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Genetics > Lecture--Chapter 7 > Flashcards

Lecture--Chapter 7 Flashcards

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Biology 101 flashcards
1
Q

Each organism contains how many genes?

A

hundreds to thousands; organisms have a small number of chromosomes compared to the estimated numbers of genes

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2
Q

Chromosomes must contain how many different genes?

A

hundreds to thousands; transmission of genes on a chromosome appears inconsistent with Mendel’s law of independent assortment

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3
Q

____ observed traits that did not assort independently:

A

Bateson and Punnett

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4
Q

Traits that did not assort independently:

A
  1. Sweet peas: flower colour and pollen shape
  2. Expected F2 generation phenotype ratios, 9:3:3:1
  3. Observed ratios (rounded), 16:1:1:5
    These genes appear to be coupled (linked)
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5
Q

allelic combination of parent(s):

A

parental

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6
Q

novel combination of alleles; recombinant

A

nonparental

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7
Q

Groups of genes that tend to be transmitted as a unit are called:

A

a linkage group; genes that are physically close together on the chromosome

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8
Q

Sweet pea ___ and ___ genes are on the same chromosome:

A

flower colour; pollen shape

they are linked.

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9
Q

Chromosomes are also ___

A

linkage groups; a group of genes that are linked together

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10
Q

The number of linkage groups in a species is:

A

the number of chromosomes in the genome of the species
(for example, in humans there are 22 autosomal linkage groups, there is an X chromosome linkage group, and there is a Y chromosome linkage group in males)

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11
Q

occurs around prophase I of meiosis:

A

recombination (crossing over)

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12
Q

homologous chromosomes form:

A

chiasmata

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13
Q

Crossing over during meiosis can:

A

separate linked genes in diploid eukaryotic species; nonparental (recombinant) phenotype combinations are produced

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14
Q

The frequency of recombinant phenotypes is proportional to:

A

distance between the genes

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15
Q

A very small frequency of recombinant offspring indicates that:

A

two loci lie very close to each other on the chromosome

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16
Q

Genes that are far apart on the same chromosome may independently assort from each other due to:

A

high frequency of cross-over events; they will appear unlinked

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17
Q

Who provided the first direct evidence of linkage?

A

T.H. Morgan

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18
Q

Studies of several traits that followed an X-linked pattern of inheritance:

A
  1. body colour
  2. eye colour
  3. wing length
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19
Q

offspring with nonparental phenotypes are the product of a crossover:

A

Morgan’s Hypothesis

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20
Q

can be used to determine whether the hypothesis is supported:

A

the Chi Square Test

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21
Q

the genes sort independently:

A

null hypothesis; provides expected results which can be calculated; genes are analysed in pairs for the test; the predicted ratio is 1:1:1:1.

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22
Q

Calculate the expected values of each of the 4 phenotypes:

A
  1. Expected ratio is 1:1:1:1
  2. 2,205 total offspring
  3. Expected number of each phenotype (combining males and females) = 1/4 x 2,205 = 551 (your E value)
  4. X^2 = ((O1 - E1)^2)/(E1) + …
    = ((O1 - 551)^2)/(551) + …
    = 2109.8
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23
Q

Interpret the calculated chi square value:

A
  1. Determine the degrees of freedom to use the chi square table
    a. the law of independent assortment states that two different genes randomly assort their alleles during gamete formation. Thus, we have two expected classes–nonrecombinant and recombinant (n=2)
    b. df = 1
  2. For a probability of 0.01 with a df=1, the chi square value is 6.635, which is < 2109.8.
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24
Q

The implications of Morgan’s data:

A
  1. The combinations of traits found in the parental generation had the highest proportions; all three genes are located on the X chromosome; therefore, they tend to be transmitted together as a unit
  2. Flies that had recombined parental traits were also obtained; genes could become unlinked.
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25
Q
  • Provided direct evidence that crossing over results in recombination.
  • One corn strain had a mutant chromosome 9 with 2 observable abnormalities.
A

Harriet Creighton and Barbara McClintock

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26
Q

located near the knobbed end:

A

kernel colour; C = coloured, c = colourless

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27
Q

Endosperm texture:

A
  1. Wx = starchy endosperm

2. wx = waxy endosperm

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28
Q

Testcrosses of the linked genes on chromosome 9 and the abnormal chromosome; inspected:

A

the chromosome structure of the nonparental recombinants with microscopes

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29
Q

could correlate the recombinant offspring with observable physical exchanges between homologous chromosomes

A

testcross results

30
Q

an undergraduate in T. H. Morgan’s lab:

A

Alfred Sturtevant

31
Q

recombination frequencies can be used to determine the map order of the genes

A

Alfred Sturtevant

32
Q

Use mapping to determine the order of genes on a chromosome

A

Alfred Sturtevant

33
Q

he created the first linkage map in 1913

A

Alfred Sturtevant

34
Q

the percentage of recombinant offspring correlates with:

A

the distance between the two genes

35
Q

Genes far apart —>

A

high recombination frequency (nonparentals are common)

