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BLGY1232 > Mendelian Genetics 2 > Flashcards

Mendelian Genetics 2 Flashcards

1
Q

What does a chi squared test do?

A

tells us whether the data are consistent with expectation

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2
Q

Chi Squared Formula

A

Χ² = Σ (O-E)²/E

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3
Q

Linkage

Definition

A

genes that are carried on the same chromosome are said to be linked they tend to segregate together, ie. they do not assort independently of each other

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4
Q

Linkage

Example

A
  • 3 loci: A, B and C each with to alleles Aa, Bb and Cc respectively
  • if not linked meiosis in the will generate gametes; ABC, abc, Abc, aBC, aBc, AbC, abC, ABcc
  • but if they are linked. e.g if A and C are on the same chromosome, then the heterozygote genotype could be either AC/ac B/b, OR Ac/aC B/b
  • for Mendelian segregation, we treat the linked alleles as if they are a single locus
    i. e. if the genotype is Ac/aC B/b, then A and c will always segregate together as will a and C
  • BUT, we also have to take into account the effects of recombination
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5
Q

Coupling

Definition

A

two alleles are said to be in coupling if they are on the same chromosome

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6
Q

Repulsion

Definition

A

two alleles are said to be in repulsion if they are on different chromosomes

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7
Q

Haplotype

Definition

A

the genotype of a chromosome

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8
Q

Recombination Frequency

Definition

A
  • recombination is more likely to occur between two linked loci if they are further apart on the chromosome than if they are closer together
  • recombination frequencies of more than 50% cannot occur as this is independent assortment
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9
Q

Linkage Mapping

A

using the frequency with which recombinant offspring occur to determine the relative positions of genes on a chromosome

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10
Q

Recombination Frequency

Equation

A

number of recombinants / total number of offspring
-multiply this by 100 to find a percentage the ercentage is also equal to the distance between the genes on the chromosome in centiMorgans

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11
Q

Recombination With 3 Point Crosses

Rules

A
  1. Identify the parental combinations of alleles (most frequent classes)
  2. Identify the middle locus (which allele has ‘changed partners’ in the least frequent classes)
  3. Calculate the recombination frequnce between each pair of loci
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12
Q

Three Point Test Cross

Random Assortment

A

if unlinked, the genes will independently segregate producing the 8 different phenotypes in a 1:1:1:1:1:1:1:1 ratio

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13
Q

Three Point Test Cross

Linkage and Recombination

A

if linked, 8 different phenotypes will be produced, but the ratio will depend on the distance between the genes on the chromosome

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14
Q

Three Point Test Cross

Linkage and No Recombination

A

if the genes are linked but no recombination occurs, then 2 different phenotypes will be produced from two different phenotypes giving a 1:1 ratio

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15
Q

Three Point Test Cross

A

A+B+C+/abc x abc/abc

homozygous recessive individual crossed with an individual heterozygous for all three genes

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16
Q

Three Point Test Cross

How to determine the distance and arrangement of the genes?

A

1) the genotype for the phenotype that is the least frequent will be the one where two recombination events have taken place, this allows you to identify the middle gene e.g. if least common is A+bC+/aBc, then B is the middle gene
2) from here calculate the distance between the middle gene and the two outside genes separately
3) add together the frequency off all the times that A+ and B+ have segregated together, divide by total number of progeny and x100 to find the distance in cM
4) do the same for B+and C+, add together the frequency of the phenotypes where B+and C+ have segregated together and divide by the total number of progeny then multiply by 100 to find the distance in cM

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17
Q

Recombination Interference

A

Cross over at one point on a chromosome reduces the chance of crossing over occurring at nearby points on the same chromosome, to lower than would be mathematically predicted

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18
Q

Three Point Test Cross

Most Frequent Progeny

A

parental combinations

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19
Q

Three Point Test Cross

Least Frequent Progeny

A

double recombinations

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20
Q

1cM

A

1cM = one centimorgan
relative measure of distance so depends on the genome and rate of recombination in that particular organism
1cM in one genome does not equate to 1cM in another

21
Q

Genotypic Sex Determination

A

based on organism having specific sex chromosomes that segregate from one another during meiosis

