Chromosomal Inheritance

Question 1

Chromosome Theory of Inheritance

Answer 1

Mendelian genes have specific loci along chromosomes and it is the chromosomes that undergo segregation and independent assortment

Question 2

What does chromosomal theory of inheritance attempt to explain?

Answer 2

Mendel’s Principles of Inheritance

Question 3

What did TH Morgan contribute?

Answer 3

Provided the first solid evidence associating a specific gene with a specific chromosome

–> Also completed work on the founding of recombination and sex-linked inheritance

Question 4

Morgan’s Experimental Model

Answer 4

Drosophila Melanogaster (fruit flies)

Question 5

Benefits of the Fruit Fly Model

Answer 5

1) Prolific breeders (one breeding could produce hundreds of progeny)

2) Cheap to grow
3) Fast generation time
4) Offered easily distinguishable phenotypes
5) Have only 4 pairs of chromosomes (homologs)

–> After Morgan’s research we know 3 are autosomal pairs and then they have a sex chromosome pair

Question 6

Wild Type vs Mutant Notation

Answer 6

MUT = letter of mutated variant
WT= (letter of mutated variant)+

Ex: w+ = wild type
w = mutant

Question 7

What was Morgan first looking for?

Answer 7

A variant/mutation in the drosophila

Question 8

What was Morgan’s first variant?

Answer 8

A white-eyed male –> (wild type is red eyes)\

–> Found after 2 years of breeding

Question 9

Wild Type

Answer 9

Phenotype for a character most commonly observed in natural populations

Question 10

Mutant Phenotype

Answer 10

Alternatives to the wild type

Question 11

Morgan’s First Cross (to make F1)

Answer 11

P Gen = True-breed Red-Eye (WT) Female + Mutant White-Eye Male

–> Produced all red-eye F1 generation (heterozygotes)

Question 12

Morgan’s Second Cross (to make F2)

Answer 12

Bred F1 offspring (heterozygotes) with each other

= 3:1 ratio of red to white eyed flies (just like Mendel’s)

BUT, they observed the white-eyed flies were only ever MALE

Question 13

Morgan’s Second Cross (to make F2) Conclusion (of the white fly males)

Answer 13

Eye color in fruit flies is linked to sex

Question 14

Female Sex Chromosomes + Gametes

Answer 14

XX

= All gametes have X

Question 15

Male Sex Chromosomes

Answer 15

XY

= 1/2 gametes have X and 1/2 have Y

Question 16

Males are _________ in sex chromosomes

Answer 16

hemizygous

–> Whatever allele is present is the one that is expressed (no second allele to mask/affect phenotype)

Question 17

Morgan’s additional crosses to exam six linkage
+
What was his conclusion?

Answer 17

1) Homo. Dom. Female + Recessive Male
–> Produced all heterozygotes (dom phenotype)

2) Homo. Rec. Female + Dominant Male
–> Produced white (rec) males

Conclusion: Allele producing eye color must reside on X chromosome which is derived from mother

Question 18

Main finding of Morgan:

Answer 18

Found that genes reside on chromosomes and that each gene resides on a SPECIFIC chromosome

Question 19

Linkage

Answer 19

Genes on the same chromosome tend to be inherited together

–> Because the chromosome is passed along as a unit

Question 20

What does linkage deviate from?

Answer 20

Mendel’s Law of Independent Assortment

Question 21

In linked genes all offspring are predicted to…

Answer 21

have a parental phenotype!!

1:1 Ratio (or 1:1:0:0)

Question 22

In unlinked genes, offspring are predicted to…

Answer 22

have either a parental or non-parental phenotype

(1:1:1:1)

Question 23

Recombinant Type

Answer 23

A new combination compared to parents

Non-Parental phenotype

Question 24

What type of cross did Morgan use to test linkage vs non-linkage phenomena?

Answer 24

Test crosses, examining two characters at a time

Question 25

Morgan’s Test Crosses

Answer 25

P Gen = Double Mutant + Double WT

F1 Gen = Double heterozygote

Crossed double heterozygote (F1) female with a double mutant male (test cross)

Offspring =
–> 2 of parental type
–> 2 of non-parental type

Found the 1:1:1:1 ratio

Question 26

Linked Genes vs Unlinked Genes Offspring Ratios

Answer 26

Linked = 1:1:0:0 (only parental)

Unlinked = 1:1:1:1 (parental + recombinant)

Question 27

Recombination

Answer 27

The production of offspring with combinations of traits that differ from those found in the parents

Question 28

Recombination is a result of

Answer 28

crossing over of non-sister chromatids in which homologs exchange DNA segments

Question 29

Crossing over in meiosis =

Answer 29

production of recombinant gametes = recombinant offspring

Question 30

Recombinant Frequency

Answer 30

(# recombinant offspring) / (total offspring) = %

Question 31

Recombinant Frequency < 50%

Answer 31

Genes assort independently = Genes linked

Question 32

Recombinant frequency = 50%

Answer 32

Genes are unlinked

(50% is the max)

Question 33

Complete linkage

Answer 33

Recombinant frequency = 0%

All offspring are identical to one parent in phenotype for the specific genes (1:1:0:0)

Question 34

Unlinked Genes (recomb. freq)

Answer 34

Recombinant Frequency = 50%

Equal # of parental vs non parental phenotypes among offspring (1:1:1:1)

Question 35

When genes on the same chromosome are far apart…

Answer 35

crossover between them is almost certain –> Recombinant frequency reaches 50% and genes appear to be unlinked

Question 36

The more distance between genes…

Answer 36

The more points there are between them where crossing over can occur (= more recombination)

Question 37

Recombinant frequency can reflect…

Answer 37

gene distances

Question 38

The greater the recombinant frequency,

Answer 38

the greater the gene distance

Question 39

How was recombinant frequency used?

Answer 39

Used to map the genome in terms of relative distances between genes