Studied by 10 people
When Mendel began his experiments, the P generation contained true-breeding individuals for contrasting traits - \n for example, one purple-flowered plant & one white-flowered plant. (5.3)
A. What does it mean that the parents are “true-breeding”?
The plants have the same phenotype and genotype as their parents before them (homozygous dominant or homozygous recessive)
When Mendel began his experiments, the P generation contained true-breeding individuals for contrasting traits - \n for example, one purple-flowered plant & one white-flowered plant. (5.3)
B. What does this mean the genotypes of these plants are?
There’s one homozygous dominant (purple) PP
There’s one homozygous recessive (white) pp
When Mendel began his experiments, the P generation contained true-breeding individuals for contrasting traits - \n for example, one purple-flowered plant & one white-flowered plant. (5.3)
C. What genotype & phenotype ratios did he observe in the F1 generation? Use a Punnett Square
1 homozygous dominant parent (PP) w/ 1 homozygous recessive parent (pp)
Genotype Ratio- 0 PP: 4 Pp: 0 pp
Phenotype Ratio- 4 Purple: 0 White
p p
P Pp Pp
P Pp Pp
When Mendel began his experiments, the P generation contained true-breeding individuals for contrasting traits - \n for example, one purple-flowered plant & one white-flowered plant.
D. When Mendel allowed the F1 generation to self-fetilize in order to produce the F2 generation, what genotype & phenotype ratios did he observe in the F2 generation? Use a Punnett Square
1 Heterozygous Parent w/ 1 Heterozygous Parent
Genotype Ratio- 1 PP: 2 Pp: 1 pp
Phenotype Ratio- 3 Purple: 1 White
P p
P PP Pp
p Pp pp
(+) Explain how Mendel’s work was able to establish rules of dominance and recessiveness in inheritance.
Mendel’s work showed that his initial prediction that the traits of 2 parents “blend” to form an intermediate wasn’t exactly correct. After his experiments, he hypothesized that if the offspring inherited two alleles (one from each parent) and the alleles differed, then only one (the dominant) had influence over the organism’s phenotype.
Describe Mendel’s 2 laws and their connection with the process of meiosis
Law of Segregation
A zygote gets 2 alleles for each gene after fertilization
The two alleles separate into haploid cells during meiosis
Law of Independent Assortment
Each pair of alleles separates w/o influencing other pairs
This leads to different combinations of traits in offspring
Due to random chromosome alignment during meiosis
Incomplete dominance
When the dominant alleles of 1 gene don’t completely determine the phenotype of a heterozygous offspring. So the heterozygous genotype creates an intermediate phenotype
Ex. Flowers
Red = RR , White= WW , Pink= RW
Codominance
When there aren’t any recessive alleles, and the heterozygous genotype creates offspring with both phenotypes
Ex. Chickens
Black= BB , White = WW , Spotted/Peppered = BW
Multiple Alleles
When there’s more than 2 alleles for 1 gene, but the offsprings still inherit only 2 copies
Ex. Blood Types
The different alleles: A, B, or O
Polygenic inheritance
When more than 1 gene contributes to 1 trait which leads to a “spectrum” of phenotypes
Ex. Skin Color
(Can google for visual examples)
Pleiotropy
When 1 gene leads to more than 1 phenotype
1 gene affecting multiple traits in the organism
Ex. A single gene that impacts sight, liver function, and heart function
→ Trait A
Single Gene → Trait B
→ Trait C
Epistasis
When alleles (even recessive) at 1 gene “override” the alleles of another gene
Ex. The labradors from Topic 5.4-5.5 Worksheet
Ext + allele leading to black fur, but a genotype of TyrP1 (-/-) canceling it out
Mitochondrial Inheritance
When mutations in the mother's mitochondrial DNA affects the mitochondrial DNA of her offspring
Since offspring only inherit the mitochondrial DNA of their mothers and not their fathers
So sons cannot pass on this mutation
Happen w/ chloroplast and plants too
Make a Punnett Square and list the genotype and phenotype ratios
A. In humans, eye color follows Mendelian inheritance. Darker eyes are dominant over light eyes. Two heterozygous people marry and have children. Use D/d
Make a Punnett Square and list the genotype and phenotype ratios
B. In pea plants, round seeds (R) are dominant to wrinkled seeds (r), and tallness (T) is dominant to \n shortness (t). A tall plant with wrinkled seeds (rrTt) is crossed with a tall plant with round seeds (RrTT). \n Complete a dihybrid Punnett, then list the phenotype ratio.
Make a Punnett Square and list the genotype and phenotype ratios
C. The gene for fur color in cats is controlled by incomplete dominance. Brown cats (B) are dominant, and white cats (b) are recessive. Orange-cats are heterozygous. An orange cat mates with an orange cat.
