[WEEK3] Part 1 Mendelian Genetics

Learning objectives

Distinguish between the following pairs of terms: dominant and recessive; heterozygous and homozygous; genotype and phenotype.  

When there's true breeding between parental generation (RR x rr) , the trait that disappears is called recessive and the trait appears to every offspring is called dominant. (Dominant gene is shown as phenotype even if it is heterozygous and the recessive trait should be always homozygous to appear as a pheonotype.) Heterozygous means that the set of chromsome has same allele on the same locus. Heterozygous means that the allele on the same locus is different. Genotype expresses all the gene in the chromosome. Phenotype is the trait at appears as the physical properties.

 

Define the terms: allele, locus

Allele is the gene in the chromosome. Locus refers to the location of the gene on the chromosome

 

Describe the law of segregation and how it relates to gamete formation

Law of segregation explains that during the meiosis (metaphase1), the set of chromosome split into each of the gamete. Diploid cell undergoes meiosis and becomes haploid by spliting homologus.

 

State Mendel’s law of independent assortment and describe how this law can be explained by the behaviour of chromosomes during meiosis.  

Law of independent assortment explains that the allele in the different chromosome undergoes meiosis independently. Allele in the different chromosome does not affect each other when they are splitting or lining up.

(But it is violated when genes are linked)

 

Explain how a testcross can be used to determine if an individual with the dominant phenotype is homozygous or heterozygous.  

When there's individual with dominant phenotype, we can know it's genotype by testcrossing, which means mating with recessive individual. When the unknown indiviual has homozygous dominant genotype, all of the offspring will show dominant trait. When it has Heterozygous genotype, dominant and recessive offspring will be produced in 1:1 ratio

 

Describe how the laws of probability govern Mendelian inheritance and solve genetics problems using the laws. 

allele of the one gene segregate into gametes independently of another gene's alleles.

When we solve the problem with mating individual with multiple alleles, we can think the probability of each allele and multiply each of the probability of allele.

 

Describe the inheritance of the ABO blood system and explain why the IA and IB alleles are said to be co-dominant. 

When IA and IB exist at the same time, it is not determined as A or B blood type. Instead, AB is the blood type for the individual possessing IA and IB. When both of the allele are shown in the phenotype, it is called co-dominant. 

 

Give an example of incomplete dominance and predict results of crosses in  these situations. 

When heterozygous has intermediate phenotype, it is called incomplete dominance

 

Explain how phenotypic expression of the heterozygote differs with complete dominance, incomplete dominance, codominance,  

For example, Black hair color cat (B) and Yellow hair color (Y) cat mate each other, 

-Complete dominance explains heterozygous shows one of the hair color to the offspring cat. Either Black or yellow whichever is dominant to other trait.

-Incomplete dominant refers to the offspring cat with heterozygous genotype shows the middle color of black and yellow, which can be brown.

-Codominance explains neither of the allele is dominant or recessive, so the offspring shows both of the trait at the same time. Heterozygous offspring has both black and yellow fur.

 

Explain pleiotropy, epistasis, polygenic inheritance, multifactorial inheritance   

Pleiotropy is when the one gene of the allele affects to the multiple traits. 

Epitstasis is when one gene in the allele affect/ alter/ control the phenotype of the other allele in the different locus.

 

Multifactorial inheritance is when more than 1 factor causes a trait or health problem, such as a birth defect or chronic illness. Genes can be a factor, but other things that aren't genes can play a part, too. These may include: Nutrition. Lifestyle.

 

Outline why sex-linked genes exhibit unique patterns of inheritance. 

-Male must get Y chromosome from the sperm of their dad. 

 

Describe 2 types of prenatal diagnostic test. 

-Aminocentesis : A sample of amniotic fluid can be taken starting at the 15th week of pregnancy

-Chorionic villus sampling : A sample of chorionic villus tissue can be taken as early as the 10th week of pregnancy

 

Construct a pedigree from information provided in a case study and determine probability of offspring inheriting trait/condition.

