Chromosomal Basis of Inheritance

Featuring fruit fly:Drosophila Melanogaster

Basic Terms/Information about Drosophila

My diploid number is 2N = 8

Mutants are non wild-type

traits

Wild-type (+) refers to the most common phenotype in fruit flies.

It’s usually dominant (but not necessarily)

Sex-Linkage or (X-linked)

XX XY

In fruit flies, (R) is the dominant wild-type gene for red eyes, and (r) is the recessive, mutant gene for white eyes. The gene is found on the “X” chromosome only. This is considered X-linked.

What would be the phenotype of this female fly?

R r

These are the X and Y chromosomes of a male fly. How is the Y chromosome different from the X?

Does the gene for eye color exist on the “Y” chromosome? Why or why not?

What would be the phenotype of this male fly?

r

The Y chromosome is shorter than the

X

No, because the gene for eye color

is found on the longer segment of

the X chromosome. This part is missing on

the YRed eye femaleWhite eye male

Sex-Linkage or (X-linked)

• When genes are sex-linked, we include the X and Y as part of their genotype. For example, the allele for red eye is not “R” but is written as XR. How would you write the allele for white eye?

Xr

Writing X-linked Genotypes

What is the possible genotype(s) for the eye color of this fly if it is a female?

What is the possible genotype for the eye color of this fly if it is male?

Answer the above questions again for this fly.

XRXR or XRXr

XRY

XrXr

XrY

Sample Problems

Example 1: What is the F1 genotypic and phenotypic ratio of a female true-breeding wild-type fly for red eyes crossed with a white-eyed male?

XXR XR

Xr

Y

XRXr XRXr

XRY XRYGenotypic Ratio

XRXr:XRY

2:2 reduced to 1:1

Phenotypic Ratio

Red-eye female:Red-eye male

2:2 reduced to 1:1

Sample Problems

Example 2: What would the genotypes and phenotypes be of the F2 generation?

XXR Xr

XR

Y

XRXR XRXr

XRY XrY

Genotypic Ratio

XRXR : XRXr : XRY : XrY

1 : 1 : 1 : 1Phenotypic Ratio

Red-eye female : Red-eye male : White-eyed male

2 : 1 : 1

X-linked disorders

Definition: diseases or disorders whose genes are found on the X-chromosome, but not on the Y.

Ex: hemophelia (Xh), color blindness (Xb), muscular distrophy (Xm)

If the disorder is recessive, more males than females will tend to have the disorder.

Why?

Take, for example, colorblindness (Xb)If you have a normal female, what is her possible

genotype(s)? _____, or _____

If you have a colorblind female, what is her genotype? ______

If you have a colorblind male, what is his genotype? ______

How many colorblind genes do males need to inherit to be colorblind? _____ Females? _____

Who does the male inherit the colorblind gene from? _____________________________________

XBXB XBXb

XbXb

XbY

1 2

His mother, who donates the X chromosome

Other traits and alleles of Drosophila melanogaster

Wild Type Traits(+)

Mutant Traits

Gray body = G+ Black body = g

Normal wings = N+ Shriveled, vestigial wings = n

Body-color and wing-type are NOT located on the sex chromosome, so they are considered autosomal

Things to think about independently….How would you confirm or test that these mutant traits are recessive?

Example 3In flies, grey bodies (G+) and normal-wing size (N+) are dominant to black bodies (g) and small wing size (n).

Predict a cross between G+gN+n and ggnn.

Predicted Cross G+gN+n x ggnn

G+N G+n gN+ gn

gn

gn

gn

gn

25% G+gN+n 25% G+gnn 25% ggN+n 25% ggnn

Surprising Results!

Actual Results

8.5%

8.5%

41.5%41.5%41.5%41.5%41.5%41.5%

41.5%41.5%

Why did this happen???

Linked GenesThe genes for body color and wing size are linked, meaning they are found on the same chromosome.

They will most likely be inherited together and will not undergo Mendel’s Law of

. cross over

segregatessegregates the linked genes

Independent Assortmentunless

G+ g

nN+

G+ g

n N+

Homologous Chromosomes

What are recombinants?

Recombinants are offspring that have different phenotypes from those of the parents.

Let’s look back at our original cross

X

8.5%8.5%

41.5%41.5%41.5%41.5%41.5%41.5%

41.5%41.5%

X

A

BC

D

Which offspring (A-D) from this cross are the recombinants?

How do we determine if two genes are linked or if two genes are

located on different chromosomes?

Answer

Calculate the recombination frequency! If the frequency is less than 50%, it is assumed that the two genes are on same chromosomes.

206185

944944965 965

X

A

BC

D

Calculate the recombination frequency for this cross!

Total Flies: 965 + 206 + 185 + 944 = 2500 flies

Total number of recombinants: 206 + 185 = 391

Recombination Frequency = recombinants / total flies = 391/2500 = .156 = 16%

Gene Mapping• Genes are mapped on

a chromosome based upon the recombination frequency.

• For ex. the distance between the genes for body color and wing type is therefore 16 “map-units” apart (16% frequency)

Black Body

Small wings

Grey Body

Normal wings

Using recombination frequencies, create a linkage map for the following a - c:between genes b and a = 10.5%; between genes c and a = 48%; between genes c and b = 37.5%

48%

a c

10.5% 37.5%

b