Make a Punnett Square and list the genotype and phenotype ratios
D. The gene for coat color in cows is controlled by codominance. Red cows (R) are dominant, and white cows (W) are also dominant. Cows who are heterozygous have red spots and white spots and are called “roan.” A roan cow mates with a white cow
Make a Punnett Square and list the genotype and phenotype ratios
E. Human blood types are controlled by multiple alleles. There are 3 alleles - A, B, and O - for one gene. Combinations of these 3 alleles determine a person’s blood type as Type A, Type B, Type AB, or Type O. A person with Type A blood (AO) has children with a person with Type B blood (BO).
Make a Punnett Square and list the genotype and phenotype ratios
F. Male pattern baldness is a human trait controlled by a gene that is inherited on the X chromosome. A dominant allele gives a person no baldness (B), and a recessive allele gives a person male pattern baldness (b). Despite being called “male” pattern baldness, females can also have this trait. A heterozygous female has children with a man with baldness.
(+) Referring to Q. 5 in the study guide (photo uploaded here)
A. Most likely Recessive, II-1 and II-2 appear as unaffected but produce an affected offspring which is a characteristic of a recessive pedigree which makes sense since II-2 had an affected parent and is therefore a carrier
B. Most likely Sex-linked, The affected mother in Gen I lead to 2 affected sons in Gen II which is a characteristic of Sex-linked recessive. Also a majority of males are affected in this pedigree which is also a characteristic of sex-linked recessive
C. (Image attached)
If the pedigree is most likely Sex-linked Recessive, we know that II-2 has a genotype of X^BX^b (their affected mother from Gen I made her a carrier of the recessive allele, and to have an affected son but be unaffected herself she must be heterozygous)
The father is also unaffected and can only carry 1 X allele which must be X^B to be unaffected
Even though they already have an affected son, the odds of having another affected child remains the same
(+) Referring to Q. 5 in the study guide (photo uploaded here)
D. Most likely recessive, the disease appears to skip a generation in Gen I, and also there are unaffected parents that have affected offspring (Gen I with II-2; Unaffected II-4 & II-5 producing affected III-5 & III-6
E. Most likely autosomal, there isn’t a pattern of affect mothers leading to affected sons (sex-linked recessive) or affected fathers leading to affected daughters (sex-linked dominant). There doesn’t appear to be a majority of affected males (a characteristic of sex-linked recessive)
F. (Image attached)
If this is most likely an Autosomal recessive pedigree, and the offsprings in Gen III are affected, the both parents must be carriers of the recessive allele if they are unaffected (aka heterozygous)
And the odds of producing another affected offspring (aa) remains the same despite already having affected offspring
(image attached)
Gen I, The affected mother → all her offspring are affected (male and female)
The Gen II-affected daughter of Gen I produced all affected offspring, but the Gen II-affected son did not produce any affected offspring
This makes sense since an affected mother leads to all affected offspring
An affected father does not pass on mitochondrial mutations to his offspring since mitochondrial DNA is only inherited from the mother
Chi-Square test Practice Problems are on the study guide sheet (flip for general formula)
The sum of ((O-E)^2)/E for each phenotype = Chi-squared
If Chi-square is less than the critical value (found using the table) then fail to reject the null hypothesis → follows Mendelian Inheritance/fits predicted phenotype ratio
If Chi-square is greater than the critical value, then reject the null hypothesis → does NOT follow Mendelian Inheritance/doesn’t fit the predicted phenotype ratio
Difference between Linked and Unlinked genes
Linked Genes
Do NOT assort independently
Are on the SAME chromosome, close together
Tend to be inherited together
Lower chance of producing recombinant gametes
gametes that are a combination of both parent’s alleles
Unlinked Genes
Do assort independently
Location
Genes that are on SEPARATE chromosomes
Genes that are on the SAME chromosome, but VERY FAR apart
Recombination Frequency (Practice on the study guide, flip for general equation)
RF = [(The number of recombinant offspring) / (Total offspring) ]x 100
Gene Mapping Tips
1 % recombinant frequency = 1 map unit/centimorgan
Large RF = further apart = more likely to cross over
Smaller RF = close together = less likely to cross over
RF “maxes out” at 50%
personal tip: If asked to draw the map:
to figure out which genes are on the ends of a chromosome, find the genes with the largest RF and map them on opposite ends of the drawing
Using the smallest RF, tells you which of the ends 1 of the genes is (ex. with the study guide problem W + Z are the smallest RF, therefore W is closest to the Z end of the chromosome)
Then check to make sure that the locations you’ve mapped match up with what the RF’s tell you
From the study guide
EX. If X + Y is 21% and X + Z is 33%, Y should be closer to X than Z is
EX. If W + Y is 13 % and W + X is 28%, Y should be closer to W than X is
Note
Flashcard