Genomics Flashcards
Allelic vs non-allelic
Same gene (allelic)
Different gene (non-allelic)
Law of segregation
The two alleles from each parent separate during meiosis
Monohybrid cross
One gene which gives phenotypic ratio of 3:1
How do you find out if a phenotype is caused by a recessive/dominant mutant gene
Cross a mutant with a wild-type individual; if the mutant is dominant then F1 will contain mutants, if if it recessive then it will be seen in F2
Law of Independent assortment
Different traits assort independently of each other
Dihybrid cross
Independent assortment; 9:3:3:1
Mutation in amino acid
Results in related proteins with some differences in function
Mutation in gene regulation
Changes the amount of gene product or/and changes when and where the gene is expressed
Mutation in intron and exon
Results in proteins with different functional domains present/absent
Recessive loss of function mutation
Partial loss of gene: leaking or hypomorphic mutation
Complete loss of gene: null mutation
Haplo-sufficient mutation
One wild-type allele provides enough normal gene to produce a phenotype because mutation is recessive
Haplo-insufficient mutation
A wild-type allele cannot provide enough normal gene product to produce a phenotype because mutation is dominant
Gain of function mutation
Increase in functional gene product - Hypermorphic mutation
New function - neomorphic mutation
Allelic series
Genes in order of dominance according to the phenotypes expressed
Conditional Mutants
Mutated phenotype occurs under certain environments
Restrictive condition: mutant phenotype
Permissive condition: wild-type phenotype
Penetrance
The proportion of individuals of a specific genotype that exhibit the corresponding phenotype
Incomplete penetrance: not everyone displays the mutant
Lethal alleles
Dominant lethal alleles
Homozygous mutant
Offspring ratio = 2:1
Recessive lethal alleles
F1 intercross of heterozygotes - ¼ of offspring will die
Offspring ratio = 1
Lethal alleles result in altered 3:1 ratios = dominant (2:1) or recessive (3:0)
Pleiotropic
A single gene with more than one phenotype
The complementation test
Determines whether recessive mutants are from the same gene or two different genes
Two homozygous mutants with the same phenotype are mated
If the F1 is the wild-type, the two mutations complement each other and represent two different genes
If the F1 is a mutant, the two mutations fail to complement and represent two different alleles of the same gene.
Gene interactions
Complementary gene interaction: the activity of both genes is needed for the final phenotype - 9:7
Duplicate gene interaction: either gene can carry out the biological process (redundancy) - 15:1
Dominant gene interaction: two genes with the same phenotype interact additively - 9:6:1
Epistasis
One gene masks the phenotype of another
Recessive epistasis: the recessive genotype of one gene blocks the phenotype controlled by another gene - 9:3:4 ratio
Mitosis
somatic cells - chromosomes replicate and divide, chromosome number is maintained
Meiosis
Germline cells - chromosomes replicate and undergo 2 rounds of division, chromosome number is halved
The chromosome theory of Inheritance
Sutton & Boveri (1902): The behaviour of chromosomes during meiosis can explain why genes are inherited according to Mendel’s Laws
Sex chromosomes
X and Y
X if greatly bigger and has more genes
Heterogametic sex: the sex with different sex chromosomes
Homogametic sex: the sex with the same sex chromosomes
Human sex chromosomes:
Biological males are hetero (XY)
Females are homogametic (XX)
ATP-binding cassette (ABC) transporters
transmembrane proteins that use ATP to transport molecules across membranes
The white gene of the Drosophila eye colour is an ABC transporters
Why do drosophila’s have white eyes
The white gene is an ABC transporter and because the Xw mutation blocks the transport of pigment precursors guanine and tryptophan, it leads to white eyes
Drosophila notation
For a recessive trait with allele e:
- Wild type allele: e+
- Homozygotes: ee, e+e+
- Hetero with genetype e+e would be the wildtype
For a dominant trait with allele B:
- Wild type: B+
- Homozygotes: BB, B+B+
- Hetero with genotype B+B would be mutant
For a sex-linked trait with allele a:
- Wild type allele Xa+
- Homo: XaXa, Xa+Xa+
- Hetero: Xa+Xa would be wildtype
- Hemizygotes with XaY would be mutant
- Hemizygotes: with only one copy of an allele present instead of two (XY males for X-linked genes). This phenotype is present regardless of dominance
Reciprocal crosses
Used to determine sex linkage: two crosses are performed, where the genotypes of the male and female parents are swapped
If offspring ratios differ, this indicates that the trait is sex-linked
Criss cross inheritance
Trait goes from mum to son
X linked
Appears more in males, mothers are carriers
Pedigrees
Family tree - female: circle, male: square
Autosomal recessive inheritance
- Individuals who have the disease are often born to parents who do not
- If only one parent has the disorder, the risk that a child will have it depends on the genotype of the other parent
- If both parents have it, all children will have it
- The gender ratio of affected offspring is expected to be equal
- The disease is not usually seen in each generation but if an affected child is produced by unaffected parents then the risk to subsequent children is ¼
- If the disease is rare, unaffected parents of affected children are likely to be related to one another
autosomal dominant inheritance
- Each individual who has the disease has at least one affected parent
- Males and females are affected in equal numbers
- Either gender can transmit the disease
- In crosses where one parent is affect, approximately half the offspring will have the disease
- Two unaffected parents will not have any children with the disease
- Two affected parents may produce unaffected children
Autosomal mutants
Autosomal dominant: one mutant allele is sufficient
Autosomal recessive: two mutant alleles result in trait
X-linked: hemizygous males express the trait
Y-Linked inheritance
- Only impact males
- Occurs in all sons of affected males
- Females will not be carriers
X linked dominant
- Trait is present in hemizygous males
- Each affected individual has at least one affected parent (except in the case of de-novo - new- mutations that may have occurred in the germline)
- Affected hetero females mated to unaffected males transmit the disease to half their sons and daughters
- Affected hemi males transmit to all their daughter
X linked recessive
- Present in hemi males
- Present in homo females
- Most affected individuals are male
- May skip generations (such as the criss cross inheritance)
- Unaffected parents with a female carrier will give no affected females but half affected males
- Affected males mated to unaffected females will have no affected offspring, but all their daughters are carriers
- Affected females mated to unaffected males will have all affected sons and no affected daughters, but their daughters are carriers
- XX carriers have half of their cells inactivated the mutant X, while half have inactivated wild-type which is why they’re ok
- If XX individuals are ‘mosaic’ for a disease, then the products of both heterozygous alleles are expressed and sufficient to prevent full manifestation of X-linked recessive traits
- Hemizygote: Only produce mutant allele in this case
- X-inactivation skewing: sometimes non-random proportion of X’s of one type are inactivated