36
Q

Genes close together —>

A

low recombination frequency (nonparentals are rare)

37
Q

Map Units: map distances are based on the frequency of recombination events:

A
  1. Frequency of recombination = (# nonparentals / total # offspring) x 100
  2. 1% frequency = 1 map unit (mu)
  3. Also called 1 centiMorgan (cM)
38
Q

determine the recombination frequencies between 5 loci on the X chromosome

A

Alfred Sturtevant’s Experiment

39
Q

Alfred Sturtevant’s Experiment: the genes: body colour

A

yellow (y); wt (y+) is gray

40
Q

Alfred Sturtevant’s Experiment: the genes: eye colour

A
  1. white (w); wt (w+) is red

2. vermillion (v); wt (v+) is red

41
Q

Alfred Sturtevant’s Experiment: the genes: wing type

A
  1. miniature (m); wt (m+) is normal

2. rudimentary (r); wt (r+) is normal

42
Q

In Alfred Sturtevant’s Experiment, 2 traits were analysed at a time, using dihybrid crosses:

A
  1. female heterozygous

2. male hemizygous recessive

43
Q

Alfred Sturtevant’s Experiment: the interpretation: some dihybrid crosses had low % of nonparental (recombinant) offspring

A
  1. only 3% recombinant offspring in crosses of v and m alleles
  2. these two genes are likely very close together (3 mu)
44
Q

Alfred Sturtevant’s Experiment: the interpretation: other dihybrid crosses had a higher % of nonparental offspring:

A
  1. 33.7% recombinant offspring in crosses of w and m alleles; these two genes are farther apart than y and w (33.7 mu)
  2. 29.7% recombinant offspring in crosses of w and v; these two genes are closer together than w and m (29.7 mu)
  3. Therefore, v is closer to m than it is to w, and the gene order is w - v - m.
45
Q

Alfred Sturtevant’s Experiment: the construction:

A

Sturtevant proposed the genetic map starting with y

46
Q

Genetic Mapping Discrepancies: the calculated distance between genes and the frequency of recombination:

A

do not always agree; for example for y - r genes, the actual map distance is 57.6, but the recombination frequency is 37.5%.

47
Q

Increasing distance means:

A

multiple crossover events may occur (but not be detected)

48
Q

Uses of genetic mapping:

A
  1. Determine linkage between genes.
  2. Provide an understanding of the complexity and genetic organisation of a particular species.
  3. Help understand evolutionary relationships.
  4. Diagnose, treat inherited human diseases.
  5. Predict likelihood of transmitting an inherited disease.
  6. Provide information for selective breeding in agriculture.
49
Q

Mapping using ____ can yield information about map distance and gene order:

A

trihybrid crosses

50
Q

What are the steps for trihybrid crosses:

A
  1. Produce heterozygotes.
  2. Perform testcross.
  3. Collect F2 generation data.
  4. Analyse the data.
51
Q

Produce heterozygotes:

A
  1. Cross two true-breeding strains that differ in three alleles
  2. F1 progeny are heterozygous in all three genes
52
Q

Perform testcross:

A
  1. F1 heterozygote females crossed with homozygous recessive males.
  2. Crossover events may occur during meiosis in F1 offspring.
53
Q

Collect F2 generation data:

A

usually requires lots of data

54
Q

Analyse the data:

A

gene order, map distances, map construction

55
Q

genetic mapping is possible to correct for:

A

double crossovers

56
Q

The discrepancy between the ____ and the ____ is due to the inability to detect double crossovers between 2 gene loci:

A

map distances; frequency of recombination

57
Q

Correction for double crossovers can be done by:

A

multiplying the number of flies (or other variable) where double crossover occurred by 2
for example: for body colour and wing shape (17.8 mu initially)
-map distance = {[179 + 2(2+1)] / 1005} x 100
= 18.4 mu

58
Q

causes the number of observed double recombinants to be lower than expected:

A

interference

59
Q

Use ___ to calculate each crossover event:

A

product rule

60
Q

one crossover event reduces the frequency of a second crossover in the same area of a chromosome

A

positive interference

61
Q

Calculating interference:

A

Interference is expressed as: 1 - C, where C = (observed # dbl crossover events) / (expected # dbl crossover events)

62
Q

crossing over can occur during mitosis:

A

mitotic recombination

63
Q

Mitosis does not involve the ___ of chromosomes to form bivalents

A

homologous pairing

64
Q

Crossing over in mitosis:

A

rarely occurs

65
Q

If mitotic recombination occurs during an early stage of embryonic development:

A

it may be observable in the somatic cells

66
Q

daughter cells with recombinant chromosomes:

A

continue to divide

67
Q

continuation of division in daughter cells can result in:

A

a patch of tissue with characteristics that are different from those of the rest of the organism

68
Q

example of mitotic recombination:

A

twin spots

69
Q

Fly strains carry X-linked genes that:

A

affect body colour and bristle morphology

70
Q

body colour:

A

yellow vs. gray

71
Q

bristle length:

A

short vs. long

Genetics (18 decks)

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