Ch3 Mendelian Genetics

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Chapter CONCEPTS• Inherited characteristics are under the control

of discrete particulate factors called genes that are transmitted from generation to generation on vehicles called chromosomes, according to rules, first described by Gregor Mendel The outcomes of crosses subject to these rules are affected by chance deviation and can be evaluated using statistical analysis. Human traits are initially studied using the pedigree method .

• 3.1 Gregor Mendel• 3.2 Monohybrid cross• 3.3 Dihybrid cross• 3.4 Trihybrid cross• 3.5 Rediscovery of Mendel’s Work• 3.6 Independent Assortment and genetic Variation• 3.7 Probability and Genetic Event• 3.8 Evaluating genetic data :Chi-square• 3.9 Human Pedigree

what organism to study? Mendel used garden pea (Pisum sativum) available cheap from local merchants easy to grow, small space, fast generation time lots of phenotypic variation could be manipulated for controlled pollination: s

elf and cross fertilization

• Genotype and Phenotype (study figure 3.1)

1. Gregor Mendel

• Mendel’s Experimental Design – Understand the difference between self-

fertilization and cross-fertilization.– Understand the organism used in Mendel’s

design.– Understand what true-breeding implies.– testcross– Understand the 7 characteristics or traits that

Mendel examined & alternative forms of each.

• Monohybrid Crosses & Law of segregation– Understand what the notations for the different

genotypes are.– Understand what monohybrid crosses are and what

reciprocal crosses are.– Understand all the terminology in this section

Understand what the Law of Segregation means and what genotypic & phenotypic ratios are.

– Natural segregation

3.2 Monohybrid cross

3.3 Dihybrid cross

• Dihybrid Crosses & Law of Independent Assortment– Understand what the Law of Independent Assortment

means and predicts in terms of dihybrid phenotypic & genotypic ratios.

– Understand how to use the branch diagram for dihybrid crosses & how to do dihybrid testcrosses.

– Understand how to predict results of trihybrid or greater crosses using the branch diagram.

How Mendel's Peas Become Wrinkled:

A Molecular Explanation • Only recently, well over a hundred years after Mend

el used wrinkled peas in his ground-breaking hybridization experiments, have we come to find out how the wrinkled 皱缩 gene makes peas wrinkled.

• The wild-type allele of the gene encodes a protein called starch-branching enzyme (SBEI). This enzyme catalyzes the formation of highly branched starch molecules as the seed matures. Wrinkled peas, which result from the homozygeus presence of the mutant form of the gene, lack the activity of this enzyme.

• The production of branch points is inhibited during the synthesis of starch within the seed.

• This in turn leads to the accumulation of more sucrose and to a higher water content while the seed develops.

• Osmotic pressure inside rises, which causes the loss of water internally and ultimately the wrinkled appearance of the seed during its maturation.

• In contrast, developing seeds that bear at least one copy of the normal gene (being either homozygous or heterozygous for the dominant allele) synthesize starch, and reach an osmotic balance that minimizes the loss of water

• the end result is a smooth-textured outer coat

• The SBEI gene has been cloned and analyzed. Interestingly, the mutant gene contains a foreign, sequence of some 800 base pairs that disrupts the normal coding sequence.

• This foreign segment closely resembles other such sequences, called Transposable elements .

• They have the ability to move from place to place in the genome of organisms.

• Transposable elements similar to this one have been found in maize (corn), parsley, and snapdragons.

• Study of the SBEi gene provides greater insight into the relationship between genotypes and phenotypes.

• We cover transposable elements greater detail in Chapter 14.

3.4 Trihybrid cross

• Three factor cross

– Understand how to use Punnett squares and the branch diagrams.

– Understand what probability is and understand when to use the product rule and sum rule of probability.

– Understand what a testcross is and the expected Mendelian phenotypic & genotypic ratios produced in testcrosses.

