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Biology > Inheritance > Flashcards

Inheritance Flashcards

1
Q

In fruit flies, the genes for body colour and wing length are linked. Explain what this means.

A

The gene for body colour and wing length are linked because:

  • The genes are on the same chromosome
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2
Q

A scientist investigated linkage between the genes for body colour and wing length.
He carried out crosses between fruit flies with grey bodies and long wings and fruit flies with black bodies and short wings.

Figure 1 shows his crosses and the results.

  • G represents the dominant allele for grey body and g represents the recessive allele for black body.
  • N represents the dominant allele for long wings and n represents the recessive allele for short wings.

Figure 1:

Phenotype of parents:

1 ) Grey body, long wings

2 ) Black body, short wings

Genotype of parents:

1 ) GGNN

2 ) ggnn

Genotype of offspring

  • GgNn

Phenotype of offspring

  • All grey body, long wings

These offspring were crossed with flies homozygous for black body and short wings.

The scientist’s results are shown in Figure 2.

Figure 2:

  • GgNn are crossed with ggnn

Number of offspring:

  • Grey body, long wings = 975
  • Black body, short wings = 963
  • Grey body, short wings = 186
  • Black body, long wings = 194

Use your knowledge of gene linkage to explain these results.

A

Explanation of results:

  • GN and gn are linked
  • GgNn individuals produce mainly GN and gn gametes
  • Crossing over produces some Gn and gN gametes
  • So there are fewer Ggnn and ggNn individuals
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3
Q

If these genes were not linked, what ratio of phenotypes would the scientist have expected to obtain in the offspring?

( A scientist investigated linkage between the genes for body colour and wing length.
He carried out crosses between fruit flies with grey bodies and long wings and fruit flies with black bodies and short wings.

Figure 1 shows his crosses and the results.

  • G represents the dominant allele for grey body and g represents the recessive allele for black body.
  • N represents the dominant allele for long wings and n represents the recessive allele for short wings.

Figure 1:

Phenotype of parents:

1 ) Grey body, long wings

2 ) Black body, short wings

Genotype of parents:

1 ) GGNN

2 ) ggnn

Genotype of offspring

  • GgNn

Phenotype of offspring

  • All grey body, long wings

These offspring were crossed with flies homozygous for black body and short wings.

The scientist’s results are shown in Figure 2.

Figure 2:

  • GgNn are crossed with ggnn

Number of offspring:

  • Grey body, long wings = 975
  • Black body, short wings = 963
  • Grey body, short wings = 186
  • Black body, long wings = 194 )

( GN and gn are linked )

A

Ratio of grey long : grey short : Black long : Black short is:

  • 1 : 1 : 1 : 1
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4
Q

Which statistical test could the scientist use to determine whether his observed results were significantly different from the expected results?

Give the reason for your choice of statistical test.

( A scientist investigated linkage between the genes for body colour and wing length.
He carried out crosses between fruit flies with grey bodies and long wings and fruit flies with black bodies and short wings.

Figure 1 shows his crosses and the results.

  • G represents the dominant allele for grey body and g represents the recessive allele for black body.
  • N represents the dominant allele for long wings and n represents the recessive allele for short wings.

Figure 1:

Phenotype of parents:

1 ) Grey body, long wings

2 ) Black body, short wings

Genotype of parents:

1 ) GGNN

2 ) ggnn

Genotype of offspring

  • GgNn

Phenotype of offspring

  • All grey body, long wings

These offspring were crossed with flies homozygous for black body and short wings.

The scientist’s results are shown in Figure 2.

Figure 2:

  • GgNn are crossed with ggnn

Number of offspring:

  • Grey body, long wings = 975
  • Black body, short wings = 963
  • Grey body, short wings = 186
  • Black body, long wings = 194 )

( GN and gn are linked )

A

A statistical test that could be used to determine whether the observed results were significantly different from the expected results is:

  • Chi squared test

The reason why the test is choosed:

  • Categorical data
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5
Q

A breeder crossed a black male cat with a black female cat on a number of occasions.
The female cat produced 8 black kittens and 4 white kittens.

Explain the evidence that the allele for white fur is recessive.

A

The allele for white fur is recessive because:

  • Parents are heterozygous
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6
Q

Predict the likely ratio of colours of kittens born to a cross between this black male and a white female.

( A breeder crossed a black male cat with a black female cat on a number of occasions.
The female cat produced 8 black kittens and 4 white kittens. )

A

Ratio of colours of kittens born to a cross between this black male and a white female is:

  • 1:1
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7
Q

The gene controlling coat colour has three alleles.
The allele B gives black fur, the allele b gives chocolate fur and the allele b^i gives cinnamon fur.

  • Allele B is dominant to both allele b and b^i.
  • Allele b is dominant to allele b^i.

Complete the table to show the phenotypes of cats with each of the genotypes shown.

Genotype:

1 ) Bb^i

2 ) bb^i

3 ) Bb

Phenotype:

1 )

2 )

3 )

A

Genotype:

1 ) Bb^i

2 ) bb^i

3 ) Bb

Phenotype:

1 ) Black

2 ) Chocolate

3 ) Black

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8
Q

A chocolate male was crossed several times with a black female.

