Question

In Drosophila (fruit flies) the genes how, dumpy and ebony are located on chromosome 3. LOF = loss of function.

• Flies homozygous for a LOF mutation (no gene product made) in ebony have dark black bodies.
• Flies homozygous for a LOF mutation (no gene product made) in dumpy have truncated (short) wings.
• Flies homozygous for a partial LOF mutation (some gene product made but significantly less than normal) in how have wings that will not fold down (held out wings; that's where the gene name is coming from, how).

Use the following allele/phenotype designations:

For how

• allele h+ = wild type/normal
• allele h = partial loss of function mutation
• h+ > h
• phenotype h+ = wild type = wings that fold, which is dominant to held out wings = phenotype h

For dumpy

• allele dp+ = wild type/normal
• allele dp = LOF mutation
• dp+ > dp
• phenotype dp+ = wild type = normal wing size, which is dominant to truncated wings = phenotype dp

For ebony

• allele eb+ = wild type/normal
• allele eb = LOF mutation
• eb+ > eb
• phenotype eb+ = wild type = normal body colour, which is dominant to black body = phenotype eb

You cross a wild type female fruit fly that is a heterozygous carrier for all three mutant alleles with a black bodied male who has truncated wings that do not fold down.

You observe their 504 offspring and note their phenotypes and the number of offspring with each phenotype:

• wings that do not fold - 129
• truncated wings - 83
• black body - 6
• wings that do not fold and a black body - 80
• truncated wings that do not fold - 10
• truncated wings and a black body - 135
• truncated wings that do not fold, and a black body - 15
• wild type - 14

Note: that non-mutant phenotypes are generally ignored. For example a phenotype of "black body" means a fly with a black body, normal wing length and wings that fold.

Question 1:

What offspring phenotypes represent the non-recombinant phenotypes? You may choose more than one.

Select one or more:

a. truncated wings with a black body

b. held out wings with a black body

c. held out wings

d. truncated wings

Question 2:

What option correctly represents the genotype of the heterozygote parent, with genes in the correct order?

Select one:

a. eb+ h dp+ / eb h+ dp

b. h dp+ eb+ / h+ dp eb

c. h dp+ eb / h+ dp eb+

d. h dp eb / h+ dp+ eb+

Question 3:

Determine the order of the genes and the distance between them. Show your work!

You can show your work by typing out your thought process, equations, and so on. Anything. This is worth 9/10 marks. Even if you don't get the correct answer you will get 9/10 for showing your work.

Hint: If you can't figure out the order, keep trying the previous question. You'll be told once you get the right answer.

Question 4:

Flies homozygous for a partial LOF mutation (some gene product made but significantly less than normal) in how have wings that will not fold down (held out wings; that's where the gene name is coming from, how).

But there is another mutation in how that is a complete loss of function (no gene product made). The complete LOF mutation is represented as h-; h- is recessive lethal. Flies homozygous for the complete loss of function mutation (h-/h-) die when trying to emerge from their pupal cases. Basically, you won't see h-/h- adult flies because they are all dead.

You mate two wild type flies who are both carriers for the dumpy LOF mutation (dp) and the complete LOF how mutation (h-). They both have the same genotype: h- dp / h+ dp+.

How many of the resulting larvae (offspring) will be h- / h-, if the genotype at dumpy doesn't matter? Remember, the complete loss of function mutant dies when coming out of its pupal case, the larval stage is before the pupal stage.

Select one:

a. 1/500

b. 1/2

c. 1/4

d. 0

Answer 1

Cross: Wildtype carrier female x male with black body, truncated wings that do not fold down

Phenotype	Genotype	No. of Progeny	Recombinant/non-Recombinant
wings that do not fold	eb+ h dp+/eb h dp	129	non-Recombinant	Parental
truncated wings	eb+ h+ dp/eb h dp	83	Recombinant	Single crossover between dp and eb
black body	eb h+ dp+/eb h dp	6	Recombinant	Double Crossover
wings that do not fold and a black body	eb h dp+/eb h dp	80	Recombinant	Single crossover between dp and eb
truncated wings that do not fold	eb+ h dp/eb h dp	10	Recombinant	Double Crossover
truncated wings and a black body	eb h+ dp/eb h dp	135	non-Recombinant	Parental
truncated wings that do not fold, and a black body	eb h dp/eb h dp	15	Recombinant	Single crossover between dp and h
wild type	eb+ h+ dp+/eb h dp	14	Recombinant	Single crossover between dp and h
		Total = 472

Question 1:

The offspring phenotypes that represent the non-recombinant phenotypes are:

a. truncated wings with a black body
c. held out wings

These are the most abundant of all phenotypic classes in the progeny, and therefore, represent gametes from the Parental mother, that did not undergo crossing over.

Question 2:

The genotype of the heterozygote parent, with genes in the correct order is:

b. h dp+ eb+ / h+ dp eb

Since the 'truncated wings with a black body' and the 'held out wings' progeny classes are the non-recombinant classes, they must have the parental chromosomes from the heterozygous parent. Therefore, one chromosome in the heterozygote is h dp+ eb+ (held out wings) and the other is h+ dp eb (truncated wings with a black body). The double crossover progeny is the progeny class with the least number of individuals, these are the 'truncated wings that do not fold' (genotype h dp eb+) and the 'black body' (genotype h+ dp+ eb) progeny classes.

Now, the h dp eb+ and the h+ dp+ eb chromosome are formed only when the parental chromosomes exchange the dp locus and the h dp+ eb+ becomes h dp eb+, while the h+ dp eb becomes h+ dp+ eb.

Since double crossovers exchange the central locus, dp is the locus in the middle. Therefore the gene order is h-dp-eb.

Question 3:

Gene order is h-dp-eb

Distance between Held-out wings and Truncated Wings (h and dp):

Map distance (h-dp) = No. of h-dp recombinants * 100 / Total no. of progeny
    = (6 + 10 + 15 + 14) * 100/472 = 45 * 100/472 = 9.534 %

The distance between the Held-out wings and the Trunated wings loci is 9.534 map units.

Distance between Ebony Body and Truncated Wings (eb and dp):

Map distance (eb-dp) = No. of eb-dp recombinants * 100 / Total no. of progeny
    = (83 + 6 + 80 + 10) * 100/472 = 179 * 100/472 = 37.924 %

The distance between the Ebony body and the Trunated wings loci is 37.924 map units.

The distance between the Ebony body and the Held-out wings loci is 47.458 map units.

Question 4:

c. 1/4

Since both parents are heterozygous for the Held-out wings LOF allele, that is their genotype is h+/h-, the probability for an individual from the progeny to be h-/h- is the same as the product of the probability of each parent passing on the h- allele.

Therefore,

Prob. h-/h- progeny = Prob. h- passed by Parent 1 * Prob. h- passed by Parent 2 = 1/2 * 1/2 = 1/4