Question

In a particular leech species, body length is determined by 3 different pairs of alleles at 3 gene loci. Every dominant allele adds 2 mm to the worm’s length. The baseline length is 10 mm.

(a) What is the length of the longest possible leech?
(b) The longest and shortest possible leeches are mated. What is the length of the F1
offspring?
(c) The F1 mate with each other and produce F2 leeches. What is the probability that
the F2 offspring will be as long as the longest parent from the P generation?
What is the probability that the F2 offspring will be the same length as one of their
parents?
(d) Write another genotype that would have the same phenotype as AaBBCc.
(e) What type of inheritance pattern is demonstrated by this example of leech body length?

Answer 1

Answer 1:

The longest possible leech will be 22mm long.

Step-by-step explanation:

As the baseline length of an leech is 10mm and six allele are coding for and each allele is adding 2 mm in it so the possible longest leech will be 22mm long.

Answer 2:

The possible length of their F1 generations will be 16mm.

Step-by-step explanation:

As we are mating the longest and shortest possible leeches so the possible genotype of the next generation will be having three dominant alleles so their length will be 16mm long.

Answer 3:

When F1 mate each other then the probability of their next generation having the off springs with P1 length is 0.015625 and the probability of F2 having the same length as parent is 0.328125.

Answer 4:

AaBbCC or AABBcc or AAbBcC

These all genotypes will have the same phenotype as AaBBCc

Answer 5:

This leech is simply showing the effect of genotype directly on its phenotype. It is also shows the phenomenon of independent assortment. Because every allele is independently showing their effect.

Answer 2

a. the length of the longest possible leech is 10mm plus 3*2*2mm = 22mm (if all of 8 alleles are dominant at 3 gene loci)

b. The genotype of the longest leech is AABBCC and the genotype of the shortest is aabbccc

P: AABBCC x aabbcc

F1:AaBbCc (3 dominant alleles 3*2mm) 10+6=16mm

c. F1: AaBbCc x AaBbCc

F2: 1/64 is the probability that F2 offspring will be as long as the longest parent from the P generation

19/64 is the probability that the F2 offspring will be the same length as one of their

parents

d. genotype that would have the same phenotype as AaBBCc is AabbCC or aaBbCC…(all with three dominant alleles)

e. The correct answer is Autosomal Dominant Inheritance.