22
Q

Sex Determination

Mammals

A 
XX = female
XY = male
23
Q

Sex Determination

Butterflies

A 
ZZ = male
ZW = female
24
Q

Sex Determination

Amphibians

A 
ZZ = male
ZW = female
25
Q

Sex Determination

Insects

A 
XX = female
XY = male
26
Q

Sex Determination

Some Plants

A 
XX = female
XY = male
27
Q

Sex Determination

Birds

A 
ZZ = male
ZW = female
28
Q

Sex Determination

Nematodes

A 
XX = hermaphrodite
X0 = male
29
Q

Hermaphrodite

Definition

A

has the reproductive organs of both the male and the female sex

30
Q

XY Sex Determination

A

50% of offspring will be male and 50% will be female

-each cell in the body has a 50:50 chance of displaying the maternal or paternal sex chromosome

31
Q

Examples of Sex Linked Traits

A
  • tortoiseshell cats

- barred plumage in chickens

32
Q

Epistasis

Definition

A

interaction between two different genes so that an allele of one of them (the epistatic gene) interacts with or inhibits the phenotypic expression of the other (the hypostatic gene)

33
Q

Sex Determination

Dominant Y Systems

A
  • the presence of the Y chromosome is what determines sex

e. g. even XXY individuals are male (Klinefelter’s syndrome)

34
Q

X Inactivation

A

-because the heterogametic sex has only one copy of the sex chromosomes whereas the homogametic sex has two copies, there has to be a mechanism to prevent overexpression of genes on the diploid chromosome

35
Q

X Inactivation in Mammals

A
  • known as dosage compensation
  • achieved by inactivation of one or other of the X-chromosomes in females
  • occurs randomly in different cells during development of the embryo
  • its effects can be seen when the X chromosomes are heterozygous for traits like coat colour
36
Q

Tortoiseshell Cats

A
  • colour gene O located on the X chromosome
  • two alleles, a recessive allele, ob which results in normal production of melanin (black coat pigment) and a dominant allele, O which results in the formation of an altered form of melanin which is orange
  • if a female cat is X0/Xob, in some cells the XO will be inactivated, cells will be black and in others, Xob ill be inactivated so the cells will be ornage
  • the sizes of the patches depend on how early in development that the X-chromosome was inactivated
  • male cats only have one copy of the X chromosome so can only be either all black or all orange
37
Q

How to determine how many genes control a phenotype?

A

-collect a population of mutants showing the SAME mutant trait
-interbreed all combinations of the mutant
-

38
Q

Heterogametic Sex

A

produces two types of gametes, each carrying a different sex chromosome

39
Q

Homogametic Sex

A

produces one type of gamete all carrying the same sex chromosome

40
Q

Lethal Alleles

Recessive Lethal

A

absence of gene product results in death of homozygotes

e.g. yellow mice

41
Q

Lethal Alleles

Dominant Lethal

A

both homozygotes and heterozygotes carrying the allele will die
e.g. Huntingdons

42
Q

Recessive Epistasis

A

Aabb and AAbb mice have the same phenotype as aabb, epistasis of bb convers A-

43
Q

Duplicate Recessive Epistasis

A

the two genes code for two proteins e.g. enzymes that take part in successive stages of a pathway i.e. both are required for the final product
-characterised by a 9:7 ratio in the progeny

44
Q

Different Mutations in the Same Gene vs Mutations in Different Genes

A
  • interbreed the mutants
  • if any offspring are wildtype, then the mutations are in different genes
  • this is the complementation test
  • the wildtype allele in one mutant compliments the muntant allele in the other and vice versa at the second locus
45
Q

Different Mutations in the Same Gene vs Mutations in Different Genes

A
  • interbreed the mutants
  • if any offspring are wildtype, then the mutations are in different genes
  • this is the complementation test
  • the wildtype allele in one mutant compliments the muntant allele in the other and vice versa at the second locus
46
Q

Complementation Group

A

all carry mutations in the same gene

47
Q

Codominance / Incomplete Dominance

A

a heterozygous individual shows a third phenotype, not the ‘dominant’ or ‘recessive’

48
Q

What is the difference between codominance and incomplete dominance?

A
  • incomplete dominance gives a blending of the ‘dominant’ and ‘recessive’ traits
  • codominance is when the ‘recessive’ and ‘dominant’ traits appear together in the phenotype of the hybrid organism

BLGY1232 (19 decks)

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