Question

In: Biology

1. There are four loci in wild sorghum that each contribute equally to the height of...

1. There are four loci in wild sorghum that each contribute equally to the height of the plant. The smallest plant is 2 feet tall. When homozygous for contributing alleles at all four loci, however, the plant stands at 10 feet tall. If one crosses a small sorghum plant (2 ft tall) with a giant sorghum plant, the F1 are all 6 feet tall. Letting the F1 self-pollinate, how often will a plant 8 feet tall be found in the F2 progeny?

A. 1/64

B. 6/64

C. 7/64

D. 20/64

2. There are four loci in wild sorghum that each contribute equally to the height of the plant. The smallest plant is 2 feet tall. When homozygous for contributing alleles at all four loci, however, the plant stands at 10 feet tall. Based on the above information, what is the individual allelic contribution to the overall height of the plant?

A. 1 foot/allele

B. 2 foot/allele

C. 4 foot/allele

D. 5 foot/allele

3. A sample population of cheetahs show that 168 individuals have a kinked tail trait (TT), 249 individuals have a partial kink (Tt), and 85 individuals have straight tails (tt). What is the frequency of the T allele in this population?

A. 0.589
B. 0.551
C. 0.411
D. 0.169

Expert Solution

Answer 1

Given information:

Smallest plant (base height) = 2 feet

Tallest plant = 10 feet

Let the alleles be A,a, B,b, C,c, D,d

Thus,

smallest plant = aabbccdd=2feet

tallest plant = AABBCCDD=10 feet (total 8 dominant alleles)

Contribution of each dominant alleles

=extra height/total no. of contributing alleles

=(10-2)/8

=8/8

=1 feet/allele

Cross1:

aabbccdd with AABBCCD

F1: all AaBbCcDd

Height = (base height + contribution of each 4 dominant alleles)

=2+4

=6 feet

Cross 2:

F1 with F1, i.e.

AaBbCcDd with AaBbCcDd

Let us make individual punnett square for each locus:

	A	a
A	AA	Aa
a	Aa	aa

	B	b
B	BB	Bb
b	Bb	bb

	C	c
C	CC	Cc
c	Cc	cc

	D	d
D	DD	Dd
d	Dd	dd

Possible genotype for a height of 8 feet will need 6 contributing alleles (6 dominant alleles).

Thus possible genotypes are:

Genotype	Probability (refer from above punnett square)
AABBCCdd	1/4×1/4×1/4×1/4	=1/64
AABBccDD	1/4×1/4×1/4×1/4	=1/64
AAbbCCDD	1/4×1/4×1/4×1/4	=1/64
aaBBCCDD	1/4×1/4×1/4×1/4	=1/64
AaBbCCDD	2/4×2/4×1/4×1/4	=4/64
AABbCcDD	1/4×2/4×2/4×1/4	=4/64
AABBCcDd	1/4×1/4×2/4×2/4	=4/64
AaBBCCDd	2/4×1/4×1/4×2/4	=4/64

On adding up all the values , we will get,

1/64+1/64+1/64+1/64+4/64+4/64+4/64+4/64

20/64

Ans1: 20/64

Ans2: 1 feet/allele (solved above)

Answer 3:

Given:

TT=168 individuals

Tt=249 individuals

tt=85 individuals

Total individuals = 502

Frequency of T allele

=No. of T alleles/ total alleles

=(168×2)+(249×1)/(2×502)

=(336+249)/1004

=585/1004

=0.58266

Ans:Frequency of T allele = 0.59

gladiator answered 1 year ago

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