Exam 2 Flashcards
(133 cards)
Mendels Law of Segregation
2 alleles of each gene separate/segregate during gamete formation, and then unite at random (1 from each parent) at fertilization
- MI separates homologs; then MII separates sisters. Each gamete ends up with 1 copy of each allele
What happens in meiosis that underlies law of segregation?
Homologous chromosomes align in metaphase I and segregate into separate daughter cells
Mendel’s Law of Independent Assortment
During gamete formation, different pairs of alleles segregate independently of each other
50% chance of receiving alleles from mother vs father
what happens in meiosis that underlies law of IA?
Homologous chromosomes align in MetaphaseI with independent orientation; the orientation of 1 tetrad does not influence the orientation of another
independent assortment on same vs different chromosomes
- alleles on different chromosomes = always independently assort
- alleles on the same chromosome may/may not independently assort
independent assortment on the same chromosomes
2 alleles on the same chromosome will assort into the same gamete unless crossing over swaps one onto the homologous chromosomes
therefore, IA on the same chromosome may or may not occur
test cross
cross recessive genotype with mystery genotype
- all dominant –> homozygote dom
- half dominant –> heterozygote dom
why do a test cross?
figure out the genotype of an individual
true-breeding/pure-breeding
homozygous individuals whose line produces the same phenotype when selfed 100% of the time
**can assume genotype is homozygous
how to figure out which is the dominant individual?
look at heterozygous
-cross 2 pure-breeding individuals to get all heterozygous F1 generation and analyze the phenotypes
monohybrid self cross
heterozygotes of 1 gene crossed with each other
Ex/ Aa x Aa
1:2:1 genotypic ratio
3:1 phenotypic ratio
/
alleles on different homologs of the same chromosomes
Ex/ A/a
;
alles on different chromosomes
Ex/ A/a;B/b
dihybrid test cross
2 genes controlling 2 traits
-heterozygotes crossed with recessive homozygotes
Ex/ A/a;B/b x a/a;b/b
genotypic: 1:1:1:1
phenotypic: 1:1:1:1
dihybrid self cross
selfing of dihybrid
genotypic: 9:3:3:1
phenotypic: 9:3:3:1
9 - both dom
3 - 1 dom; 1 rec
3 - 1 rec; 1 dom
1 - both rec
product rule
AND
- the probability that 2 or more independent events occurring together is the product of the probabilities that each will occur by itself
sum rule
OR
- the probabilities of 2 mutually-exclusive events occurring is the sum of their individual probabilities
a scientific hypothesis makes — predictions and is —-.
testable and is falsifiable.
*null hypothesis must make a testable prediction.
ex/ IA will occur.
null hypothesis
there is no significant difference between the observed and expected frequencies
must be very certain that you can reject the null hypothesis
5%
chi-square tests
determine p-value using a formula
total = (observed - expected)^2 /expected
compare values in chart
p value
represents the probability that the null hypothesis is TRUE
p > 0.05
fail to reject the null hypothesis
p < 0.05
reject the null hypothesis with 95% certainty
- there is a greater than 95% chance that the null hypothesis is not true