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Solutions to New Generations
Mr. D Patterson
Contents (Hold Ctrl and click on an Activity to jump to that page)
Activity 32
Activity 42
Activity 52
Activity 63
Activity 74
Activity 135
Activity 165
Activity 176
Activity 186
Activity 198
Activity 219
Activity 3
1. (ReproductionAsexualSexualInternalExternalBuddingRegenerationBinary FissionVegetative Spores)
2.
a-9b-6c-10 (should be Binary fission, not budding)
d-8e-7f-3
g-5h-4i-1
j-2
Activity 4
2. Gametes
3. Eggs formed in ovaries. Sperm formed in testes.
4. Internal fertilisation is when sperm and eggs fuse (come together) inside the body. Humans and dogs use internal fertilisation.
5. External fertilisation is when sperm and eggs fuse (come together) outside the body. Fish and coral use external fertilisation.
6. External fertilisation requires more eggs. The chance of successful fertilisation and survival of the offspring is much lower outside the body, so need more eggs to increase survival rate.
7. Internal fertilisation requires special organs. Offspring will grow inside the body thus need a way to receive oxygen and nutrients from the mother.
Activity 5
2. The features of both mother and father can be passed onto the offspring.
3. Variation gives a greater chance of the species surviving. If the environment changes, some of the species may have features better suited to surviving.
Activity 6
(Fertilisation occurs in the fallopian tubes (oviducts). NOT in the uterus)
Activity 7
(Pollen is not well labelled on the diagram. It is the small grains at the end of the anther.) (OvaryFilamentPetalStigmaPistilStyleStamenOvulesAnther) (To aid in the reproduction of the plant ) (263) (451)
Activity 13
1. Because two haploid (cells (ie: sperm and egg) will combine to form a new individual with the normal amount of chromosomes.
2. 23 (half)
3. 46 (normal chromosome number or "diploid")
4. 23
5.
16
8
60
30
196
98
48
24
22
11
1262
631
Activity 16
1. i) RW (or WR)ii) RRiii) WW
2a.i) SSii)CCiii) SC (or CS)
Parents:
SC
SC
Gametes:
S
C
S
C
Offspring
SS
SC
SC
CC
2b.
Hair type: 1 straight, two wavy, 1 curvy
3a. Intermediate inheritance (because offspring show a blending of parent characteristics)
Parents:
WW
BB
Gametes:
W
W
B
B
Offspring
WB
WB
WB
WB
3b.
All grey, so no chance for black
3c. Two out of four (or ½ or 50%)
3d. Black
4a. LL
4b. RR
4c. Oval
Parents:
LR
LR
Gametes:
L
R
L
R
Offspring
LL
LR
LR
RR
4d.
4e.
Parents:
LR
LL
Gametes:
L
R
L
L
Offspring
LL
LL
LR
LR
Offspring will be half long and half oval
Activity 17
1. Tall (T)
2. Short (t)
3. Hybrid is the same as heterozygous. It means the organism has different alleles (eg: Tt)
4. Dominant/recessive inheritance means one characteristic will dominate over another. There are no intermediate effects.
2. Bb (both parents need a recessive allele so otherwise white (recessive) offspring would not be possible
3. BB and Bb
4. bb
Activity 18
1.
Cross
Genotypes (parents)
Genotypes (offspring)
1. pure brown x pure white
BB x bb
All Bb
2. hybrid brown x hybrid brown
Bb x Bb
BB, Bb, bb
3. hybrid brown x pure white
Bb x bb
Bb and bb
4. pure brown x hybrid brown
BB x Bb
BB and Bb
5. pure brown x pure brown
BB x BB
All BB
6. pure white x pure white
bb x bb
All bb
2. a) Homozygous means the alleles are the same. Heterozygous means the alleles are different
b) homozygous brown = BBc) heterozygous = Bb
3) As it is dominant/recessive inheritance, a heterozygous (hybrid) black cow means that black is the dominant allele. (ie : Bb cow is black, so B (black) is dominant over red (b))
a)
b
b
B
Bb
Bb
b
bb
bb
50% Bb, 50% bb
b) 2 different phenotypes (black and red)
c) 50% black (due to Bb) and 50% red (due to bb)
4) a) and b) need to be thought of together to understand how to get the answer.
Parents with a recessive phenotype (genotype = dd) can only have offspring with a recessive phenotype.
Parents that are both hybrid (Dd) can have both phenotypes show in offspring. This can be shown with a punnet square.
D
d
D
DD
Dd
d
Dd
dd
Dimples (D) must be dominant to no dimples (d).
Parents are hybrid (Dd) and produce an homozygous non dimpled child (dd)
c) i) 25%ii) 75%
5.
T
t
T
TT
Tt
t
Tt
tt
Possible genotypes = TT, Tt and tt
Possible phenotypes = Tall (TT, Tt) and short (tt)
6.
B
b
b
Bb
bb
b
Bb
bb
50% chance of bb = 50% chance of non barking
Activity 19
Problems involving intermediate inheritance
1a.
Individual
Genotype
1
CC
2
SS
3
SC
4
SC
5
SC
6
SC
7
SS
8
SS
9
SC
10
SC
11
CC
12
SC
13
SC
1b. 100%1c. 0%1d. 50%1e. 0%
2a. MM:MN:NN = 1:2:12b. 25%
Problems involving Dominant/recessive inheritance
1a,b,c. 4 black, 4 white. The black coats will all be heterozygous (no pure bred). (predicted from genotypic ratios).
2a. Ll2b. Ll2c. Must be Ll x Ll to produce recessive ll.
3. Check with your teacher.
4a. 10 and 11= rr (recessive homozygous - lefthanded).
4b. Rr (must be heterozygous to produce recessive offspring)
4c. Rr (Shows dominant phenotype but produces recessive offspring)
4d. 2, 3, 4, 7, 11 and 13.
4e. 8, and 9 are impossible to predict with certainty
4f. 50% (based on genotypic ratios)
Activity 21
1. Individual 2 and individual 4
2. Female
3. 7, 8,9 and 11
4. 25% (use punnet square)
5. 0% (use punnet square. Is equal to 0% chance regardless of mother’s genotype.)
1. Answers (X = clean chromosome, Xh = diseased chromosome)
a. XY
b. XXh
c. XhY
d. XXh
e. XX or XXh
f. XhY
2. This individual is not affected by the disease but ‘carries’ it, passing it on to offspring.
3. Males who have a diseased X chromosome do not have another X chromosome to dominate the diseased one. A clear Y chromosome does not dominate a diseased X.
4. 25% chance (use punnet square)
Colour blindness
1. X (more genes than Y to hold disease)
2. 2 (XcXc)
3. A. Women needs two diseased chromosomes, so father must pass on colour blind gene. Hence XcY.
4. B. Women needs two diseased chromosomes, so mother must pass on colour blind gene. Hence XcX OR XcXc (if mother is colour blind too).
C. Man only needs one diseased X chromosome. This could come from mother OR father. So mother may be clear, carrier or affected. XX, XcX or XcXc.