Genetics Lecture 6, 9/10: H substance, Lethal alleles

Here is the audio (if you see a time in parenthesis, it indicates that you can find that specific content at that point in the recording.)

• • A second gene that’s involved in blood type.
• o H substance. The majority of the population has the H substance.
• o You produce a gugar that is linked to galactose that is linked to fructiose linked to N acetylglucosamine. It then secretes the sugar onto the surface of the red blood cell.
• o The Ia ib allele code for enzymes that cleave the H substance to make the respective antigens (A and B)
• o People with the O blood type just have the original H substance.
• o 1952 Bombay there was a women test for the A and B antigen. She came back O. However, one of her parents is AB. Genotypically this is not possible. This person can only be A, B, or AB. It was discovered that her genotype was B → how do you get the O phenotype. She had a rare recessive mutation for the H locus. Recessive individuals do not secrete the sugar molecule. Because of this they cannot produce the A or B antigens (ending up phenotypically O).
• o The Parents (8:20)
• • Mom – AB (heterozygous for the secreter locus)
• • Dad – BO (heterozygous for secreter locus)
• • AB x BO
• ¼ AB
• ½ B
• ¼ A
• • Hh x Hh
• ¾ H_
• ¼ hh
• • ¼ AB
• ¾ H_ → 3/16 ABH_ (AB Blood type)
• ¼ hh → 1/16 ABhh (O Blood type)
• • ½ B
• ¾ H_ → 3/8 BH_ (B Blood type)
• ¼ hh → 1/8 Bhh (O blood type)
• • ¼ A
• ¾ H_ → 3/16 AH_ (A blood type)
• ¼ hh → 1/16 Ahh (O blood type
• • Results of Blood types: 6 B, 4 O , 3 A, 3 AB
• • LETHAL ALLELES
• o A number of genes which are essential to the survival of an organism. If mutate these genes . . . death of the individual with the mutation.
• o Mutant alleles can be dominant or recessive. Predominantly recessive.
• o Typically in a recessive lethal the homozygous recessive individual will die. The time of death corresponds to when that gene product is essential. Most often death occurs during embryonic period.
• o Typically heterozygotes have some phenotypical alteration.
• • Ex. Coat color in mice. Normal color grey (agouti) variant is yellow
• • Cross two agouti mice all of the offspring are agouti.
• • When you cross two yellow mice you get: 1/3 agouti and 2/3 yellow.
• • If yellow was recessive you would expect to cross two yellow mice and get 100% yellow.
• • In this case the recessive allele is lethal (22: 45)
• • AA → agouti
• • Aa → yellow
• • Aa → dead
• • Crossing two agouti mice you get all AA (agouti)
• • Cross two yellow mice
• ¼ AA → agouti
• ½ Aa → yellow
• ¼ aa → dead. Occurs during development. No live birth
• Final ratio 1/3 agouti and 2/3 yellow.
• o Dominant recessive alleles (rare)
• • Both the homozygous dominant and the heterozygote die.
• • Huntington’s disease (wildtype for the recessive trait(35:20)) – caused by a homozygous dominant allele. Onset of the disease occurs in the early 40’s
• • Mom hh. Dad Hh
• • Affected individual has offspring with unaffected individual then 50% of the offspring end up with the disease.
• o (36:00)Mice – two yellow (agouti is dominant) mice where recessive is lethal. Second trait is Fat vs. Skinny. Fat is dominant. Normal mendelian inheritance.
• o Yellow (Aa), fat (FF) x yellow, fat
• o Aa x Aa
• • ¼ agouti
• • ½ yellow
• • ¼ dead
• • SO 1/3 agouti and 2/3 yellow
• o Ff x Ff (3/4 F_ and 1/4ff)
• • ¼ FF
• • ½ Ff
• • ¼ ff
• o 1/3 agouti
• • ¾ fat = 3/12 agouti, fat
• • ¼ skinny = 1/12 agouti skinny
• o 2/3 yellow
• • ¾ fat = 6/12 yellow, fat
• • ¼ skinny = 2/12 yellow, skinny
• • Phenotypes affected by more than one Gene (42:35)
• o Mendels work was based on the idea that you have a single gene and it is passed on. There are times when two gene products interact to create a phenotype. This doesn’t necessarily mean that the proteins interact.
• o Oftentimes one gene product can mask a second gene product. Often cellular processes are linear. They are referred to epistasis interactions.
• o Example of epistasis (Bombay phenotype)
• • H locus is epistatic to the ABO blood locus
• • If an individual is Hh this will mask the genotype at the ABO locus. hh AO genotype → phenotype → O blood type.

Genetics Lecture 5, 9/8: Chi-squared statistics, extensions of Mendel

Only half of the audio of today's lecture is recorded due to tech mess-up.

