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28 Terms

1

Draw and label a simplified structure of a nucleotide

Must have: diagram with ribose /dexoyribose / sugar, phosphate and base labelled and connected Correct linkage and spartial arrangements (phosphate linked to sugar and base linked to sugar) Just easy shapes, remember linkage

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2

A genetic cross was made between pure-breeding snapdragon plants with red flowers and pure-breeding snapdragon plants with white flowers. The cross produced F1 offspring that had only pink flowers. When the F1 plants self-polinated, the resulting F2 generation had some red, some white and some pink flowers.

Identify the relationship between the red and white alleles for flower colour.

Co-dominance (both alleles equally expressed in the phenotype)

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3

A genetic cross was made between pure-breeding snapdragon plants with red flowers and pure-breeding snapdragon plants with white flowers. The cross produced F1 offspring that had only pink flowers. When the F1 plants self-polinated, the resulting F2 generation had some red, some white and some pink flowers.

Deduce the genotype of the F1 plants

RW (both dominant)

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4

A genetic cross was made between pure-breeding snapdragon plants with red flowers and pure-breeding snapdragon plants with white flowers. The cross produced F1 offspring that had only pink flowers. When the F1 plants self-polinated, the resulting F2 generation had some red, some white and some pink flowers.

Construct a Punnett grid to show the cross between two F1 plants

Grid with gametes shown correctly (RW/C^R and C^W) All offsprings' genotypes shown

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5

A genetic cross was made between pure-breeding snapdragon plants with red flowers and pure-breeding snapdragon plants with white flowers. The cross produced F1 offspring that had only pink flowers. When the F1 plants self-polinated, the resulting F2 generation had some red, some white and some pink flowers.

Deduce the portion of the different phenotypes of the F2 offspring

1 red, 2 pink, 1 white Codominance

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6

Discuss two advantages of genetic screening

can prevent genetic disease frequency of harmful alleles reduced allows for early diagnosis/detection can predict genetic disease

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7

State two properties of the fragmented pieces of DNA which allows them to be separated in gel electrophoresis.

electrical charge size (smaller DNA moves more than large)

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8

The diagram below shows a DNA profiling of a family with five children. Segments of the DNA inherited by some members of the family are shown as two dark band in each column. The DNA fragments are labeled A to F.

Determine which DNA fragments Son 2 inherited from his mother and which from his father.

From mother: B From father: E

Read off the diagram and see which adds up.

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9

Identify the child that genetically most resembles one of the grandparents.

Son 1

Read off the diagram and see who adds up.

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10

Apart from determining family relationships, outline one other application of DNA profiling.

Criminal investigation to confirm suspects Rape cases Tracking individuals in populations

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11

Define sex linkage

A gene/allele/trait on a sex-determining chromosome (X and Y)

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12

State one example of sex linkage

Hemophilia Hunter's syndrome (both in males)

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13

Define the term co-dominance

Pair of alleles that both affect the phenotype (when presented in a heterozygote) Both alleles are expressed and recognized in the phenotype

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14

Discuss the potential benefits and possible harmful effects of genetic modification.

Benefits: more specific breeding, faster, selective breeding cannot produce desired phenotypes, increased productivity of food production, less use of chemicals, food production possible in extreme conditions, human insulin engineered so no allergic reactions

Harmful effects: Some gene transfers can cause harm to organisms, release of genetically engineered organisms in the free environment, could spread and compete with naturally occurring varieties, reduces genetic variation and biodiversity

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15

State what type of sugar lactose is

Disaccharide

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16

State a function of lactose

Provide energy (for young mammals)

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17

Explain the production of lactose-free milk

Lactase added to milk / immobilised Lactose broken down into glucose and galactose Increases sweetness of milk Lactose intolarant can drink it now

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18

Outline the effect of increasing calcium levels in the water on calcium levels in the tissue of bivavles

Leads to an increase in calcium tissues

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19

Outline the effect of increasing calcium levels in the water on metals other than calcium in the tissue of bivavles

Leads to a decrease in other metals in tissues

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20

Suggest a reason for the effects of calcium on the levels of other metals in the tissues.

Competitive inhibition Uptake of other metals is prevented Export of metals in cells when calcium enters

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21

Evaluate the implications of these results for monotoring water quality in regions where bivavles are harvested

Can indicate amount of metals in the tissue To measure water quality Posinous level of heavy metals

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22

Primary structure involves the sequence of amino acids that are vonded together to form a polypeptide. State the name of the linkage that bonds the amino acids together.

Peptide bonds

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23

Beta pleated sheets are an example of secondary structure. State one other example

Alpha helix

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24

State one type of bond that stabilizes the tertiary structure

Ionic bonds Hydrogen bonds

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25

Outline the quaternary structure of proteins

Linking together of polypeptides to form a single protein uses some bonding as tertiary structure. Linking of a non-polypeptide structure.

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26

State three activities that occurr in part A of the cell cycle

Protein synthesis DNA replication Cell growth Respiration Transcription

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27

Signficance of complementary base pairings

replication of DNA requires it Produces identical copies of DNA Each new cell gets a copy of each DNA molecule Mutations can lead to cancer

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28

Distinguish between the structures of different types of fatty acids in food

fatty acids can be saturated or unsaturated; can be monounsaturated or polyunsaturated; saturated fats have no double bonds/have maximum number of hydrogen atoms;ORunsaturated fatty acids have «at least one» double C=C bond;ORpolyunsaturated fatty acids have more than one double bond / OWTTE; d cis-form has hydrogen atoms on same side of carbon double bond;ORcis-form has bend at carbon double bond; e trans-form has hydrogens on opposite sides of carbon double bond;ORtrans-form makes a straight carbon chain; f length of hydrocarbon chain can vary; ORposition/number of carbon double bonds can vary;

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