MENDELIAN ANALYSIS: 1 n GENES

3 methods of working out expected outcomes of controlled breeding experiments:1. Punnet square2. tree method (long) – genotypes . tree method (short) – phenotypes3. (a+b)N

MENDELIAN ANALYSIS: PUNNET SQUARE

all pairings of ♀ and ♂ gametes

= probabilities of all pairings some pairings occur >1 different Ps for different

genotypes & phenotypes 1 gene, 2 x 2 = 4 cells 2 genes, 4 x 4 = 16 cells 3 genes, 8 x 8 = 64 cells...

too much work !

MENDELIAN ANALYSIS: LONG TREE1 gene, alleles A, a GENOTYPE PHENOTYPE

¼ A/A 1/4 A/A 3/4 A

½ A/a 1/2 A/a

¼ a/a 1/4 a/a 1/4 a

MENDELIAN ANALYSIS: LONG TREE2 genes, alleles A, a, B, b... GENOTYPE PHENOTYPE

¼ B/B 1/16 A/A B/B 9/16 AB¼ A/A ½ B/b 1/8 A/A B/b

¼ b/b 1/16 A/A b/b 3/16 Ab¼ B/B 1/8 A/a B/B

½ A/a ½ B/b 1/4 A/a B/b¼ b/b 1/8 A/a b/b¼ B/B 1/16 a/a B/B 3/16 aB

¼ a/a ½ B/b 1/8 a/a B/b¼ b/b 1/16 a/a b/b 1/16 ab

MENDELIAN ANALYSIS: SHORT TREE2 genes, alleles A, a, B, b... PHENOTYPE

¾ B 9/16 AB ¾ A

¼ b 3/16 Ab

¾ B 3/16aB ¼ a

¼ b 1/16 ab

much easier... can extend to >2 genes

formulae for n genes, dominance, n hybrid crosses …

1. # possible different types of gametes = 2n

2. # possible different genotypes of progeny = 3n

3. # frequency of least common genotype = ¼ n

4. # possible different phenotypes of progeny = 2n

MENDELIAN ANALYSIS: 1 n GENES

STATISTICS: BINOMIAL EXPANSION

diploid genetic data suited to analysis (2 alleles/gene) product and sum rules apply examples... coin flipping

define symbols...

use fomula: (p+q)n = n

[n!/(n–k)! k!](p)n–k (q)k = k=0

From 3.7 Probability and Genetic Event

e.g., outcomes possible from monohybrid cross

1 offspring, n = 1, (p+q)1 = 1

(1!/1!0!)(¾)1(¼)0 = 3/4 (1!/0!1!)(¾)0(¼)1 = 1/4 = 1

2 offspring, n = 2, (p+q)2 = 1

(2!/2!0!)(¾)2(¼)0 = (1)9/16 (2!/1!1!)(¾)1(¼)1 = (2)3/16 (2!/0!2!)(¾)0(¼)2 = (1)1/16 = 1 16/16

3 offspring, n = 3, (p+q)3 = 1

(3!/3!0!)(¾)3(¼)0 = (1)27/64 (3!/2!1!)(¾)2(¼)1 = (3)9/64 P of 2A_ + 1aa (3!/1!2!)(¾)1(¼)2 = (3)3/64 (3!/0!3!)(¾)0(¼)3 = (1)1/16 = 1

STATISTICS: BINOMIAL EXPANSION

Q: True breeding black and albino cats mate and have a little of all black kittens. If these kittens grow up and breed among themselves, what is the probability that at least two of the F2 kittens will be albino?

STATISTICS: BINOMIAL EXPANSION

Q: True breeding black and albino cats mate and have a little of all black kittens. If these kittens grow up and breed among themselves, what is the probability that at least two of the F2 kittens will be albino?

A: First, define symbols and sort out basic genetics... one character – black > albino, one gene B > b ...

STATISTICS: BINOMIAL EXPANSION

P1: black × albino genotype: BB bb

gametes: P(B) = 1 P(b) = 1 F1: black × black

genotype: Bb Bb gametes: P(B) = ½ , P(b) = ½ P(B) = ½ , P(b) = ½

expected F2: P(black) = P(B_) = (½ )2 + 2(½ )2 = ¾ = p

P(albino) = P(bb) = (½ )2 = ¼ = q

STATISTICS: BINOMIAL EXPANSION possible outcomes...