Test Content 30-40 pts. mult. choice and short answer from chapter two.

Chi-square analysis

• Mendel's Ratios
• Monohybrid 3:1
• Dihybrid 9:3:3:1
• Ratios are based on assumptions:
• each allele is either dominant or recessive
• normal segregation
• independent assortment
• fertilization is random
• Chi square analysis allows you to test your assumptions
• Does your data match your assumptions over a number of cases?
• Begin with a null hypothesis (Ho). What should your data look like based on your assumptions. This is determined by working out your problem.
• EX. (8:00)
• Yellow, round x green, wrinkled
• 9/16 yellow, round -->587
• 3/16 yellow, wrinlked--> 197
• 3/16 green, round --> 168
• 1/16 green, wrinkled --> 56
• Total of offspring ------>1,008
• Multiply the ratios by the total number of offspring to determine the number of expected offspring

• 	# of offspring	expected (E)	(D) Deviation	D^2	D^2/E
  	587	567	20	400	.71
  	197	189	8	64	.34
  	168	189	-21	441	2.33
  	56	63	-7	49	.78
  Total	1008				x^2=sum of d^2/E which equals = 4.16

• --
• What does the x^2 value tell us?
• Determine the degrees of freedom by taking n (where n is = to the number of unique individuals) and subtracting 1.
• In the above problem df=4-1 . . . . 3.
• You can calculate the P-value. A P-value of .05 --> 1 Fail to reject your original hypothesis. A p-value below .05 you reject your original hypothesis. What you are saying is that there is a greater than 95% chance that your variation is random.
• We failed to reject the Ho because 4.16 is less than 7.82.
• 
  

Chap 4 - EXTENSIONS OF MENDEL

• hw 8,11,19,24,28
• Concepts
• Wildtype --> the typical phenotype for a particular trait in a population. Typically a wildtype trait is dominant.
• Mutant - trait caused by a mutation. typically recessive
• New letters to be used
• e+/e
• e+ = wildtype trait (most often dominant)
• e = mutant trait
• multiple alleles: I^a I^v I^o
• Incomplete dominance - two alleles for a trait and neither allele overtakes the other one completely. "A form of intermediate inheritance in which heterozygous alleles are both expressed, resulting in a combined phenotype." (http://biology.about.com/bldefincomdom.htm)
• Snapdragon flowers R=red r=white
• homozygous red, homozygous white, heterozygous pink
• Typical F1 cross: Rr x Rr
• 1/4 RR (red 1/4), 1/2 Rr (pink 1/2), 1/4rr (white 1/4)
• Tay - sach's disease
• Homozygous recessive. fail to produce enzyme hexosaminidase. Individuals die by age 3.
• Heterozygous 50% of the WT level of the hexosaminidase your phenotype is normal. (Het. are the carriers)
• Homozygous dominant you will make 100% of the WT level of enzyme and are phenotypically normal.
• Codominance - Two alleles of a trait and the heterozygote expresses both alleles. the heterozygote displays the phenotypic characteristics of both alleles. MN blood type is governed by two alleles, M and N. Individuals who are homozygous for the M allele have a surface molecule (called the M antigen) on their red blood cells. The gene that governs ABO blood types has three alleles: I^A, I^B, and I^O. I^A and I^B are codominant, but I^O is recessive.
• Example is the MN blood groups in humans. a glycoprotein found on the surface of red blood cells
• M and N are different versions of this glycoprotein.
• L^m Lⁿ. Cross a heterozygote by M blood group
• L^m Lⁿ x L^m L^m.
• 1/2 L^m L^m --> M blood group
• 1/2 L^m L^m --> M + N blood groups
• ABO bloodtypes you can have more than 2 alleles in a population. The gene that governs ABO blood types has three alleles: I^A, I^B, and I^O. I^A and I^B are codominant, but I^O is recessive.
• I^a --> presence of the antigen
• I^b --> presence of the B antigen
• I^o --> absence of antigen
• Jen is "O" Ben is "B" blood type. Could they produce an offspring with "O" blood type?
• Jen is I^o I^o Ben is I^bI^o or I^bI^b
• If ben is I^bI^b then all offspring are I^bI^o
• I^bI^o x I^o I^o then 1/2 (O blood type) and 1/2 I^bI^o (B blood type)
• AB x AB. results:
• 1/4 aa is ab blood
• 1/2 ab which is ab
• 1/4 bb is b blood

Genetics Lecture 4, 9/5: Trihybrid Crosses , Binomial Theorum

Audio for today's lecture



What happens when you examine how two traits are passed from one generation the next.

