Mendelian Genetics (Lec 1 and 2)

Question 1

Father of Modern Genetics

Answer 1

Gregor mendel

Question 2

Why were garden peas used in Mendel’s experiments? (5)

Answer 2

Monoecious - Can self or cross pollinate
Short Generation time
Many offspring
Easy to care for
Many paired characteristics

Question 3

Given a cross between two heterozygous parents, what is the genotypic and phenotypic ratio expected from an autosomal dominant trait? Let A and a be the alleles of the gene.

Answer 3

Genotypic - 1:2:1 AA:Aa:aa
Phenotypic - 3:1 A_:aa

Question 4

Give the Genotypic and Phenotypic ratio of the ff, let A and a be the dominant and recessive allele respectively:

AA x AA
AA x Aa
Aa x aa
AA x aa
aa x aa

Answer 4

AA x AA - G: 1 AA ; P: 1 A_
AA x Aa - G: 1:1 AA:Aa ; P: 1 A_
Aa x aa - G: 1:1 Aa:aa ; P: 1:1 A_:aa
AA x aa - G: 1 Aa ; P: 1 A_
aa x aa - G: 1 aa ; P: 1 aa

Question 5

What is a test cross? How does it do what it does?

Answer 5

A test cross is used to determine the genotype of an unknown dominant phenotype.

This is done by crossing it with a homozygous recessive phenotype with a known genotype and determining based on mendelian ratios

Question 6

If a test cross of an unknown Purple flower was crossed with a white flower with genotype pp, what is the genotype of the purple flower if in 423 offspring, 215 were purple and 208 were white? Assume autosomal dominance

Answer 6

The unknown genotype is Pp

Since the ratio is ~50% purple and 50% white

Question 7

What is the law of segregation?

Answer 7

Alleles for each trait segregate or separate during gamete formation such that there is only one factor from each pair.

Two alleles for one trait seperate; have a 50-50 chance for expression

Question 8

Given a dihybrid cross of AaBb x AaBb, what are the expected genotypic and phenotypic ratios?

Answer 8

Phenotypic - 9:3:3:1 A_B_:A_bb:aaB_:aabb

Genotypic - 1:2:1:2:4:2:1:2:1
1 YYRR 2 YYRr 1 YYrr 1:2:1
2 YyRR 4 YyRr 2 Yyrr 2:4:2
1 yyRR 2 yyRr 1 yyrr 1:2:1

Tip: For genotypes, you can set it up first as
1:2:1 AA:Aa:aa and 1:2:1 BB:Bb:bb
Then multiply it like a polynomial, FOIL method

1x1 = 1AABB; 1x2 = 2AABb; 1x1 =1AAbb
2x1 = 2AaBB; 2x2 = 2AaBb; 2x1 =2AAbb
1x1 = 1aaBB; 1x2 = 2aaBb; 1x1 =1aabb

Question 9

What is the Law of Independent Assortment?

Answer 9

Pairs of factors for one trait segregate independently of the factors of other traits

Question 10

What are the 5 mendelian assumptions?

Answer 10

Each trait has a corresponding pair of factors

Alleles are found in nucleus

One allele (dominant) will always mask the effect of the other (recessive) on the same locus.

Trait from each locus is fully expressed

No mutations have occured

Question 11

What is Achondroplasia, what causes it?

Answer 11

Dwarfism, autosomal dominant mutation in fibroblast growth factor receptor gene 3 (FGFR3) on 4p16.3

Question 12

What is Huntington disease and what causes it?

Answer 12

Muscle locking (seizure-like) symptoms due to nerve cell degeneration.

Caused by trinucleotide repeat expansion (CAG) in the Huntingtin (Htt) Gene on 4p16.3

Question 13

What is cystic fibrosis and what causes it?

Answer 13

Mucous accumulation in airways causing decreased gas exchange

Caused by mutation in CFTR gene found in 7q31.2

Question 14

Problem Solving Hehe

Given: Htt Gene for Huntington disease is an Autosomal Dominant Trait; Both parents are heterozygous; they have 4 kids.

Required: Probability of all kids being affected

Answer 14

Solution:
Since it is Hh x Hh, and we are looking the phenotypic ratios are 3:1 Affected:Unaffected, therefore, the answer is

P(4H_) = (3/4)^4 = 81/256 = 0.316

Question 15

Given: Htt Gene for Huntington disease is an Autosomal Dominant Trait; Both parents are heterozygous; they have 4 kids.

Required: Probability of all kids being unaffected

Answer 15

Solution:
Since it is Hh x Hh, and we are looking the phenotypic ratios are 3:1 Affected:Unaffected, therefore, the answer is

P(4hh) = (1/4)^4 = 1/256 = 3.906e-3

Question 16

Given: Htt Gene for Huntington disease is an Autosomal Dominant Trait; Both parents are heterozygous; they have 4 kids.