F2 * Probability 3 black + 0 albino P ( ) = (¾ )3 × (¼ )0 = 27/64 P ( ) = (¾ )2 × (¼ )1 = 9/64

P ( ) = (¾ )2 × (¼ )1 = 9/64 2 black + 1 albino P ( ) = (¾ )2 × (¼ )1 = 9/64

P ( ) = (¾ )1 × (¼ )2 = 3/64

P ( ) = (¾ )1 × (¼ )2 = 3/64 1 black + 2 albino P ( ) = (¾ )1 × (¼ )2 = 3/64

0 black + 3 albino P ( ) = (¾ )0 + (¼ )3 = 1/64 Total 64/64

STATISTICS: PASCAL’S TRIANGLESample Frequency Possible Outcomes* Combinations

1 × 1 + 1 2 orders

2 × 1 + 2 + 1 3 orders

3 × +++ 4 orders

4 × 1 + 4 + 6 + 4 + 1 5 orders

5 × 1 + 5 + 10 + 10 + 5 + 1 6 orders

6 × 1 + 6 + 15 + 20 + 15 + 6 + 1 7 orders

STATISTICS: BINOMIAL EXPANSION binomial expansion of (p + q)n =

... if p = ¾ and q = ¼ , ... expansion of (p + q)3 = (¾ + ¼)(¾ + ¼ )(¾ + ¼ ) =

n

[n!/(n–k)! k!](p)n–k (q)k = k=0

n

[n!/(n–k)! k!] (¾ )n–k (¼ )k = k=0

at least two albinos…

(3!/3!0!) (¾ )3 (¼ )0 = × 27/64 0 albinos

(3!/2!1!) (¾ )2 (¼ )1 = × 9/64 1 albino

(3!/1!2!) (¾ )1 (¼ )2 = × 3/64 2 albinos

(3!/0!3!) (¾ )0 (¼ )3 = × 1/16 3 albinos > 10/64 = 5/32 64/64

Binomial Expansion• 1 、遗传分析中后代组合的概率推算Aa X aa 杂交——二个后代出现 Aa 和 aa 的组合如

下：No1. No2 概率Aa Aa ¼Aa aa ¼Aa Aa ¼aa aa ¼分布方式 1 ： 2 ： 1 ，正是（ p+q)2=1p2+2pq+1q2

[(Aa) + (aa)]2 = 1(Aa)(Aa):2(Aa)(aa):1(aa)(aa)

Aa X aa 产生三个后代• （ p+q)3 = (1/2 + ½ )3

基因型及组合 概率 Aa aa 3 0 P3=(1/2)3=1/8 2 1 3p2q=3(1/2)2(1/2)=3/8 1 2 3pq2=3(1/2)(1/2)2=3/8 0 3 q3=(1/2)3=1/8一般说，设 p 为某一基因型（或表型）出现的概率，

q 或（ 1-p) 为另一基因型（或表型）出现概率，那么有 p+q=1 。这些事件每一特定组合的概率都可以用二项式展开说明，二项式展开的系数表示在总共可能出现的组合数中的比率。

• 杨辉三角：则为二项式展开的系数 1 1 1

1 2 1 2

1 3 3 1 3

1 4 6 4 1 4

1 5 10 10 5 1 5

1 6 15 20 15 6 1 6

n!/s!(n-s)psqn-s

• 如果不考虑出现顺序，可有如下通式计算表型或基因型某一特定组合的概率：– n—— 子代数目– S—— 某一基因型（或表型）数– n-s—— 另一基因型（或表型）数– P—— 某一基因型（或表型）出现的概率– q—— 另一基因型（或表型）出现的概率

