Exam 4 - GENE 3200

Sexual reproduction

_____________= the mixing of genomes via meiosis and fusion of gametes

Dioecious species

_____________= either male or female reproductive structures are present in each individual

Monoecious species

_______________= both male and female reproductive structures are present in each individual

Hermaphrodites

________________= individuals that contain both male and female sex organs

Sex determination

______________= the mechanism by which the sexual phenotype of an individual is established in a given species

Sex chromosome

_______________= a chromosome involved with determining the sex of an individual

Autosome

______________= a chromosome not involved with determining the sex of an individual

Chromosomal sex

_____________= determined at fertilization and is the presence of chromosomes characteristic of each sex

Phenotypic sex

_______________= the internal and external morphology of each sex, and results from differences in gene expression

are not

Most of the sequences of X and Y chromosomes ________ (are/are not) homologous

Pseudoautosomal regions (PAR1 and PAR2)

The X and Y chromosomes have two small regions of homology, known as the ______________

- X/A ratio (# of x-chromosomes to # of sets of autosomes).

- Female= 1.0 ratio, males= 0.5 ratio

In drosophila, sex is determined by ______________.

- Pink= Two X chromosomes, & two of each autosome... So 2:2= 1 ratio which make it a female!

- Blue= One X and one Y, & two of each autosome... So 1:2= 0.5 ratio which makes it a male!

What is the sex of the following drosophila?

ZW (instead of X and Y); Females= ZW, males= ZZ

___________ sex determination is used by birds, butterflies, some reptiles, some amphibians, and some fish

SRY on Y chromosome

DMRT1 on Z chromosome

Sex determining gene for mammals is ___________

Reciprocal crosses

___________________ can be used to infer sex-linked inheritance

A= Xw X+

B= X+ Y

C= X+ X+ or Xw X+

D= X+ Y

E= Xw Y

What would the F1 and F2 look like?

A= Xw X+

B= X+ Y

C= Xw X+

D= Xw Xw

E= X+ Y

F= Xw Y

What would the F1 and F2 look like?

A) Wing phenotype: F1 progeny of both sexes have full wings and there is 3:1 segregation of full vs. vestigial in both sexes of F2 – wing trait is autosomal, full is dominant to vestigial

Body color: male and female F1 have different phenotypes and male F1 have the same phenotype as female parent and there is 1:1 segregation of yellow vs. gray in both sexes of F2 – body color trait is X-linked, gray is dominant to yellow

B) Look at next slide

Answer the following questions

ok

Answer to part B from the previous slide

X-linked recessive

Hemophilia is caused by an _________________ mutation in humans

sons

In X-linked recessive mutations, when mothers have the disorder, all _________ will have the disorder

A) Because Frank is XaY, Sally and Frank's mother must have been XAXa This means Sally has 1/2 chance of being XAXa If Sally is XAXa, there is 1/2 chance that her son will be XaY Total probability is 1/2 x 1/2 = 1/4

B) Since the son has RP, Sally must be XAXa . So there is a 1/2 chance the daughter will be a carrier

Answer the following questions

This is NOT X-linked inheritance. In the left cross, F1 females have the same phenotype as theirfathers (crisscross inheritance) and in the F2, cameo and blue is present in both sexes (1:1:1:1). In the right cross, F1 males and females are all blue, and in the F2, all males are blue and females are blue or cameo in 1:1 ratio. These results are consistent with Z-linked inheritance (female birds are ZW and male birds are ZZ), with blue dominant to cameo.

Is this an X-linked or Z-linked inheritance?

A)

- 25% barred males, 25% black males

- 25% barred females, 25% black females

B)

- 50% barred males, 50% black females

Dosage compensation

__________________= accounts for different copy numbers of X-linked genes in XX vs XY individuals

X-inactivation

_______________= in placental mammals, one X-chromosome in all somatic cells of XX individuals is silenced at the transcriptional level, i.e. with a few exceptions, none of the genes on the entire chromosome are transcribed

monoallelic expression

X-inactivation results in _________________ of X-linked genes (biallelic expression occurs for most autosomal genes)

Barr body

The inactive X chromosome is called the _______________

no. X chromosomes - 1

No. Barr bodies= _________________

mosaic

Because of X-inactivation, somatic cells of adult female mammals are ____________ for expression of alleles of X-linked genes

aX, iX

You can use ________ for the active X, and ________ for the inactive X

inactivated

In females that are heterozygous for X-linked mutation (X+ Xm): One-half of somatic cells have the wild-type allele _____________ (iX+aXm) One-half of somatic cells have the mutant allele _______________ (aX+iXm)

heterozygous

XX cats are ______________ for alleles of an X-linked gene that cause black or orange hair color

Chi-square goodness-of-fit test: (compares how well the observed results fit the expected results)

In an experimental cross, how do you know if the observed values are consistent with expected values?

Null hypothesis

______________= any deviation between observed and expected values is the result of chance

Look at the picture

What is the equation for chi squared?

n - 1 (n = the number of phenotype classes; this is the number of independent variables)

df or degrees of freedom= ____________

P-value

___________ = probability that the deviation from expected numbers had occurred by chance

- 0.05

- If P< 0.05 reject the null hypothesis.

