eyes
179
172.5
0.326
Chi square value = 0.905+ 0.418+ 0.639+ 0.326 = 2.288 = 2.3 (rounded upto 1 decimal place)
Degrees of freedom = (k-1) – (r-1) where k is number of phenotype & r is number of allele
Compare chi square value for the relevant degree of freedom from the standardised chart of chi square values; if chi square value ( 2.3) is less than tabulated value, null hypothesis holds true.
Hence gene assortment occurs independently.
PHENOTYPE
Observed number (O)
Expected number(E)
(O-E)2/E
Brown, red eyes
185
172.5
0.905
Brown, green eyes
164
172.5
0.418
Black, red eyes
162
172.5
0.639
Black, green eyes
179
172.5
0.326
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