They produced

  • 11 black kittens
  • 2 chocolate kittens
  • 5 cinnamon kittens.

Using the symbols in part ( b ), complete the genetic diagram to show the results of this cross.

( The gene controlling coat colour has three alleles.
The allele B gives black fur, the allele b gives chocolate fur and the allele b^i gives cinnamon fur.

  • Allele B is dominant to both allele b and b^i.
  • Allele b is dominant to allele b^i. )

Parental phenotypes:

1 ) Chocolate male

2 ) Black female

Parental genotypes:

1 )

2 )

Gametes:

1 )

2 )

Offspring genotypes:

1 )

2 )

Offspring phenotypes

1 ) Black

2 ) Chocolate

3 ) Cinnamon

A

Parental phenotypes:

1 ) Chocolate male

2 ) Black female

Parental genotypes:

1 ) bb^i

2 ) Bb^i

Gametes:

1 ) b b^i

2 ) B b^i

Offspring genotypes:

1 ) Bb, Bb^i

2 ) bb^i

3 ) b^ib^i

Offspring phenotypes

1 ) Black

2 ) Chocolate

3 ) Cinnamon

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9
Q

The breeder had expected equal numbers of chocolate and cinnamon kittens from the cross between the chocolate male and black female.
Explain why the actual numbers were different from those expected.

( A chocolate male was crossed several times with a black female.

They produced

  • 11 black kittens
  • 2 chocolate kittens
  • 5 cinnamon kittens. )
A

The actual numbers were different from those expected of chocolate and cinnamon kittens because:

  • Fusion of gametes are random
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10
Q

The breeder wanted to produce a population of cats that would all have chocolate fur.
Is this possible?
Explain your answer.

( The gene controlling coat colour has three alleles.
The allele B gives black fur, the allele b gives chocolate fur and the allele b^i gives cinnamon fur.

  • Allele B is dominant to both allele b and b^i.
  • Allele b is dominant to allele b^i. )
A

A population of chocolate fur cats is:

  • Possible if the parents are homozygous bb
  • However the chocolate cats can be either bb or bb^i
  • So two chocolate cats can produce cinnamon fur cats
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11
Q

A student investigated the monohybrid inheritance of eye shape in fruit flies.
Two fruit flies with bar ( narrow ) eyes were crossed.
Of the offspring, 1538 had bar eyes and 462 had round ( normal )
eyes.

Using suitable symbols, give the genotypes of the parents.
Explain your answer.

A

Genotypes of parents:

  • Bb

Explanation:

  • Both parents have bar eyes, but they have some offspring with round eyes, so the parents must be carriers of the recessive allele for round eyes
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12
Q

The ratio of bar-eyed flies and round-eyed flies in the student’s results were not the same as the ratio she had expected.
What ratio of bar-eyed to round-eyed flies was the student expecting?

( A student investigated the monohybrid inheritance of eye shape in fruit flies.
Two fruit flies with bar ( narrow ) eyes were crossed.
Of the offspring, 1538 had bar eyes and 462 had round ( normal )
eyes. )

A

Ratio of bar-eyed to round-eyed flies that the student was expecting is:

  • 3:1
  • ( 1538 : 462 = 3.33 : 1 )
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13
Q

Suggest two reasons why observed ratios are often not the same as expected ratios.

( A student investigated the monohybrid inheritance of eye shape in fruit flies.
Two fruit flies with bar ( narrow ) eyes were crossed.
Of the offspring, 1538 had bar eyes and 462 had round ( normal )
eyes. )

A

Observed ratios are often not the same as expected ratios because:

  • Fusion of gametes are random
  • A small sample is used
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14
Q

The student wished to test her results with the ones she had expected.
Which statistical test should she use?

( A student investigated the monohybrid inheritance of eye shape in fruit flies.
Two fruit flies with bar ( narrow ) eyes were crossed.
Of the offspring, 1538 had bar eyes and 462 had round ( normal )
eyes. )

A

The statistical test she should use is:

  • Chi squared
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15
Q

This fruit fly has another characteristic controlled by a pair of codominant alleles, W^N and W^V.
What is meant by codominant alleles?

A

Codominanat alleles means:

  • Both alleles are expressed in the phenotype
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16
Q

There were 850 fruit flies in one population.
In this population, 510 fruit flies had the genotype W^NW^N, 255 had the genotype W^NW^V and 85 had the genotype W^VW^V.
Calculate the actual frequency of the allele W^V.
Do not use the Hardy-Weinberg equation in your calculation.

A

The frequency of the allele W^V is:

  • 0.25
17
Q

In another population of 950 fruit flies, the frequency of the W^V allele was 0.2.
Use the Hardy-Weinberg equation to calculate the number of insects that would be expected to have the genotype W^NW^V.

