Genetics Final Lab Review

Genetic linkage leads to the formation of non-recombinant gametes which are …

Significantly more frequent than predicted by chance

Syntenic genes

Genes which are located on same chromosome

The percentage of people when producing gametes with crossing over is higher when …

Related genes are separated by a large distance

When crossing two AaBbCc, the expected phenotype ratio among the offspring is …

(3:1) x (3:1) x (3:1)

How to solve a 3 point cross/mapping problem (Definitely on Lab Final on Monday)

1. Identify if the observation #s deviate from the expected #s

2. Identify the parental allele combinations → biggest #s and the recombinant allele combinations → smallest #s

3. Identify which gene is in the middle by comparing the parental and recombinant allele combinations and seeing which allele changes → get 2 possible gene orders from this

4. Calculate the recombination frequencies for both possible gene orders by doing single crossovers and then using numbers associated with result from crossover into formula:

   1. single crossover 1 # + single crossover 2 # + double crossover #s / total x 100%

5. Solve for interference (I=1-C_oC_o)

   1. Find expected double crossover = (calculated rec freq/map distance G-T) (calculated rec freq/map distance L-T) (total)

   2. Find observed double crossover = sum of recombinant freq

   3. Divide observed double crossover / expected double crossover to get C_oC_o

   4. Plug into eqn: I=1-C_oC_o

Autosomal dominant

• Gene is on autosomes, and one copy is enough to show the trait.

• Affects males and females equally

• Appears in every generation

• No carriers (heterozygotes are affected)

• Affected parent → ~50% of children affected

• Genotypes:

  • AA or Aa → affected

  • aa → normal

Autosomal recessive

• Gene is on autosomes, and two copies needed to show the trait

• Affects males and females equally

• Can skip generations

• Parents often unaffected carriers

• More common with consanguinity

• Genotypes:

  • AA → normal

  • Aa → carrier

  • aa → affected

Dihybrid cross phenotypic ratio (AaBb x AaBb)

9:3:3:1

Epistasis

Gene interactions in which an allele of one gene modifies or prevents expression of alleles of another gene

Test cross phenotypic ratio (AaBb x aabb)

1:1

Monohybrid cross phenotypic ratio (Aa x Aa)

3:1

Incomplete Dominance

Phenotype of both allele is partially expressed

Example of Polygenic Inheritance

• Human eye colors are found on a spectrum that ranges from completely pigmented (dark brown eyes) to a complete lack of pigment (light blue eyes)

  • b/c multiple genes are involved in determining eye color

The eukaryotic cell cycle is divided into what two principal phase?

Interphase and M phase

If two recessive mutations are on the same gene, then crossing them …

Will not restore function → offspring stays mutant

If recessive mutations are on different genes …

Each parent provides a working copy of the other gene → offspring are normal

Interphase

• A period prior to mitosis in which the cell undergoes a period of growth

• During this phase, DNA is synthesized and cell organelles are produced

S phase

The phase of interphase where DNA is copied

The number of chromosomes _____ from G1 to G2

Do NOT change

Centromere

This is a specific region of DNA on a chromosome that holds two sister chromatids together. It acts as the physical foundation and anchor point for the chromosome.

Steps to calculate X² value for Chi-square analysi

1. Calculate the difference between the observed (O) and expected (E), then square it and divide by E

2. Sum for each outcome class

• Then, find where X² falls on a normal distribution to determine p-value

• Need to also know degrees of freedom (df = n-1)

• If fail to reject = they assort independently