- If P> 0.05 accept the null hypothesis

In the table provided on the exam (pictured) which confidence limit should you use?

A) Total = 105 + 45, appears to be a 3:1 segregation (monohybrid intercross), white is recessive to purple

B) Total = 105 + 45 = 150

150 x 3/4 = 112.5

150 x 1/4 = 37.5

C) 2 phenotype classes in F2, df = 2 - 1 = 1

Answer the following questions

- (105-112)^2/112= 0.44

- (45-37)^2/37= 1.73

- df=1

- X^2= 0.44+ 1.73= 2.2

- P value > 0.05 (between 0.5 and 0.1), so the hypothesis cannot be rejected.

Now use the chi squared equation for the previous question, as a reminder the numbers are included in the image.

- If long tail is recessive to short tail, then expect 3:1 segregation in short tail x short tail crosses.

- Cross 3 produced a total of 80 pups, so if 3:1 segregation, expect 60 pups with short tail and 20 pups with long tail.

- Chi-square= (52-60)^2/60 + (28-20)^2/20= 4.27

- P value < 0.05 (between 0.5 and 0.025), so the hypothesis can be rejected.

Do a chi-square to test whether there is 3:1 segregation of tail phenotypes in cross 3

Complete dominance

____________= Heterozygous phenotype is the same as the phenotype for one of the homozygotes

Incomplete dominance

_________________= Heterozygous phenotype is intermediate between the two homozygous phenotype

Codominance

________________= Heterozygous phenotype includes both of the homozygous phenotypes

neither

With incomplete dominance and codominance ____________ allele is dominant or recessive to the other

another allele of the same gene

Dominance is always relative to _____________

functional

More than one pattern of dominance may exist between different alleles of a gene... This can reveal _________________ differences between gene products encoded by different alleles

incomplete (heterozygous phenotype isn't always half-way between the two homozygous phenotypes)

With ______________ dominance, the phenotype of the heterozygous condition is between that of the two homozygous conditions

F1= all incomplete dominance (Aa)

F2= 1:2:1; 50% incomplete dominance(AA, Aa, aa)

With incomplete dominance, the mix between a homozygous recessive allele and homozygous dominant allele would give us what ratio for F1? What about F2?

codominance

With ________________, the heterozygous phenotype includes both of the homozygous phenotypes

I^A, I^B, & i

What are the 3 alleles in ABO blood types?

I^A and I^B

Four blood types of the ABO system ______ and _______ alleles are codominant but each is completely dominant to the i allele

- N-acetyl-galactosamine

- Galactose

- I^A allele... glycosyltransferase adds ______________

- I^B allele, glycosyltransferase adds ____________

L^M, and L^N

What are the 2 alleles of the MN system? they are codominant!

ok

Know these antigens??

Penetrance

____________= percentage of a population with a particular genotype that shows the expected phenotype (All or none... Has it or doesn't)

Expressivity

_______________= degree or intensity with which a particular genotype is expressed in a phenotype

dominant

Both incomplete penetrance and variable expressivity occur with dominant and recessive phenotypes, but are easiest to observe with ____________ phenotypes

Allelic series

_____________= Multiple alleles of a gene can exist in natural populations, in domesticated animals, and in lab strains

Null mutation

___________= Function of gene product is completely abolished

Hypomorphic mutation

____________= Function of gene product is partially abolished

Hypermorphic mutation

______________ = Function of gene product is increased or expressionof gene occurs in ectopic sites in the body

Mendel's 1st law

____________ law applies to crosses involving multiple alleles

A) F2:

1/4 MRM (restricted)

1/4 MRmd (restricted)

1/4 M md (mallard)

1/4 mdmd (dusky)...

So we would have 1/4 mallard, 1/2 restricted, and 1/4 dusky.

B) Possible F1 parents: M^R M and MM, or M^R m^d and MM, or M^R M and M m^d

Answer the following questions

A- C C, C c^ch, C c^h, & C c

B- c^ch c^ch, c^ch c^h, & c^ch c

C- c^h c^h, & c^h c

D- cc

What are the different allele combinations that would give us the different rabbit colors?

C> c^b, c^s > c (c^s and c^b alleles are either codominant or have incomplete dominance to each other)

How do the alleles in this relationship work?

A) Aa & a^t a

B) 1/4 x 20= 5 tonkinese

1/4 x 20= 5 siamese

1/4 x 20= 5 burmese

1/4 x 20= 5 albino

Answer the following questions

Answer on next slide!

Answer the following

ok

Answer to previous question

Pleiotropy

____________= one gene affects multiple characteristics, that may or may not be obviously related to each other

recessive inheritance

Almost all lethal mutations have __________________ (only the homozygotes are inviable; viability is dominant to lethality)

2:1

With lethal alleles, Aa x Aa gives a ____________ segregation.

E.g.