Probability -Autosomal dominant -Autosomal recessive - X linked dominant -X linked recessive -Y linked

 

PPT

 

Gregor Mendel

biologist Jean-Baptiste Lamarck

Mendel's Experiments

pure breeding 순종 hybrids 잡종

*Hybridisation = mating two contrasting true-breeding varieties

-> P generation (two true- breeding) -> F1 (hybrid off spring) self pollinate -> F2 (produced with 3:1 ratio)

대립형질에 관여하는 유전자가 1쌍인 경우에는 그 교잡종을 단인자잡종(monohybrid) 또는 단성잡종이라 하고, 2쌍 유전자인 경우에는 2인자잡종(dihybrid) 또는 양성잡종이라 한다

-Wrinkled trait disappeared in the F1 generation -> so it is genetically recessive

(Recessive trait doesn't appear in the heterozygote)

Dominant = 우성 / Recessive = 열성

Homozygous = 순종 / Heterozygous = 잡종

*In heterozygote, the dominant trait is expressed

*Principle of independent assortment : the genes for seed shape and seed color assort independently, because they are located on different chromosomes

검정교배(영어: test cross)는 유전학에서 유전자형을 결정하기 위해 우성 표현형을 가진 유전자형을 모르는 생물체와 열성호모를 교배하는 것을 말한다.

 

 

 

Mendel's Law

 

Terminology

-Back cross 

두 계통 A와 B를 교배하여 F1을 얻고 여기에 다시 양친인 A나 B를 교배하는 것이다. 즉 (A X B) X A 또는 B이다

(잡종과 부모중 하나 또는 부모와 유전적으로 유사한 개체를 교배하여 부모의 유전적 정체성에 더 가까운 자손을 얻는 것.)

-Monohybrid/ Dihybrid

A monohybrid cross is defined as the cross happening in the F1 generation offspring of parents differing in one trait only. A dihybrid cross is a cross happens F1 generation offspring of differing in two traits.

 

Probability and Mendel

-The rules of probability : the alleles of one gene segregate into gametes independently of another gene's alleles

-The multiplication rule : the probability that two or more independent events will occur together is the product of their individual probabilities

-The addition rule : the probability that any one of two or more exclusive events will occur is calculated by adding together individual probabilities

- 1/4 x 1/2 x 1/2 =1/16

-1/8 : aaBbcc +  aaBBcc => (1/4 x 1/2 x 1/2) + (1/4x 1/2 x 1/2) = 1/8

 

Extending Mendel's Rules

A polygenic trait is a characteristic, such as height or skin color, that is influenced by two or more genes. 

-Intermediate phenotype => incomplete dominance

-phenotype of bothe allele => codominance 

Pleiotropy

상위성(epistasis)은 한 유전자가 발현형질로 나타날 때 다른 유전자가 영향을 주는 현상에 대한 집단유전학의 개념이다.

The major difference between the two is that pleiotropy is when one gene affects multiple characteristics (e.g. Marfan syndrome) and polygenic inheritance is when one trait is controlled by multiple genes

-Yes, depending on the area where they live, and the extent of sun expose can make their skin color different. Skin color is affected by multiple genes so it is also a polygenic trait

-It increases the number of phenotypes?

 

 

 

Pedigrees

consaguineous 혈연의

-Achondroplasia : 연골무형성증

-Huntington's disease : 무도병으로도 알려진 헌팅턴병(HD)은 뇌 세포의 죽음을 초래하는 유전 질환이다. 초기 증상은 종종 기분이나 정신 능력에 미묘한 문제가 있다. 일반적인 조정의 부족과 불안정한 걸음걸이가 종종 따라온다.

-Why a female needs two copies of the allele for a recessive X-linked trait to be expressed?? Why not it can have only one allele in homologous (heterozygous)? XaXa 만 왜가능 XAXa 라도 남자가 XaY이면 XaXa 가능하지 않나??

Hemizygous는 유전자에 대립 유전자가 하나만있는 상태입니다. 따라서 이러한 유전자는 일반적으로 두 개의 유전자 사본을 소유하지 않습니다. 

Hemophilia is usually an inherited bleeding disorder in which the blood does not clot properly. 혈우병

-X-linked recessive allele

-X-linked dominant traits rarely skip generation 

: affected male has all affected daughters but no affected sons

1. She always had cold, she had chest infection, her bowel movements were bulky and offensive, lower height and weight ranges for her age,

Symptoms of cystic fibrosis

• a persistent cough that sometimes produces thick mucus.
• difficulty breathing.
• wheezing.
• frequent lung infections.
• salty sweat – salt loss in hot weather may produce muscle cramps or weakness.
• tiredness, lethargy or reduced ability to exercise.
• poor growth or weight gain.

2. A sweat test measures the amount of chloride in sweat. Chloride is part of the salt that's found in sweat. The test can diagnose cystic fibrosis (CF) because people with CF have higher levels of chloride in their sweat. 

4.5.6 1/4, 1/2, 1/16

 

 

 

 

 

Self paced quiz 

4번 5번 안함

Complete dominance is when the dominant trait shown in the phenotypes of both in homozygotes and heterozygotes.

Incomplete dominance is when heterozygote shows the intermediate trait in its phenotype.

