L3Biol261F2014Dihybrid and Meiosis

8/9/2019 L3Biol261F2014Dihybrid and Meiosis

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Lecture 3


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In a species of birds with a ZW sex determination system, bluebeak is a sex-linked recessive trait and red beaks are dominant.If a blue beaked male is mated to a red beaked female, which of thefollowing would you would expect to find among theiroffspring ?

a) All progeny are red beaked. b) All males are red beaked and all females are blue beaked. c) All males are red beaked, half the females are blue beaked.d) All females are red beaked, "of the males are red beaked,

one half of the males are blue beaked.e) Half the females are red beaked, half the males are red

beaked


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Lecture 3:(1)

The chromosomal theory of inheritance provides amechanism that explains Mendels Laws and theirapplication in pedigree analysis, for example.

(2) When sample sizes are small or breeding designsareinconvenient or impossible to use, the inheritance ofclassical polymorphisms are most easily deduced in a familypedigree, when breeding is monitored.

(3)

Use the expected outcome of single gene crosses to predictand interpret (independent) multiple gene crosses.

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AA x aa= 1 genotype possible Aa, 1 phenotype (pointed)AA x Aa= 2 genotypes : 0.5AA, 0.5 Aa 1 phenotype (pointed)

Aa x aa= 2 genotypes : 0.5Aa, 0.5 aa2 phenotypes (pointed, round 0.5 each)

Aa x Aa= 3 genotypes : 0.5Aa, 0.25 aa, 0.25AA2 phenotypes: 0.75 pointed, 0.25round

aa x aa = 1 genotype (aa), 1 phenotype (round)

AA x AA = 1 genotype (AA), 1 phenotype (pointed)

Memorize, Understand and Apply These Single gene, 2 allele CrossRatios

A_-dominant pointed leaf, aa round leaf

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Pedigrees:

Using Mendelian genetics to deduce thepattern of inheritance in a pedigree.

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Affected vs. non affected-a classical

Mendelian character (categorical).

proband or propositus The firstindividual affected in the pedigree .

Roman numerals: I, II, etc = generation

Arabic numerals: 1,2,3,ID children

males (square)and females (round).

The primary information is:

(1)affected sex, (2)affected proportionof offspring, and (3) generation.

Expected Mendelian proportions are a

guide but small sample sizes means

hi h variation.

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Figure 2-28

Pedigree symbols

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Note: it is assumed that

these married

individuals are not

closely related.

In the absence ofother information,

assume, the person

outside the pedigree

does not carry rare

genes or theprobability of being a

carrier is no higher

than the general

population.6

Inherited diseases are typically rarein the general population,but not within families hence the usefulness of a pedigree.


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Exact Mendelian (expected) ratios are generally not observed, any ratio

is possible and all are seen.

But, you can use the expected ratios to interpret frequencies in a

pedigree.

If you know the relationship between individuals, you can use expected

ratios to estimate probabilities of one outcome or another.

(1?) (.5)A (.5)a

(.5)A .25AA .25Aa

(.5)a .25aA .25aa

Genotypes ?

If unknown,it is a

question of

probability

If genotypes are known, it is not a question of probability.

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malefemale

parents

siblings

Generation I

Generation II

1 2 3

Using Mendelian ratios to interpret genotypes and predict

human inheritance - pedigrees

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parents

grandparents

children

Dominant or Recessive, Autosomal or Sex -Linked ?

9

II

I

III

1 2 3 4

1 2

Assume full dominance-recessive trait, rare condition


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(1) An autosomal Mendelian recesssive pattern. The key observations are: (1) the disease

appears in the progeny of unaffected parents, or the effect may skip generations. (2) the

affected progeny are both males and femalesand they are affected in equal proportions

(2) Autosomal Dominantdisorders are implicated when (1) the (mutant) phenotype appears

in every generation; (2) affected fathers and mothers transmit the character to both sons and

daughters - in equal proportions.

