Term 2 Practice question: Achondroplasia (genetic inheritance)

Question 1

Question

Answer 1

A little person is one of the more than 200 medical conditions known as dwarfism. Dwarfism is a medical or genetic condition that usually results in an adult height of4’10” or shorter, although in some cases a person with a form of dwarfism may be slightly taller than that.

One type of dwarfism Schmid-type metaphyseal chondrodysplasia appears in where only one parent has Dwarfism.

a. With the assumption that Schmid- type metaphyseal chondrodysplasia is rare, is this type of dwarfism inherited as a dominant or recessive trait? Explain your reasoning.

b. On the basis of your answer for part a, what is the expected ratio of children with dwarfism to children with normal stature in the families given in Table 1.? Use a chi-square test to determine if the total number of children of each phenotype is significantly different from the number expected.

c. Use chi-square tests to determine if the numbers of children of each phenotype in family C and in family D are different from the numbers expected on the basis of your proposed mode of inheritance. How would you explain any deviations from the overall ratio expected?

Question 2

Based on a paper

Answer 2

An Achondroplasic Mutation and the Nature of its Inheritance
F.E. Stevens, University of Utah, Published in 1943
based on a large Mormon family from the 1800’s
Accurate and extensive genealogy

Question 3

Part a

Answer 3

The trait is inherited when only one parent is affected and is therefore a dominant trait - then expand to explain dominant inheritance

Question 4

Part b

Answer 4

Phenotypic Ratio expected is 1:1 affected to unaffected children as only one parent has the dominant trait. The affected parent is heterozygous for the dominant allele. If they were homozygous then every child would be achrondroplasic. If both parents were heterozygous for the dominant trait it would be a 3:1 ratio and both parents would display the Dwarfism phenotype.

Phenotype/ Observed/ Expected (1:1)
Normal stature / 52 / x 92 = 46
Dwarfism/ 40/ x 92 = 46
Total/ 92 / 92
= 0.78 + 0.78 = 1.56

Degrees of freedom = 2-1 =1
(no. of phenotypes – 1)

P value for 1 degree of freedom = 1.56 = 0.975 < p < 0.2
There is no significant difference between observed and expected phenotype numbers of normal stature and little people offspring. Any difference is due to chance.

Question 5

Part c

Answer 5

Family C

Phenotype/ Observed/ Expected (1:1)
Normal stature/ 1 / x 7 = 3.5
Dwarfism/ 6 / x 7 = 3.5
Total/ 7/ 7

= 1.78 + 1.78 = 3.57

Degrees of freedom = 2-1 =1
(no. of phenotypes – 1)

P value for 1 degree of freedom = 0.1 < p < 0.05There is no significant difference between observed and expected phenotype numbers of normal stature and little people offspring. Any difference is due to chance.

Family D

Phenotype/ Observed/ Expected (1:1)
Normal stature/ 6 / x 8 = 4
Dwarfism/ 2 / x 8 = 4
Total/ 8/ 8

= 1 + 1 = 2

Degrees of freedom = 2-1 =1

P value for 1 degree of freedom = 0.2 < p < 0.1There is no significant difference between observed and expected phenotype numbers of normal stature and little people offspring. Any difference is due to chance.

CONCLUSION
In Families C and D, although the numbers look unusual, there is no statistical difference between the observed and expected phenotypes. Predicting the outcomes of a genetic cross is based on probability, not a certainty. Every time Family C or D had a child, there was a 50% chance of it being affected by achondroplasia, but the outcome was always down to chance. Note : If a condition is rare is doesn’t necessarily mean that its recessive. You have to look at the phenotypes of the parents to establish the pattern of inheritance.