• 例：某医院一天出生 6 个婴儿，三男三女的概率是多少？– （ 6 ！ /3 ！ 3 ！）（ 1/2 ） 3 （ 1/2 ） 3=5/16

2 、孟德尔遗传的分离比率的二项式• F2 基因型比为（ 1∶2∶1 ） 2 ，即（ 1/4+2/4+1/4 ） 2 三

项式展开式的各项系数；• F2 表型比为（ 3∶1 ） 2 ，即（ 3/4+1/4 ） 2 二项式展开

式的各项系数。（ 3 ： 1 ） n （ n— 基因对数），既（ a+b)2=a2 + 2ab + b2

• 以 3 代 a, 以 1 代 b ，则有，– 当两对基因时

(3+1)2=(3)2 + 2(3)(1)+ (1)2

– 当三对基因时(3+1)3=(3)3 + 3(3)2(1) + 3(3)(1)2 + (1)3

9:9:9 3:3:3由此可见任何对基因杂种自交后代表型都可以用二项式展开计算。

N 对基因的二项式展开• 随着基因对数的增加，二项式展开将变

得复杂。当有 n 对基因时：• (a+b)n=Cn

0 an+Cn1 a n-1 b + Cn

2a n-2 b2…Cnk

a n-k bk +Cnnbn

• Cnk = n!/K!(n-k)!

同样可以用杨辉三角推得

3.5 Rediscovery of Mendel’s Work

MENDELIAN ANALYSIS: 2 LAWS

1. particulate hereditary determinants (genes)2. alleles different phenotypes; dominance3. alleles of a gene segregate into different gametes

4. gametes receive 1 allele with equal probability5. union of gametes at fertilization is random6. alleles of different genes* assort independently

into different gametes * gene pairs on different chromosomes

STUFF TO THINK ABOUT

Mendel’s choice of characters was critical did he chose 7 characters that appeared to reside on

different chromosomes by chance? what would happen if some were linked? all exhibited dominant/recessive relationships

3.6 Independent Assortment and genetic Variation

• One major consequence of independent assortment is the production by an individual of genetically dissimilar gametes. Genetic variation results because the two members of any homologous pair of chromosomes are rarely if ever genetically identical. Therefore, because independent assortment leads to the production of al1 possible chromosome combinations, extensive genetic diversity results.

• We have seen that the number of possible gametes, each with different chromosome compositions, is 2n, where n equals the haploid number. Thus, if a species has a haploid number of 4, then 24 or l6 different gamete combinations can be formed as a result of independent assortment.

• Although this number is not high, consider the human Species, where n = 23, if 223 is calcu1ated, we find that in excess of 8 x l06. or over 8 million, different types of gametes are represented. Because fertilization represents an event involving only one of approximately 8 x l06 possible gametes from each of two parents. each offspring represents only one of (8 x 106)2. or 64 x 1012, potential genetic combinations!

• No wonder that except for identical twins, each member of the human species demonstrates a distinctive appearance and individuality. This number of combinations is far greater than the number of humans who have ever lived on earth !

• Genetic variation resulting from independent assortment has been extremely important to the process of evolution in all organisms.

3.7 Probability and Genetic Event

review essential Mendel genetic ratios & rules statistics

chi-square binomial expansion Poisson distribution

pedigree analysis to some front page

3.8 Evaluating genetic data :Chi-square

• Mendel's 3:1 monohybrid and 9:3:3:1 dihybrid ratios are hypothetical predictions based on the following assumptions:

• (1) Each allele is dominant or recessive;• (2) segregation is operative; • (3) independent assortment occurs; and• (4) fertilization is random. The last three assumptions a

re influenced by chance events and are therefore subject to random fluctuation. This concept, called chance deviation, is most easily illustrated by tossing a single coin numerous limes and recording the number of heads and tails observed.

• Chi-square test.– Understand what a null hypothesis is and

how to state a testable hypothesis.– Understand what degrees of freedom are.– Understand how to calculate a chi-square

value and how to interpret what it means.– Understand how to determine whether to

accept or reject your hypothesis.– Understand how to read a chi-square table to

interpret the probability of getting a given chi-square value.

