GTS 161 Practical 2 Memo 2014
Additional Questions
A1.
As these chromosomes are in interphase before the S-phase, each consists of only one
chromatid.
b. non-homologous chromosomes
d & e. homozygotic alleles
d & f.
heterozygotic
alleles
t
H
c. locus
h
t
c. locus
a. homologous chromosomes
a. homologous chromosomes
OR:
A2.
No. They are the same size, have the same centromere positions and carry the same loci /
genes in the same order but can have different alleles at specific loci. This will mean they can
differ slightly in their DNA nucleotide sequences.
Pierce Chapter 3
7.
Parental genotype
Aa × Aa
Aa × aa
AA x Aa
AA × AA
aa × aa
AA × aa
Genotypic ratio of progeny
1Aa : 2Aa : 1aa
1Aa : 1aa
1AA : 1Aa
Uniform progeny – All AA
Uniform progeny – All aa
Uniform progeny – All Aa
9.
According to the principle of independent assortment, genes for different characteristics
(different loci) segregate independently of one another, if the two genes are located on different
chromosomes.
The principle of segregation indicates that the two alleles at a locus separate; this is true for
every gene as the two chromosomes of a homologous pair separate at anaphase I of meiosis.
The principle of independent assortment indicates that the separation of alleles at one locus is
independent of the separation of alleles at other loci, because each pair of homologous
chromosomes lines up independently of all other pairs on the metaphase I equator, and
separates independently of other pairs at anaphase I.
1
10.
In anaphase I of meiosis, each pair of homologous chromosomes segregate independently of all
other pairs of homologous chromosomes. The assortment is dependent on how the homologs
line up during metaphase I. This random assortment of homologous pairs explains how genes
located on different chromosomes will separate independently of one another.
Anaphase II results in the separation of sister chromatids and subsequent production of
gametes carrying single alleles for each gene locus as predicted by Mendel’s principle of
segregation.
Mendel’s principles of independent assortment and segregation do not apply to mitosis, which
produces cells genetically identical to each other and to the parent cell.
15.
Yes, they could be sisters. Albinism is inherited as an autosomal recessive trait, meaning that a
person must receive two copies of the albino allele, one from each parent, implying that both
parents had to be heterozygous (carriers of this allele). These parents could therefore have
produced normal pigmented children (being homozygous dominant or heterozygous) or albino
children (being homozygous recessive).
16.
a.
The F1 generation contains all tan seed progeny and is the result of crossing a tan seed plant
with a red seed plant. The F1 result suggests that the tan phenotype is dominant to red. In the
F2 generation, the ratio of tan to red seed plants is about 3.9 to 1, which is similar but not
identical to a 3 to 1 F2 ratio expected for monohybrid cross involving dominant and recessive
alleles. The F2 ratio therefore suggests that the F1 parents which show the dominant tan
phenotype are heterozygous.
b.
Define the dominant tan allele as “R” and the recessive red allele as “r.”
Tan seed ♀ parent: RR
Red seed ♂ parent: rr
F1 tan seed offspring: Rr
F2 tan seed offspring: RR or Rr
F2 red seed offspring: rr
18.
IA (type A) > iB (type B)
IA IA = type A
IA iB = type A
iB iB = type B
a.
♂ type A × ♀ type B → 4 type A and 3 type B kittens
Because the female parent has blood type B, she must have the genotype i BiB. The male parent
could be either IAIA or IAiB. However, as some of the offspring are kittens type B, the male
parent must have contributed an iB allele to these kittens. Therefore, the male must be
genotype IAiB.
b.
♂ type B × ♀ type B → 6 type B kittens
Because type B is caused by the recessive allele iB, both parents must be homozygous for the
recessive allele or iBiB. Each contributes only the iB allele to the offspring.
c.
♂ type B × ♀ type A → 8 type A kittens
The male with type B blood must be iBiB. A female with type A blood could have either the IAIA
or IAiB genotypes. Because all of her kittens have type A blood, this suggests that she is likely to
be homozygous for the for IA allele (IAIA) and contributes only the IA allele to her offspring. It
could be possible that she is heterozygous for type A blood, but if so it is unlikely that chance
alone would have produced eight kittens with blood type A.
20.
The black coat colour is likely recessive.
When Sam was crossed with a black cat, one-half the offspring were white and one-half were
black. This 1:1 ratio potentially indicates that one of the parental cats is heterozygous, while the
other parental cat is homozygous recessive (typical testcross ratio).
The interbreeding of the black kittens produced only black kittens, indicating that the black
kittens are likely to be homozygous, and thus the black coat colour is the recessive trait.
If the black allele was dominant, we would have expected the black kittens to be heterozygous,
containing a black coat colour allele and a white coat colour allele. Under this condition, we
would expect one-fourth of the progeny from the interbred black kittens to have white coats.
Because this did not happen, we can conclude that the black coat colour is likely recessive.
2
25.
a.
b.
c.
d.
28.
a.
b.
c.
d.
1/6.
P(rolling either a 1 OR a 2) = 1/6 + 1/6 = 2/6 = 1/3
P(rolling either a 2 OR a 4 OR a 6) = 1/6 + 1/6 + 1/6 = 3/6 = ½
P(rolling a 1 OR a 2 OR a 3 OR a 4 OR a 5) = 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 5/6
OR 1 – P(rolling a 6) = 1 – 1/6 = 5/6
Let P (normal, no PKU) > p (PKU)
PP = normal
Pp = normal
pp = PKU
As two normal parents produce a child with the disease, they both have to be carriers
(heterozygous), i.e. Pp genotype
Pp x Pp → GR: ¼ PP : ½ Pp : ¼ pp
PR: ¾ normal : ¼ PKU
½
½
P(child with PKU) = P(pp offspring) = P(p-allele ♂ and p-allele from ♀) = ½ × ½ = ¼
OR directly from progeny calculated for cross: P(pp offspring) = ¼
P(Pp offspring) = P[(P-allele ♂ and p-allele from ♀) or (p-allele ♂ and P-allele from ♀)]
= (½ × ½) + (½ × ½) = ½.