(3) Sex - related characters: X linked recessive( 1) more males than females show the character

because they arehemizygous - only 1 X chromosome (2) The male offspring of an affected

male and his descendents are not affected. (3) Half the sons of carrier daughters are affected,or the effect skips generations. X - linked color blindness, hemophilia and Duchenne

Muscular Dystrophy are well known examples.

(4) X-linked Dominantdisorders: they have the following characteristics: (1) affected males

pass the character to their daughters but not their sons. (2) affected heterozygous females

married to unaffected males pass the condition to half their sons and daughters. (3) thus

more females than males will show the trait.

(5) Y - linked traitsonly show up in males and are expressed every generation. Many of

these traits are related to male sex expression. Otherwise there are not many genes on the Y

chromosome and they are not implicated in well known diseases.

(6) Mitochondrial traits are passed from mother to daughter and son, but an affected

son never transmits the trait to any of his offspring.

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Figure 3-19

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Mitochondriainherited maternally (almost always)Chloroplasts Angiosperm generalization inherited maternally, but

there are lots of exceptions (paternal, bi parental). Conifers -paternal


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Figure 2-31 part 1

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What is the pattern of inheritance in the pedigree above ?

a)

Autosomal dominantb) Autosomal recessivec) X linked dominant d) X linked recessivee) None of the above


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16


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Fig. 4.2418


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Genes assort independently, therefore you can calculate expectedprobabilities for them independently.AA x aa= 1 genotype possible Aa, 1 phenotype (A_)

AA x Aa= 2 genotypes : 0.5AA, 0.5 Aa 1 phenotype (A_)Aa x aa= 2 genotypes : 0.5Aa, 0.5 aa 2 phenotypes (Aa, aa 0.5)

each)Aa x Aa= 3 genotypes : 0.5Aa, 0.25 aa, 0.25AA

2 phenotypes: 0.75A_, 0.25aa

aa x aa = 1 genotype, 1 phenotypeAA x AA = 1 genotype, 1 phenotype

Use these expectations to calculate more complicated 2 and 3locus, multiple allele expectedphenotypeandgenotyperatios.

e.g. In the cross Aa Bb x Aa Bb what is the expected frequencyof the 2 gene dominantphenotype? (A_ = 0.75) * (B_ = 0.75) =0.5625= 9/16

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Mendels second law-independent assortmentof different genes on different chromosomes

.Allele pairs oftwo(or more)differentgenes(on differentchromosomes)assort (into gametes)independently (of each other).

Mendels second law states that more than onegene pairs that areon differentchromosomes assort independently at meiosis.

Segregationof the members of any pair of alleles for one geneis independentof thesegregationof alleles for other genes in the formation of of reproductive cells.

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Nomenclature: R/R . y/y

separate different genes, using a space R y or a dotR.y

Separate allelesusing / which is shorthand for //

// is an abstraction of paired homologous chromosomes

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Dihybrid breeding design


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Breeding Design - Dihybrid Cross

2 inbred strains R/R.yy r/r.YY

Each strain shows 2 different characters (shape, color),implying at least 2 genes.

Each character has 2 character states (dominant orrecessive), implying 2 alleles.

F1expected ?


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Figure 3-3 step 3

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Dihybrid cross

Check ?

Inbreed F1


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(2) Rr YY x rr yy?

(3) RR YY x rr yy?

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Breeding Design - Test Cross

(1) Frequency of: R, Y alleles ?


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Mendels breeding designthat produced a 9 : 3 : 3 : 1phenotype ratio

315/556= 0.567, ~ 9/16 (0.562)

108/556=0.194, ~3/16(0.188)

32/556=0.058, ~1/16(0.0625)

101/556=0.182, ~3/16(0.188)

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Inbreed


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Punnett square illustrating the 9 : 3 : 3 : 1phenotype ratio and underlying genotypes

(2 genes*2 alleles) gamete combinations

that heterozygotes produce.