STATISTICS: CHI-SQUARE ANALYSIS

4 phenotypes ~ roughly 9:3:3:1 ratio

hypothesis what does it

mean ? test hypothesis

STATISTICS: CHI-SQUARE ANALYSIS

O : E O-E (O-E)2 (O-E)2/E YR 315 9 313 2 4 0.013 Yr 108 3 104 4 16 0.154 yR 101 3 104 -3 9 0.087 yr 32 1 35 -3 9 0.257 556 16 556 2c = 0.511

P .995 .99 .975 .95 .9 .75 .5 0.25 .1 .05 .01 .001

df

1 .000 .000 .001 .004 .016 .102 .455 1.32 2.71 3.84 6.63 10.8 2 .010 .020 .051 .103 .211 .575 1.39 2.77 4.61 5.99 9.21 13.8 3 .072 .115 .216 .352 >2c> .584 1.21 2.37 4.11 6.25 7.81 11.3 16.3 4 .207 .297 .484 .711 1.06 1.92 3.36 5.39 7.78 9.49 13.3 18.5 5 .412 .554 .831 1.15 1.61 2.68 4.35 6.61 9.24 11.1 15.1 20.5

STATISTICS: CHI-SQUARE ANALYSIS

express this as: 0.95 > P(2c = 0.511) > 0.90

the data do not deviate significantly from a 9:3:3:1 ratio we do not reject our H0 (alternatively we could reject) “seed color and seed shape fit an unlinked, 2 gene cla

ssic Mendelian model with complete dominance” the probability of deviation by chance from this model l

ies between 90 and 95% (in other words, the biological explanation is supported by data)

3.9 Human Pedigree

– Understand what a proband is in a pedigree.– Understand how to recognize dominant and rec

essive traits in pedigrees.– Understand how to predict the probability of an

affected individual appearing in a future generation.

• Work the problems at the end of the chapter

• Mendelian Genetics in Humans– Understand the major symbols used in pedigrees.

• Open squares and open circles represent males and females unaffected by whatever condition is being studied in the pedigree.

• Shaded squares and shaded circles represent males and females affected by the condition studied in pedigree.

• See figure 10.18 (p. 272) for other major symbols

PEDIGREE ANALYSIS

order of events for solving pedigrees 1. identify all individuals according to number and letter 2. identify individuals according to phenotypes and

genotypes where possible 3. for I generation, determine probability of genotypes 4. for I generation, determine probability of passing allele 5. for II generation, determine probability of inheriting

allele 6. for II generation, same as 3 7. for II generation, same as 4… etc to finish pedigree

• There are 2 major systems used in genetics to symbolize two alternative forms (alleles) of genes.

– Corn Notation System (used by Mendel)• Upper case letter is used to symbolize the

dominant allele.• Lower case form of the same letter is used to

symbolize the recessive allele.• Since there are more than 26 traits in any

given species, more than one letter may be used to designate the alleles.

• Example 1: let R = round seeds and r = wrinkled seeds. Therefore, genotypes are: RR or Rr for round and rr for wrinkled.

• Example 2: let PU = purple flowers and pu = white flowers. Therefore, genotypes are: PUPU or PUpu for purple & pupu for white.

– Drosophila Notation System (used by Morgan)• Uses same case of a letter(s) for both dominant and

recessive alleles.

• If mutant is recessive to normal, lower case letters are used for both alleles, and + symbol is added to the normal (non-mutant) allele.

• If mutant is dominant to normal, upper case letters are used for both alleles, and + symbol is added to the normal (non-mutant) allele.

• Example 1: Let b = black body and b+ = normal body color. Therefore, the genotypes are b+b+ or b+ b for normal and bb for black body. Black body is recessive to normal body color.

• Example 2: Let B = bar eyes and B+ = normal eyes. Therefore, the genotypes are BB or B B+ for bar eyes and B+B+ for normal eyes. Bar eyes are dominant to normal eyes.