• P generation- true breeding
• yellow, round YYRR
• true breeding
• green, wrinkled yyrr
• F1 generation
• yellow, round. YyRr
• yellow is dominant to green, round is dominant to wrinkled.
• Y=yellwo
• R=round
• y=green
• r=wrinkled
• F2 generation
• YyRr x YyRr
• dihybrid cross

• 	YR	Yr	yR	yr
  YR	YRYR	YRYr	YRyR	YRyr
  Yr	YrYR	YrYr	YryR	Yryr
  yR	yRYR	yRYr	yRyR	yRyr
  yr	yrYR	yrYr	yryR	yryr

• To do a dihybrid cross, d the monohybrid cross for each trait
• seed color Yy x Yy

• 	Y	y	1/4YY
  Y	YY	Yy	1/2Yy
  y	Yy	yy	1/4yy

• Genotype ratio- ratio of genotypes in the offspring
• phenotype ratio - ratio of phenotypes in the offspring.
• seed shape Rr x Rr

• 	R	r
  R	RR	Rr
  r	Rr	rr

• How do we put the mono-hybrid crosses together>
• Product Law 12:00 - probability of two events occurring simultaneously is equal to the product of the individual probability
• multiply the individual probabilities
• 3/4 are yellow multiplied by
• 3/4 round = 9/16 yellow and round
• 1/4 wrinkled = 3/16 yellow, wrink
• 1/4 green multiplied by
• 3/4 round = 3/16 green and round
• 1/4 wrinkled = 1/16 green wrinkled
• 9 yellow, round; 3 yellow, wrinkled; 3 green round; 1 green wrinkled
• Mendels 4th postulate (18:00)
• during gamete formation segregating pairs of unit factors (genes) assort independently of each other. Law of independent assortment.
• Trihybrid cross: watching 3 traits be passed from generation to generation (call the genes ABC)
• P generation (21:10)
• AABBCC x aabbcc
• F1 generation
• anytime you cross two homozygous individuals (one dominant, one recessive) you get a heterozygous for all three traits; AaBbCc
• F2 generation
• cross two heterozygotes; AaBbCc x AaBbCc
• do the three monohybrid crosses for Aa x Aa; Bb x Bb; Cc x Cc

• 	A	a	1/4AA	3/4A_
  A	AA	Aa	1/2Aa
  a	aA	aa	1/4aa	1/4aa

• genotype ratio
• 1/4 AA multiplied by
• 1/4 BB = 1/16 AABB multiplied by
• 1/4CC = 1/64 AABBCC (homozygous dominant for all three traits)
• 1/2Cc = 1/32 AABBCc
• 1/4cc = 1/64 AABBcc
• 1/2Bb (heterozygous) = 1/8 AABb multipied by
• 1/4 CC = 1/32 AABbCC
• 1/2Cc = 1/16 AABbCc
• 1/4cc =1/32AABbcc
• 1/4 bb (homozygous recessive) = 1/16 AAbb multiplied by
• 1/4 CC = 1/64 AAbbCC
• 1/2 Cc = 1/32 AAbbCc
• 1/4 cc = 1/64 AAbbcc
• 32 min explains what he looks for on tests
• Phenotype ratio is easier than genotype ratio - two phenotypes and three genotypes for each trait
• Probablilty laws influence genetic outcome
• Product law: The likelihood of two or more events occurring simultaneously is equal to the product of the individual probabilities.
• ex. 1/4 chance blue eyes and 1/8 chance of blonde hair gives you a 1/32 chance of having blue and blonde hair
• Binomial Theroum (38:55)
• can be applied to any situation where there are only 2 outcomes (ex. You are either male or female, right handed or left handed)
• you can multiply out probabilities over a number of events.
• You can use this for family composition. If you have "x" number of children what is the likelihood you will have "x" number of boys and "x" number of girls
• Bob and Judy want all boys what is the likelihood of this happening?
• (a+b)^n . . . . (a+b)^3
• general --> a^3 + a^2b + ab^2 + b^3
• for this problem --> 1a^3 + 3a^2b + 3ab^2 + 1b^3
• 1a^3 - 3 girls; 3a^2b - 2girls, 1 boy; 3ab^2 -2 boys, 1girl; 1b^3 - 3 boys
• (1/2)^3 = There is a 1 in 8 chance they will have all boys
• If they waned 2 boys and a girl
• 3(1/2)(1/2)^2
• 3 in 8 chance of 2 boys and 1 girl
• Jen and ben want a boy. They want 5 kids. If she has 5 kids what is the likelihood shell have at least one boy?
• 1a^5 5a^4b 10a^3b^2 10a^2b^3 5ab^4 1b^5
• A quick way to do this is to calculate the probability of having 5 girls (1a^5).
• (1/2)^5. . . 1 in 32 chance you'll have 5 gilrs. Therefore 1-(1/32)=31/32 chance of having at least one Boy