| A2 Level | Unit 1 | Unit 2 | Unit 3 | Unit 4 | Unit 5 | Unit 6 | Unit 7 | Unit 8 | Unit 9 | Unit 10 | Unit 11 | Unit 12 | Unit 13 | AS Level | Unit 1 | Unit 2 | Unit 3 | Unit 4 | Unit 5 | Unit 6 | Unit 7 | Unit 8 | Unit 9 | Unit 10 | Unit 11 | Unit 12 | Unit 13 |
A2 Human Biology Unit 1
Text questions
| P8/Q1 |
Because they control the same characteristics. |
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| P9/Q2 | a) |
Chromosomes become shorter and fatter; or Nucleolus disintegrates; or Nuclear envelope disintegrates |
| b) |
In prophase of meiosis (I) homologous chromosomes
pair up; (they don't in mitosis). |
|
| P10/Q3 |
Anaphase of meiosis (I) |
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| P12/Q4 |
223 |
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| P13/Q5 |
The possible number of genetically different cells
will be higher (possibly infinite). |
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| P14/Q6 | a) |
Haploid →
possessing half of the normal/somatic number of
chromosomes (23 for a human); → possessing a chromosome from each pair of homologous chromosomes. |
| b) |
Homogametic →
producing gametes which have the same set of
chromosomes (eg. one X - sex chromosome, no Y chromosome). |
|
| P16/Q7 |
Dominant |
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| P16/Q8 |
Both allelles (R-red, W-white) show up (codominance),
(red + white = pink). |
|
| P18/Q9 | a) |
When Io is paired up with any other
allelle (eg Ib), then it is always «switched off», and the other allelle determines the
antigen present (ie. : Ib). |
| b) |
When IA and IB are paired up,
they both determine the phenotype (antigen A &
antigen B are both present on the plasma membrane of
RBC). |
|
| P19/Q10 | a) |
3 : 1 (for monohybrid with dominance) |
| b) |
9 : 3 : 3 : 1 (dihybride with dominance) |
|
| P22/Q11 |
Father never passes chromosome X (locus of Xh is
there) to his son! |
|
| P23/Q12 | a) |
Individual 4 is a female, her son is a sufferer, so
she must be a carrier (XHXh). |
| b) |
Individual 10 is a female, only her father is a
sufferer (his X chromosome is Xh), so she
must have inherited Xh, hence she must be
XHXh. |
Assignment
| P27/Q1 |
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P27/Q2 |
a) |
Typica WTWT Intermediate WTWR Annulate WRWR |
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| b) |
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P28/Q3 |
a) |
¯Order of decreasing dominance WB WM WAWC WH |
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| b) |
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Examinations
| P28/Q1 | a) |
(i) Meiosis →
Reduction of number of chromosomes from 2n to n.
(ii) Mitosis →
Number of
chromosomes unchanged. |
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| b) |
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c) |
Males develop from unfertilised eggs → males inherit all genes from their mother as they are |
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| P29/Q2 | a) |
(i) XBXb (ii) Because a male has only one X chromosome
→ there is only
one allele (either XB or Xb). |
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| b) |
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P30/Q3 |
a) |
Male with normal sight (nn). |
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| b) |
Nn → she is a
sufferer (so must have gene N). She is not a
homozygote (NN) because some of her children are not
sufferers → she
must be a heterozygote then. |
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| c) |
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