AP Biology—Unit 6 Name:
Mendelian Genetics Notes
I. Genetic Info.
A. DNA
i. Chromosome: coiled when cell divides
ii. Chromatin: long, thin, uncoiled to unzip (when not dividing)
iii. Chromatid: two sister strands of DNA (2 chromatids=1 chromosome)
B. Chromosomes-2 types
i. Autosomes-come in homologous pairs (1 from mom, 1 from dad)- alike in size, info form same type of protein
ii. Sex chromosome- you have 2…could be homologous (XX) or not (XY)
C. Karyotype-photograph of chromosomes
i. Tells us: correct number of chromosomes, gender, correct length (couldn’t tell us about nucleotide changes
or about specific genes)
ii. Why?
· Test for disorders like Down’s Syndrome (Chromo. 21)-couldn’t survive w/extra or missing chromosome like #1 b/c it’s big and has too much info.
· Also may use it for sport testing-women can have a condition in which they have an XY (Androgen Insensitivity Syndrome), these women tend to be tall, thin, athletic, etc.
iii. Dark/light areas = banding—helps pair chromosomes up (just means they coil more/less tightly)
iv. Look up karyotypes on internet for examples
D. Gene-section of DNA w/info. for 1 protein
i. In humans, we have 24 chromosomes of genes b/c we have to count each sex chromosome
ii. How many genes do we have? Not sure but # is getting smaller b/c 1 string of amino acids could code for more than 1 protein (~20,000 genes in 24 chromosomes)
iii. Alleles: pair of genes for the same protein
II. Meiosis-see picture and notes accompanying notes
III. Gametogenesis-see picture and accompanying notes
IV. Variation in Meiosis—see picture and accompanying notes
A. Segregation: ½ sex cells get 1 of each chromosome and other ½ sex cells get other 1 of each chromosome (separation)
B. Independent Assortment: position of chromatids @ Metaphase I is random; one pair does not influence what happens to other pairs
2n (n=number of pairs) possible gamete combinations
C. Crossing Over:
i. in prophase I, 2 non-sister chromatids form a chiasma (crossing over region) and exchange equal bits of chromosomes
ii. produces recombinant chromosomes (combines genes from both parents)
D. Random Fertilization
i. 223 for egg * 223 for sperm= 70 trillion possibilities PLUS crossing over
E. Non-Disjunction
i. Homologous pairs don’t separate properly in Anaphase I OR…
ii. Chromatids don’t separate properly in Anaphase II
iii. Result=aneuploidy (incorrect # of chromosomes)
· Monosomic (missing 1 chromosome)
· Trisomic (extra chromosome in a pair)
· Polyploidy (extra entire set of chromosomes- only happens in plants-sometimes on purpose i.e. seedless plants)
V. Mendelian Genetics-see problems
A. Monohybrid cross (1 trait)-
i. Dominant- a trait that is expressed (G=green peas)
ii. Recessive- only expressed when dominant is not present (g=yellow peas)
iii. Homozygous- alleles are identical (GG or gg)
iv. Heterozygous- alleles are not identical (Gg)
v. Phenotype- a description of the trait-use words (green or yellow peas)
vi. Example: Gg*Gg
½ G / ½ g½ G / ¼ GG / ¼ Gg
½ g / ¼ Gg / ¼ gg
vii. Example: Gg*gg
½ Gg
½ gg
viii. Example: GG*gg
1Gg
If both homozygous then all same
If 1 heterozygous ½ and ½
If both heterozygous, ¼, 2/4, ¼
Not ½ because we need to do ratios (for genotype)
Can say ½ for phenotype
B. Dihybrid Cross (2 traits)-
R-round G-green
r-wrinkled g-yellow
Genotypes of…
~heterozygous round, yellow? Rrgg
~wrinkled, homozygous green? rrGG
Phenotype of GgRr? Green and round peas
Examples:
1. Rrgg x RrGg
Look at each gene pair separately…
Rrgg x RrGg
¼ RR ½ Gg
2/4 Rr ½ gg
¼ rr
Now…multiply the gene pair probabilities together. We call this the law of multiplication. It would be a good idea to watch Bozeman if you’re confused…
Genotypic Ratio Phenotypic Ratios
1/8 RRGg 3/8 Round and green peas
1/8 RRgg 3/8 Round and yellow peas
2/8 RrGg 1/8 wrinkled and green peas
2/8 Rrgg 1/8 wrinkled and yellow peas
1/8 rrGg
1/8 rrgg
2. wrinkled yellow x homozygous round, heterozygous green
rrgg x RRGg
1 Rr ½ Gg
½ gg
Genotypic Ratio Phenotypic Ratio
½ RrGg ½ Round and green peas
½ Rrgg ½ Round and yellow peas
3. Chance of Rrgg from RrGg x RrGg
Chance of getting Rr = ½
Chance of getting gg = ¼
Multiply the two chances together…and the answer is… there is a 1/8 chance of getting Rrgg.
NOTE: I could reduce 2/4 to ½ for the Rr because we aren’t looking at ratios in this problem.
Add T-tall and t-short to others…
TTRrGg x ttrrgg
Give genotypic and phenotypic ratios for above cross
1Tt ½ Rr ½ Gg
½ rr ½ gg
Genotypic Ratios Phenotypic Ratios
¼ TtRrGg ¼ Tall, round and green
¼ TtRrgg ¼ Tall, round and yellow
¼ TtrrGg ¼ Tall, wrinkled and green
¼ Ttrrgg ¼ Tall, wrinkled and yello