OR directly from progeny calculated for cross: : P(Pp offspring) = ½
31.
B (bitter) > b (sweet)
BB = bitter
Bb = bitter
bb = sweet
a.
P
F1
F1 x F1
F2
S (yellow spots) > s (no spots)
SS = yellow spots
Ss = yellow spots
ss = no spots
BB SS × bbss
GR: All Bb Ss
PR: All bitter fruit, yellow spots
Bb Ss × Bb Ss.
PR: 9/16 bitter fruit, yellow spots
3/16 bitter fruit, no spots
3/16 sweet fruit, yellow spots
1/16 sweet fruit, no spots
b.
Backcross of F1 plant (Bb Ss) x bitter, yellow-spotted parent (BB SS)
↓
GR: ¼ BBSS : ¼ BBSs : ¼ BbSS : ¼ BbSs
PR: all bitter, yellow-spotted offspring.
c.
Backcross of F1 plant (Bb Ss) x sweet, nonspotted parent (bb ss)
↓
GR: ¼ BbSb : ¼ Bbss : ¼ bbSs : ¼ bbss
PR: ¼ bitter fruit, yellow spots : ¼ bitter fruit, no spots :
¼ sweet fruit, yellow spots : ¼ sweet fruit, no spots
34.
The easiest procedure for determining the proportion of a particular genotype in the offspring is
to break the cross down into simple crosses, and consider the proportion of the offspring for
each cross.
Aa Bb Cc dd Ee × Aa bb Cc Dd Ee
Locus 1: Aa × Aa → ¼ AA : ½ Aa : ¼ aa
Locus 2: Bb × bb → ½ Bb : ½ bb
Locus 3: Cc × Cc → ¼ CC : ½ Cc : ¼ cc
Locus 4: dd × Dd → ½ Dd : ½ dd
Locus 5: Ee × Ee → ¼ EE : ½ Ee : ¼ ee
b.
d.
Aa bb Cc dd ee: ½ (Aa) × ½ (bb) × ½ (Cc) × ½ (dd) × ¼ (ee) = 1/64
AA BB CC DD EE: Will not occur. The Aa Bb Cc dd Ee parent cannot contribute a D allele
(offspring genotype DD not possible), and the Aa bb Cc Dd Ee parent cannot contribute a B
allele (offspring genotype BB not possible).
3
Additional Questions
A3.
A (absence of moles) > a (moles present)
AA = absence of moles
Aa = absence of moles
aa = presence of moles
For this couple:
Aa x Aa → GR: 1/4 AA : 1/2 Aa : 1/4 aa (or 1 AA : 2 Aa : 1aa)
PR: 3/4 absence of moles : 1/4 presence of moles (or 3 no moles :1 moles)
Let
p = P(absence of moles) = ¾
q = P(presence of moles) = ¼
a.
(p + q)5 = p5 + 5p4q + 10p3q2 + 10p2q3 + 5pq4 + q5
b.(i)
P (child with moles) = q = ¼
(ii)
P (5 without moles) = p5 = (3/4)5 = 0,237
(iii)
P(1st moles and 2nd moles and 3rd no moles and 4th no moles and 5th no moles)
= (1/4)2(3/4)3 = 0,026
(iv)
P(3 without and 2 with) = 10p3q2 = 10(3/4)3(1/4)2 = 0,264
c.
This is an example of conditional probability. As the child in question has no moles (dominant
phenotype), his genotype cannot be aa, he must be either AA or Aa. As the parents are both
heterozygous, the expected offspring ratio is 1 AA : 2 Aa : 1aa. However if you only consider
those offspring with the dominant phenotype, the ratio is 1 AA : 2 Aa.
So the probability for the child without moles to be heterozygous (Aa) is ⅔
4
A4.
a.
Consider each locus separately, and convert numbers to ratios.
Leaves:
122 + 118 = 240 wavy; 38 + 41 = 79 smooth
240 : 79 → approximately 3:1 ratio
Typical ratio resulting from cross between 2 individuals both heterozygous for a
dominant trait.
Let leaf shape be controlled by the A locus, with alleles A and a.
A (wavy) > a (smooth)
AA = wavy
Aa = wavy
aa = smooth
Genotypes of parents for A-locus: Aa x Aa
Pollen:
122 + 28 = 160 round; 118 + 41 = 159 long
160 : 159 → approximately 1:1 ratio
Typical ratio where one parent is heterozygous for dominant trait, and the other
is homozygous recessive.
Let pollen shape be controlled by the B locus, with alleles B and b.
B (round) > b (long)
OR:
B (long) > b (round)
BB = round
BB = long
Bb = round
Bb = long
bb = long
bb = round
Genotypes of parents for B-locus: Bb (round) x bb (long) OR Bb (long) and bb (round) (there
is not enough information available to enable you to decide which allele is dominant).
b.
Parents:
c.
AaBb x Aabb
wavy round wavy long
OR
Aabb
x AaBb
wavy round wavy long
AaBb x Aabb
Ab
ab
AB
AABb
AaBb
Ab
AAbb
Aabb
aB
AaBb
aaBb
Ab
Aabb
aabb
GR:
1 AABb :
2 AaBb :
1 AAbb :
2 Aabb :
1 aaBb :
1 aabb
1 aabb
5
PR:
3 A_B_
3 wavy round
3 A_bb
3 wavy long
1 aaB_
1 aabb
1aabb
1 smooth round
1 smooth long
1 smooth long