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Breeding Design - Dihybrid Cross


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The expected phenotype ratioin a classical dihybrid crossis9:3:3:1.

e.g Parentals AA bb x aa BB):

F2(9A_ B_; 3A_ bb; 3aa B_; 1 aabb)

How many different genotypes are expected, and in what ratios?

3 methods : (other than memorizing)Punnett square- reliable, visual but awkward and time-consuming.

Line Diagram reliable, direct, but time consuming.

Use the product rule- reliable and quick

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Arrow, Line or Tree Diagrams ofa Classical 2 gene, 2 allele dihybrid cross

phenotypes

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0.25RR*0.25YY

0.25RR*0.5Yy

0.5Rr*0.5Yy


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Product Rule: if A andB are independent, the probability they co-occur is the product of their independent probabilities

or p(A) x p(B) (key word and

outcome Aand B)

For example: A blue -eyed male and a brown-eyed female whosefather had blue eyes, plan to have children.What is theprobability of having a blue-eyed daughter as a first child ?

Xb

XB

Xb

XBXb

XbXb

Y

YXB

YXb

Probability of having adaughter = 0.5 (column 1)Probability of having blue eyes

= 0.5 (row2)

Probability of a child that isblue-eyed and a girl = 0.25Probability of having two blue-eyed daughters in sequence =

0.25 * 0.25

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0.5 0.5

0.5

0.5


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In the cross AaBb X AaBb, what is the probabilitythat theprogenywill show the recessive phenotypefor at least onegene?

Possibilities: A_ bb = 3/4 (A_) * 1/4 (bb) = 3/16, or

aaB_ = 1/4 (aa) * 3/4 (B_) = 3/16, oraabb = 1/4 (aa) * 1/4 (bb) = 1/16

Probability of any 3/16 +3/16 +1/16 = 7/16

What is the probability of a boy or a girl ?

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Addition rule: the probability of occurrence ofany two mutually exclusive events is the sum of the

probabilities of the individual events. Key wordor


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Find the probability that each of 3 children in a familywill be of the same sex.

a) 1/2b) 1/4c) 1/8d) 1/16

e) 3/8

What is the probability of having two girls and aboy ?(GGB or GBG or BGG)

a) 1/2b) 1/4c) 1/8d) 1/16e) 3/8

In genetics you often use both multiplicative and additive rules in oneproblem


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MEIOSIS


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Prophase -Metaphase chromosome

Interphasechromosomes


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To represent the cellular behavior of chromosomes at the cytological

scalein meiosis we use the following model of chromosome structure,

a double stranded DNA molecule with associated histone proteins.

The centromere is part of a DNA molecule, it is hundreds of kilobases

long, with no functional genes. Count chromosomes by counting

centromeres.


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Figure 2-15

Key stages of meiosis and mitosis

Homologouschromosomes

Sister chromatids

Sister Chromatids


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Stages of Prophase I

Prophase I

Fig. 2.13a

(1)Chromosomes condenseand coil - Leptotene

Nuclear membrane begins tobreaks down

(2) Zygotenehomologouschromosomes pair:

Meiosis differs from mitosis inearly prophase 1, whenhomologous chromosomes arelinked in a protein(synaptonemal) complex


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Crossover, recombinationand repair has finished

in (4)Diplotene, thechromosomes are beginningto repel and the chiasmata(points of recombination)

are beginning to migrate tothe telomeres.

In (5) Diakinesis,chiasmata have migrated to

the chromosome ends(telomeres)


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Metaphase 1: (a) recombined chromosomes(b)random segregationof maternal and paternal

chromosomes and alleles


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Box 2-2

Stages of Meiosis

Meiosis 1, (firstreduction division)

Meiosis 2, (second reductiondivision)

diakinesis (centromeres repel andchromosomes separate).

Homologous chromosomes

segregate

Sisterchroma

tids

segregate


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Metaphase and Anaphase 2

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L3Biol261F2014Dihybrid and Meiosis