- Cross: yellow (AY A) x yellow (A^Y A) gives 48 yellow (A^Y A), 27 agouti (AA)

2:1, after

- ________ ratios of viable yellow and agouti mice (when you cross 2 yellow mice A^YA x A^YA)

- Lethal alleles can be truly dominant, but lethality occurs ___________ the heterozygotes have produced offspring

Cross is Bb Cp/+ x Bb Cp/+

A) 1/2 Bb x 2/3 Cp/+ gives 2/6 blue with short legs

B) 1/4 bb x 1/3 +/+ gives 1/12 black with normal legs

Answer the following questions

Gene interaction in pathways, epistasis

What are 2 different ways gene interactions can occur?

Epistasis

______________= an allele of one gene modifies or prevents the phenotype caused by alleles of a different gene

two, four

Interactions between ______ genes for pigment synthesis determines _______ fruit colors in peppers. This 9:3:3:1 segregation is of four phenotypes for the same character (color), rather than four combinations of phenotypes for different characters

Dominance is the relationship between two alleles of the same gene. Epistasis is the interaction between alleles of different genes.

Why is epistasis is not the same as dominance?

epistatic, hypostatic

In epistasis an allele of one gene overrides (masks) the phenotype of another gene; The overriding gene/allele is ____________ to the masked gene/allele... And the masked gene/allele is ___________ to the overriding gene/allele

9:3:4 & 9:7

What are the ratios for these epistatic genotypes?

- aa, B;

- 9/16= A_B_,

- 3/16= A_bb,

- 3/16= aaB_ + 1/16= aabb so 4/16 have aa phenotype

Recessive epistasis: 9:3:4 phenotype ratio in F2 progeny of dihybrid intercrosses: ________ is epistatic to all genotypes for the ______ gene. List where the phenotype ratios come from

- Blue= W_M_

- Pink= W_mm

- colorless= wwM_ & wwmm

What are the phenotypes and corresponding genotypes for each F2 progeny?

A= BBYY

B= bbyy

C= BbYy

D= B_Y_

E= bbY_

F= B_yy & bbyy

Answer the following question

B= BbYy, bbYy

C=

Start with:

- bb x Bb = 1/2 Bb and 1/2 bb

- Yy x Yy = 3/4 Y_ and 1/4 yy.

Next find all phenotype proportions:

- 1/2 Bb x 3/4 Y_ = 3/8 Bb Y_ (lavender)

- 1/2 bb x 3/4 Y_ = 3/8 bb Y_ (blue)

- 1/2 Bb x 1/4 yy = 1/8 Bb yy (yellow)

- 1/2 bb x 1/4 yy= 1/8 bb yy (yellow)

- 1/8 +1/8= 2/8 (yellow)

Answer the following question

A= Testcross to bbee.

B= Testcross to yellow.

C= Testcross to bbEE.

Answer the following question

O

Regardless of genotype at the ABO locus, an hh individual is blood type _____ (i.e. hh is epistatic to IA and IB)

9:7 (9/16 A__B__, 3/16 A__bb, 3/16 aaB__, 1/16 aabb) Homozygous recessive for either gene results in the same phenotype as homozygous for both genes!

Duplicate recessive epistasis: ________ phenotype ratio in F2 progeny of dihybrid intercrosses

Complementary gene interaction

Duplicate recessive epistasis is also called what?

A)

- 9/16 not deaf (A_ B_) = 45,

- 7/16 deaf (3/16 aa B_, 3/16 A_ bb, 1/16 aa bb) = 35.

B)

- 3/8 not deaf (1/8 Aa BB and 2/8 Aa Bb)

- 5/8 deaf (1/8 Aa bb deaf, 1/8 aa BB, 2/8 aa Bb, 1/8 aa bb)

Answer the following question

The phenotypes of the mutant parents are caused by mutations in two different genes; the two mutations are NOT allelic; applied only with recessive, not dominant, phenotypes; can only be seen in experimental crosses as well as in human pedigrees

Genetic complementation

The phenotypes in the two mutant parents are caused by different alleles of the same gene; the two mutations ARE allelic

No genetic complementation

Have the same phenotype but are caused by mutations in different genes

Heterogeneous traits

sex

Some phenotypes can be influenced by the _________ of the organism

Sex-limited traits

__________= Inheritance is autosomal but the trait appears in only one sex.

Sex-influenced traits

____________= Inheritance is autosomal but is much more frequent in one sex.

males

Read carefully and fill in the blank for sex LIMITED traits

A= have beards

B= Have beards

C= Beardless

D= Have beards

E= Beardless

F= Beardless

Fill in the chart for the sex-influenced trait

Cross is BbBb male X B+B+ female. F1s are BbB+ bearded males X BbB+ beardless females.

A) Males with beards are BbBb and BbB+... 1/4 BbBb x 1/2 males = 1/8, 1/2 BbB+ x 1/2 males = 1/4. SO 1/8+ 1/4= 3/8, 3/8 x 80= 30.

B) Females with beards are BbBb... 1/4 BbBb x 1/2 females = 1/8. SO 1/8 x 80 = 10.

Answer the following