-allele : the gene in the chromosome which determines the trait different

-locus : the location of the gene on the chromosome

-homozygote : when the homologous chromosome own same allele

-heterozygote : when the homologous chromosome own different allele for a particular gene

a) Spotted is the dominant characteristic. Only the dominant allele can be shown in the heterozygote offsprings

b) 2/3 of the spotted rabbit should be heterozygous. All of solid rabbits would be homozygous. 

c) Conduct a testcross with those spotted rabbits. If it were homo, all of the offspring would be dominant. If it were hetero, spotted and solid colored rabbits will be born in 1:1 ratio

-Y linked disproof : If it is Y linked, G2 3 individual should be affected as it got the Y chromosome from G1 1, dad's sperm. And also, G2 2 and 7 are affected but they are female. Female cannot be affected as they cannot receive Y chromosome from their dad.

 

-X linked recessive disproof : If it X linked recessive, G3 1,2 should be XaXa to be affected but it is impossible. Genotype of G2 1 is XAY, and G2 2 is XaXa. G3 1,2 are female so they must receive XA chromosome from their dad G2 1. If they own even one XA, it makes them unaffected in phenotype

 

=> If this is X linked recessive, genotype of G1 1 should be XaY and G1 2 should be XAXa. Genotype of G2 2 should be XaXa to be affected, and the mate G2 1 should be XAY. Their offsprings should be XAXa or XaY. So their offspring should be unaffected female or afffected male. G3 1,2 can be the disproof of this mode as they are affected female.

 

-X linked dominant disproof : If it is X linked dominant, G2 3 individual should own XaY as a genotype as it must recieve Y from G1 1 and Xa from G1 2 individual. However in the pedigree it is expressed that is is affected but XaY cannot be affected.

=> If this is X-linked dominant, then IND2 in Gen1 would has the genotype XaXa. Therefore all of her son must be unaffected. However, IND5 in Gen2 is affected male. Therefore this cannot be X linked dominant.

 

-Autosomal recessive

=> all the genotype is possible

 

-Autosomal dominant 

 

-Y linkage : If it is Y linked, G1 would pass Y chromosome to all male offspring, so all of male in G2 should be affected. G2 3 is not affected so, this is incorrect.

 

-X linked dominant : If it is X linked dominant, Genotype of G1 1 should be XAY and G1 2 should be XaXa. From this parents, all male should be unaffected and all female should be affected. When you see G2 1 and G2 4, there are affected male (XAY) and unaffected female (XAXa) each. Their genotypes should be XaY and XAXa instead. 

 

-X linked recessive : If it is X linked recessive, Genotype of G1 1 should be XaY and G1 2 should be XAXa. From their genotype, Genoytype of G2 7 and G2 8 are XAY and XaXa. They can only have unaffected female and affected male as an offspring. So G3 7 and 8 can be used to disproof this mode as they are unaffected male and affected female.

 

The woman will have IB and i genotype and the man would have IA and i. To be Blood group O, it should own homozygous allele, so both of the parents should own i in their genotype.

a) yes, IAi + IAi -> ii

b) yes, IAi + IBi -> ii

c) no, there is no IB

d) no, there is no way to combine ii together as AB doesn't have i

e) yes, IA from AB and i from B combine togther

BW x BW -> BB,BW,BW,WW (1:2:1 ratio of genotypes) (blue, mixed, white 1:2:1 ratio phenotypic ratio)

a) TTRR x ttrr 

b) TtRr

c) Genotype RrTt , Phenotype long tongue, red eyes

d) RT, Rt, rT, rt

e) RR, Rr, Rr, rr x TT, Tt, Tt, tt (F1)  ->

Genotype : RRTT, RRTt, RRTt, RRtt / RrTT, RrTt, RrTt, Rrtt/ RrTT, RrTt, RrTt, Rrtt/ rrTT, rrTt, rrTt, rrtt 

Phenotypic ratio 9:3:3:1 (R_T_:R_tt:rrT_:rrtt)

f) RrTt x rrtt -> Rr, Rr, rr, rr x Tt, Tt, tt, tt

Genotype : RrTt, RrTt, Rrtt, Rrtt/ RrTt,RrTt, Rrtt,Rrtt / rrTt, rrTt, rrtt,rrtt / rrTt,rrTt,rrtt,rrtt

RrTt, Rrtt, rrTt, rrtt

Phenotypic ratio -> (R_T_: R_t_: r_T_:r_t_) 4:4:4:4 = 1:1:1:1

Woman genotype : XAXa

Male genotype : XaY

Son's possible genotype : XAY or XaY

XaY -> color blind

XAY -> normal 

 

 

 

 

 

 

 

Summary