Genetics Inheritance Flashcards
Monohybrid cross
Aa x Aa = AA : Aa : aa 1:2:1
Dihybrid cross
AaBb x AaBb = A_B_: a_B_ : A_b_: aabb 9:3:3:1
Test cross AaBb x aabb = 1:1:1:1
Linked test cross: more parental recombinants
Linked genes
And crossing over
Are found on the same chromosome
Examine the percentage of recombinant (COV) frequency. The greater the COV, the greater the distance between genes as there is greater chance of crossing over.
Explain difference between expected and observed ratio
The expected ratio of 1:1:1:1 would occur if alleles of each character were assorted independently and randomly segregated,
Whereas observed ratio arose because both genes were linked on the same chromosome
B and C tend to segregate together into the same daughter cell, and b and c tend to be segregate together into the same cell
Resulting in higher frequency of parental types and less recombinant types, and their phenotypes
Recombinant types occur less frequently because crossing over is a chance event and is determined by distance between the genes.
Standard ratio
9:3:3:1 of F2 phenotypic ratio indicates that indicates that alleles of each character are found on different chromosomes and are inherited independently. Alleles are independently assorted as alignment of homologous pairs on metaphase plate is independent of one another during meiosis.
How do certain recombinant phenotype form
During prophase I of meiosis I, non-sister chromatids of homologous chromosomes under crossing over where at chiasma, portion of chromatid containing allele A breaks and rejoins to portion of chromatid containing b
Resulting in formation of new linkage group where A is linked with b on same chromosome
Gametes formed with A linked with b fused with another gamete containing ____ thus = phenotype
Locus
The position of a gene on a chromosome. Different alleles of the same gene occupy the same locus and different genes occupy different loci.
Dominant and recessive allels
Allele T codes for functional enzyme that synthesises
Allele t codes for non-functional enzyme/does not code for any production of enzymes
In heterozygote, expression of T masks phenotypic effect of t
Since 1 T allele present is sufficient to produce enzyme
Epistasis
Allele A codes for functional enzyme A
while allele B codes for functional enzyme B.
Genotype aa masks phenotypic expression of B_ genotype
Since lack of A will result in this phenotype despite presence of functional enyme B
Chi-squared test
Since p < 0.05 at 5% level of significance, we reject the null hypothesis that observed results = expected results. The difference between observed and expected results is significant and not due to chance.
H0 = There is no significant difference between O and E results and any difference is due to chance alone.
Or external factors: dying from other reasons, or small sample size
Variation
Continuous variation
Range of phenotypes is due to slight phenotypic differences which vary along a continuum indicating continuous variation. This indicates polygenic inheritence where multiple genes are involved
There is an additive effect of genes where each gene has an overall effect, as well as environmental factors
Why pea plants
1) Grow and reproduce quickly allowing for faster results
2) Have many observable traits with contrasting forms
3) Ease of manipulating pollination (self and cross)
4) Produce many offspring in one cross