( There were 850 fruit flies in one population.
In this population, 510 fruit flies had the genotype W^NW^N, 255 had the genotype W^NW^V and 85 had the genotype W^VW^V. )

A

The number of insects that would be expected to have the genotype W^NW^V is:

  • 304

( p^2 + 2pq + Q^2 = 1 )

18
Q

Meiosis results in cells that have the haploid number of chromosomes and show genetic variation.
Explain how.

A

Meiosis results in cells that have the haploid number of chromosomes and show genetic variation because:

  • Homologous chromosomes pair up
  • Crossing over occurs
  • Which produces a new combination of alleles
  • Chromosomes separate
  • Randomly
  • Which produces varying combinations of chromosomes
19
Q

In mice, two genes affecting coat colour are on different chromosomes.
One gene controls whether there is any black pigment in the hairs. The dominant allele of this gene, B, results in black fur.
The recessive allele, b, results in white fur.
The second gene controls banding of the fur.
The dominant allele, A, causes a yellow band to develop
on each hair.
The resulting coat colour is called agouti.
The recessive allele, a, results in hairs with no bands on them.
This gene has no effect on mice with white fur; white mice do
not develop bands, even if they have the A allele.

Breeders performed many crosses in which agouti mice were crossed with white mice, homozygous for both genes.
They expected agouti, black and white mice in the offspring in
a 1 : 1 : 2 ratio.

Complete the genetic diagram to show how this ratio of phenotypes would be produced.

Parental phenotypes:

1 ) Agouti

2 ) White

Parental genotypes:

1 )

2 )

Gamete genotypes:

1 )

2 )

Offspring genotypes:

1 )

2 )

Offspring phenotypes:

1 )

2 )

A

Parental phenotypes:

1 ) Agouti

2 ) White

Parental genotypes:

1 ) BbAa

2 ) bbaa

Gamete genotypes:

1 ) BA, Ba, bA, ba

2 ) ba

Offspring genotypes:

1 ) BbAa, Bbaa

2 ) bbAa, bbaa

Offspring phenotypes:

1 ) Agouti, Black

2 ) White, White

20
Q

The actual numbers of offspring with each phenotype were

Agouti 34

Black 35

White 51

The x^2 test can be used to test the hypothesis that there is no significant difference between these results and the expected 1 : 1 : 2 ratio.
Complete the table to calculate the value of x^2 for these results.

Colour of offspring:

1 ) Agouti

2 ) Black

3 ) White

Observed ( O ):

1 ) 34

2 ) 35

3 ) 51

Expected ( E ):

1 )

2 )

3 )

( O - E ):

1 )

2 )

3 )

( O - E )^2:

1 )

2 )

3 )

( O - E )^2 / E:

1 )

2 )

3 )

Sum of ( O - E )^2 / E =

A

Colour of offspring:

1 ) Agouti

2 ) Black

3 ) White

Observed ( O ):

1 ) 34

2 ) 35

3 ) 51

Expected ( E ):

1 ) 30

2 ) 30

3 ) 60

( O - E ):

1 ) 4

2 ) 5

3 ) 9

( O - E )^2:

1 ) 16

2 ) 25

3 ) 81

( O - E )^2 / E:

1 ) 0.53

2 ) 0.83

3 ) 1.35

Sum of ( O - E )^2 / E = 2.71

21
Q

The table shows values for x^2 at different levels of probability and for different degrees of freedom.

Degrees of freedom: Probability, p:
0.2 0.1 0.05 0.02 0.01

1 ) 1.64 2.71 3.84 5.41 6.64

2 ) 3.22 4.61 5.99 7.82 9.21

3 ) 4.64 6.25 7.82 9.84 11.35

4 ) 5.99 7.78 9.49 11.67 13.28

5 ) 7.29 9.24 11.07 13.39 15.09

What should the breeders conclude about the significance of their results?
Explain your answer.

A

The breeders can conclude that the significance of their results is because:

  • Expression is due to genotype
  • Expression is due to the enviroment
22
Q

What is meant by the term phenotype?

A

Phenotype is:

  • The expression of a gentype
  • An expression due to the enviroment
23
Q

The inheritance of fruit colour in summer squash plants is controlled by two genes, A and B.
Each gene has two alleles.

The diagram shows the interaction of these two genes in controlling fruit colour in summer squash plants.

         aa              B
           |                |
     Enzyme 1   Enzyme 2
           |                |
White - > Green - > Yellow
           ^                ^
           | Inhibition | No functional enzyme
          A               bb

Name the type of gene interaction shown in the diagram above.

A

the type of gene interaction shown in the diagram above is:

  • Epistasis
24
Q

What fruit colour would you expect the following genotypes to have?

AAbb

aaBB

( The inheritance of fruit colour in summer squash plants is controlled by two genes, A and B.
Each gene has two alleles.

The diagram shows the interaction of these two genes in controlling fruit colour in summer squash plants.

         aa              B
           |                |
     Enzyme 1   Enzyme 2
           |                |
White - > Green - > Yellow
           ^                ^
           | Inhibition | No functional enzyme
          A               bb )
A

AAbb:

  • White

aaBB:

  • yellow

Biology (28 decks)

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