Question

2. The Aldose gene is 10 map units from the fructose gene (A = dominant, can produce aldose; a recessive, cannot produce aldose; F= dominant allele, can produce fructose; fe recessive allele, cannot produce fructose). You cross worms with the chromosomes shown below. What is the probability that an offspring will be able to produce both Aldose and Fructose? A = donninant Af a = recessie af Af af AF Af af af F= dominant af af Aaff Aaff aaff aaff forecessive af hace haft aap Flaaff B1 0.1 af Aaff Raff capf aaff C) 0.2 aflaar Bacaaff agF D) 0.4 E) 0.45 Af x afta 20.05 Aaff Xaaff 1/20 A 3. species of frogs normally has yellow spots on its belly. Frogs from line A have a mutation in the A gore. When examined more closely 'aa' frogs have no spots on their bellies. Frogs from the Bline have a mutation in the B gene. When examined more closely 'bb' frogs have white spots on their bellies. The A and B genes follow a recessive epistasis inheritance pattern and A is epistatic to B. If two frogs heterozygous for both A and B are mated, what is the probability that an offspring will have no spots? A) 9/16 (B)4/16 03/16 D) 1/16 E) O

work for 2 and 3 please, dont understand it.

Answer 1

Answer:

2). A). 0.05

Explanation:

Distance between genes (m.u.)= Recombination frequency (%)

Af/aF genotype produces four types of gametes as below:

Recombinant Gametes (10%):

AF= 5%

af= 5%

Non-recombinant gametes (90%):

Af = 45%

aF = 45%

af/af genotype produces only one type of gametes, af.

The probability of an offspring will be able to produce both aldose and fructose = AF/af = 5%=0.05

3).

B). 4/16

Explanation:

AaBb x AaBb --- parents

Aa x Aa = A_(3/4)& aa(1/4)

Bb x Bb = B_(3/4)& bb (1/4)

Probability of an offspring with no spots = aaB_ + aabb = 1/4*3/4 + 1/4*1/4= 4/16