Both Bobz and Suzie mer-people have a single tail fluke. All the mer-babies that
ID: 149095 • Letter: B
Question
Both Bobz and Suzie mer-people have a single tail fluke. All the mer-babies that Bobz and Suzie have possess a double tail fluke. One of Bobz and Suzie's children is planning for mer-babies with a no-tail fluke mer-person. If the tail fluke genes are linked 22 map units apart on the same chromosome, what is the probability that this couple will produce a) A no-tail fluke mer-baby. b) A single tail-fluke mer-baby c) A double tail fluke mer-baby Tail Fluke (duplicate interaction) F-G- F-gg ffG- ffgg Double Single NoneExplanation / Answer
Answer:
Fg/Fg x fG/fG----mer-people
Fg/fG-----mer-baies (double tail)
Fg/fG x Fg/fG----
The followings frequencies of gametes are produced.
FG(0.11)
Fg (0.39)
fG (0.39)
fg(0.11)
FG(0.11)
FG/FG (0.0121)
FG/Fg (0.0429)
FG/fG (0.0429)
FG/fg (0.0121)
Fg (0.39)
FG/Fg (0.0429)
Fg/Fg (0.1521)
Fg/fG (0.1521)
Fg/fg (0.0429)
fG (0.39)
FG/fG (0.0429)
Fg/fG (0.1521)
fG/fG (0.1521)
fG/fg (0.0429)
fg(0.11)
FG/fg (0.0121)
Fg/fg (0.0429)
fg/fG (0.0429)
fg/fg (0.0121)
F_G_ = 0.0121 + 0.0429 + 0.0429 + 0.0121 + 0.0429+0.0429+0.0121+0.1521+0.1521 = 0.5121
F_gg = 0.1521+0.0429+0.0429 = 0.2379
ffG_ = 0.1521+0.0429+0.0429 = 0.2379
ffgg = 0.0121
a). A no-tail fluke mer-baby =0.0121 = 1.21%
b). A single tail-fluke mer-baby = 0.2379+0.2379 = 47.58%
c). A double tail fluke mer-baby = 0.5121 = 51.21%
FG(0.11)
Fg (0.39)
fG (0.39)
fg(0.11)
FG(0.11)
FG/FG (0.0121)
FG/Fg (0.0429)
FG/fG (0.0429)
FG/fg (0.0121)
Fg (0.39)
FG/Fg (0.0429)
Fg/Fg (0.1521)
Fg/fG (0.1521)
Fg/fg (0.0429)
fG (0.39)
FG/fG (0.0429)
Fg/fG (0.1521)
fG/fG (0.1521)
fG/fg (0.0429)
fg(0.11)
FG/fg (0.0121)
Fg/fg (0.0429)
fg/fG (0.0429)
fg/fg (0.0121)
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.