Required: Probability of having 3 kids unaffected and 1 kid affected

Answer 16

Solution:
Follow 3:1 phenotypic ratio
Using binomial theorem:
P = nCrP(A)^(r)P(B)^(n-r)

P(3 unaffected and 1 affected):
4C3(1/4)^(3)3/4^(1) = 4 * (1/64) * (3/4)
P(3hh,1H_) = 3/64

Question 17

Given: Htt Gene for Huntington disease is an Autosomal Dominant Trait; Both parents are heterozygous; they have 4 kids.

Required: Probability of having the first 3 kids unaffected and last kid affected

Bonus: What is the diff with having 3 kids unaffected and 1 kid affected

Answer 17

Solution:
Follow 3:1 phenotypic ratio
Multiply probability for each child

P(1-3 hh, 4H_) = (1/4)^(3)*3/4^(1) = 3/256

Bonus: Since there is a specific arrangement asked, no need to multiply by combination to account for the different sequences possible

Question 18

Given: Htt Gene for Huntington disease is an Autosomal Dominant Trait; Both parents are heterozygous; they have 4 kids.

Required: If the first kid is already affected, what is the probability that they are Homozygous?

Answer 18

Solution:
Follow 1:2:1 genotypic ratio
Since its known they are affected, remove the aa from the count

1:2 AA:Aa

Therefore your probability
P(AA) = 1/3

Question 19

What is the product rule of probability?

Answer 19

Keyword: “AND”, the combined probability of two independent events is equal to the product of their probabilities

Question 20

What is the sum rule of probability?

Answer 20

Keyword: “OR”, combined probability of two mutually exclusive events is equal to sum of their individual probabilities

Question 21

What is the combined probability of two non-mutually exclusive events?

Answer 21

Let A and B be 2 events.
The sum of probability is
P(AnB)=P(A) + P(B) - P(AUB)

where P(A) is the probability of event A
P(B) is the probability of event B
and P(AUB) is the Union or the overlap of A and B.

Correction, its -P(AUB)

Question 22

A man and a woman, both of whom are heterozygote normal for cystic fibrosis (CF), decide to have 5 children.

How many combinations are possible where two are normal and three are with CF?

Answer 22

Since heterozygote normal, Cc x Cc (Dihybrid Cross); Phenotypic ratio is 3:1::Unaffected:Affected

5 children, take 2 or 5C2
5C2 = 5!/(3!*2!) = 20/2 = 10

Question 23

A man and a woman, both of whom are heterozygote normal for cystic fibrosis (CF), decide to have 5 children.

What is the probability that all children will be normal?

Answer 23

Since heterozygote normal, Cc x Cc (Dihybrid Cross); Phenotypic ratio is 3:1::Unaffected:Affected

P(5A_) = (3/4)^5 = 243/1024

Question 24

A man and a woman, both of whom are heterozygote normal for cystic fibrosis
(CF), decide to have 5 children.

What is the probability that three children will be normal and two will have
CF?

Answer 24

Since heterozygote normal, Cc x Cc (Dihybrid Cross); Phenotypic ratio is 3:1::Unaffected:Affected

P(2A_,3aa) = 5C3 * (3/4)^3 * (1/4)^2
P(2A_,3aa) = 135/512

Question 25

A man and a woman, both of whom are heterozygote normal for cystic fibrosis
(CF), decide to have 5 children.

What is the probability that the first child is normal, the second is affected with CF, the third is normal, the fourth is affected, and the fifth is normal?

Answer 25

Since heterozygote normal, Cc x Cc (Dihybrid Cross); Phenotypic ratio is 3:1::Unaffected:Affected

Just multiply the probabilities for each child

3/4 * 1/4 * 3/4 * 1/4 * 3/4 = 27/1024

Question 26

A man and a woman, both of whom are heterozygote normal for cystic fibrosis
(CF), decide to have 5 children.

Given that their first child is normal, what is the probability that the child is a carrier?

Answer 26

Since heterozygote normal, Cc x Cc (Dihybrid Cross); Genotypic ratio is
1:2:1::AA:Aa:aa

Since we know the child is unaffected, the only choices are
1:2::AA:Aa -> 2/3 Aa Therefore, the probability they are a crarrier is 2/3

Question 27

A cross between two purple-flowered and yellow-seeded pea plants yielded the following results:

purple flowers, yellow seeds: 587
purple flowers, green seeds: 197
white flowers, yellow seeds: 168
white flowers, green seeds: 56

A. What are the most probable genotypes of the parental strains?

B. Prove your hypothesis by conducting a Chi square test.

Answer 27

A. Check the ratios of the results. For this example the ratios are roughly 9:3:3:1 for
P_Y_:P_yy:ppY_:ppyy therefore it is a Dihybrid cross. The parents are both PpYy.

B. Check this sheets na lang:

https://docs.google.com/spreadsheets/d/1WXPyd0dlYTV3we4bHSwdTvQBMZ7iOq2prKBuXRYUpL4/edit?usp=sharing

tl:dr acceptable hypothesis.

Question 28

Hydration check

Answer 28

Gl