6.2.6 Predicting Inheritance: Identifying Linkage Flashcards
What is autosomal linkage?
Dihybrid crosses and their predictions rely on the assumption that the genes being investigated behave independently of one another during meiosis
However, not all genes assort independently during meiosis
Some genes which are located on the same chromosome display autosomal linkage and stay together in the original parental combination
Linkage between genes affects how parental alleles are passed onto offspring through the gametes
When writing linked genotypes it can be easier to keep the linked alleles within a bracket
For example, an individual has the genotype FFGG.
However, if there is linkage between the two genes, it would be written as (FG)(FG)
How can you identify autosomal linkage from phenotypic ratios?
In the following theoretical example, a dihybrid cross is used to predict the inheritance of two different characteristics in a species of newt
The genes are for tail length and scale colour
The gene for tail length has two alleles:
Dominant allele T produces a normal length tail
Recessive allele t produces a shorter length tail
The gene for scale colour has two alleles:
Dominant allele G produces green scales
Recessive allele g produces white scales
A newt heterozygous for a normal tail and green scales is crossed with a newt that has a shorter tail and white scales
Parental phenotypes: normal tail, green scales x short tail, white scales
Parental genotypes: TtGg x ttgg
Parental gametes: TG or Tg or tG or tg
Without Linkage:
The outcomes for this dihybrid cross if the genes are unlinked are as follows:
Predicted ratio of phenotypes in offspring = 1 normal tail, green scales : 1 normal tail, white scales : 1 short tail, green scales : 1 short tail, white scales
Predicted ratio of genotypes in offspring = 1 TtGg : 1 Ttgg : 1 ttGg : 1 ttgg
With Linkage:
However, if the same dihybrid cross is carried out but this time the genes are linked, we get a different phenotypic ratio
There would be a 1 : 1 phenotypic ratio (1 normal tail, green scales : 1 short tail, white scales)
This change in the phenotypic ratio occurs because the genes are located on the same chromosome
The unexpected phenotypic ratio, therefore, shows us that the genes are linked
The explanation for this new phenotypic ratio is given in the worked example below:
In reality, the genes for tail length and scale colour in this particular species of newt show autosomal linkage
Parental phenotypes: normal tail, green scales x short tail, white scales
Parental genotypes: (TG)(tg) (tg)(tg)
Parental gametes: (TG) or (tg) (tg)
Predicted ratio of genotypes in offspring = 1 (TG)(tg) : 1 (tg)(tg)
Predicted ratio of phenotypes in offspring = 1 normal tail, green scales : 1 short tail, white scales
What is sex-linkage?
Sex-linked genes are only present on one sex chromosome and not the other
This means the sex of an individual affects what alleles they pass on to their offspring through their gametes
The presence of sex linkage can be identified using pedigree diagrams and Punnett squares
When a gene is sex-linked the phenotypes are not spread evenly across the sexes
In the case of a gene that causes a sex-linked disease, one sex will be disproportionately affected
If the gene is on the X chromosome males (XY) will only have one copy of the gene, whereas females (XX) will have two
Because males only have one X chromosome, they are much more likely to show sex-linked recessive conditions (such as red-green colour blindness and haemophilia)
Females, having two copies of the X chromosome, are likely to inherit one dominant allele that masks the effect of the recessive allele
A female with one recessive allele masked in this way is known as a carrier; she doesn’t have the disease, but she has a 50% chance of passing it on to her offspring
If that offspring is a male, he will have the disease
For sex-linked genes that occur on the X chromosome, there are three phenotypes for females: ‘normal’, ‘carrier’ and ‘has the disease’, whereas males have only two phenotypes: ‘normal’ or ‘has the disease’
Worked example of sex-linkage:
Haemophilia is a well known sex-linked disease
There is a gene found on the X chromosome that codes for a protein called factor VIII. Factor VIII is needed to make blood clot
There are two alleles for factor VIII, the dominant F allele which codes for normal factor VIII and the recessive f allele which results in a lack of factor VIII
When a person possesses only the recessive allele f, they don’t produce factor VIII and their blood can’t clot normally
The genetic diagram below shows how two parents with normal factor VIII can have offspring with haemophilia
Parental phenotypes: carrier female x normal male
Parental genotypes: XFXf XFY
Parental gametes: XF or Xf XF or Y
Predicted ratio of phenotypes in offspring = 1 female with normal blood clotting : 1 carrier female : 1 male with haemophilia : 1 male with normal blood clotting
Predicted ratio of genotypes in offspring = 1 XFXF : 1 XFXf : 1 XFY : 1 XfY
