Zadanie 8, Paper 2, maj 2022, TZ1, Biologia IB HL
Wzorowane na: Biology HL Paper 2, May 2022, TZ1, pytanie 8. Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
Polecenie
Nucleic acids underlie inherited biological variation.
(a) Draw and label one RNA nucleotide.
(b) Describe how DNA profiling can be used to investigate paternity.
(c) Explain why human adult heights vary.
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Odpowiedź z klucza
(a) Ribose is drawn as a pentagon and labelled. ✔
A nitrogenous base is connected to the correct anomeric carbon (1′) and labelled. ✔
A phosphate group is connected through the 5′ position and labelled. ✔
(b) DNA samples are collected from the child and potential parents. ✔
A suitable source is used, such as cheek cells, saliva or blood cells. ✔
PCR amplifies the selected DNA regions. ✔
Short tandem-repeat regions are analysed. ✔
Repeat numbers differ between individuals. ✔
Matching at every tested variable locus is unlikely for unrelated individuals. ✔
Electrophoresis separates fragments according to length. ✔
A characteristic pattern of bands is obtained. ✔
The band pattern constitutes the DNA profile. ✔
At each tested locus the child’s inherited bands/alleles must be accounted for by the two parents. ✔
(c) Environmental conditions influence height. ✔
Nutrition affects growth OR another valid environmental example. ✔
Inherited alleles also influence height. ✔
Many genes contribute, making the trait polygenic. ✔
This produces continuous variation. ✔
A large population commonly has an approximately bell-shaped height distribution. ✔
Different alleles can increase or decrease height. ✔
Many combinations of these alleles are possible. ✔
Specific mutations can produce unusually short or tall stature. ✔
Meiosis produces new combinations of alleles. ✔
Mutation introduces new genetic variants. ✔
Sex-related differences contribute to variation in average adult height. ✔
Ageing can reduce standing height. ✔
Quality: the response is clear on first reading, concise and contains little repetition or irrelevant material. ✔
Schemat punktowania
Part (a): 1 per correctly drawn and labelled feature, maximum 3; the phosphate–ribose connection must be at the 5′ position and the base at 1′. Part (b): any 5 distinct points, maximum 5. Part (c): any 7 distinct points, maximum 7. Award 1 independent quality mark for the stated communication criterion. Total maximum 16.
Umiejętności w zadaniu
- Environment-responsive phenotypic plasticityĆwicz podobne
- Recognizing molecular changes in genesĆwicz podobne
- Interpreting DNA profilesĆwicz podobne
- Amplifying DNA and separating fragmentsĆwicz podobne
- Genetic diversity generated by meiosisĆwicz podobne
- Parts of a nucleotideĆwicz podobne
- Polygenes, environment and continuous variationĆwicz podobne