Download (1) Free ear lobe is dominant.

TOPIC: BIOLOGY (Principles of Inheritance & Pedigree Analysis)
Prac Sheet - 2 KEY
1.
(1)
(2)
(3)
Free ear lobe is dominant.
Parents of family A are heterozygous for the character because family A has 2 children of
attached ear-lobes and 3 children of free ear lobes.
(i)
Squares indicate Male.
(ii)
Circles indicate Female.
2.
(i)
(ii)
It is autosomal dominant trait.
It is autosomal recessive trait.
3.
(1)
(2)
(3)
(4)
The affected phenotype is determined by a recessive allele.
Inheritance in 4 generations has been shown.
After mating in I generation, 4 offspring are produced, i.e. 2 females and 2 males.
III generation shows the marriage between close relatives and thus recessive allele shows
phenotype in IV generation.
4.
(i)
Yes
5.
As red hair appears in 50% members of the second generation and not all, it cannot be due to
homozygous dominant allele. The same is the condition of dark hair. Therefore, out of the two
parents, one is homozygous recessive while the other is heterozygous (Aa×aa = 50%Aa + 50%aa).
The children are either heterozygous dominant (50%) or homozygous recessive (50%) only two of
the four children marry. One of them is dark haired while the other is red haired. Both marry dark
haired partners. The red dark haired couple produces both red haired and dark haired children while
the dark haired one give rise to dark haired children .The former is possible when one of two partners
is heterozygous dominant while the other is homozygous recessive. The second is possible if dark
haired husband is homozygous. In the above pedigree both possibilities of red hair being recessive
trait as well as dominant (heterozygous here) are possible. Therefore, the data is insufficient to
decide the issue
6.
Since the shaded symbol appears in all offspring,
father must be homozygous dominant (AA × aa = all Aa) while the mother homozygous recessive,
because in all other cases this possibility is absent
(Aa × aa = 2Aa + 2aa,
aa × AA = all Aa,
aa × Aa = 2aA + 2aa )
(ii)
No
(iii)
Yes
(iv)
Yes
(v)
No
All the members of II generation will, therefore be heterozygous (Aa) This is further confirmed by
marriage of II-1 with homozygous recessive (Aa × aa = 2Aa, 2aa) and bearing children of both the
parental types. Marriage of II-3 with the homozygous recessive can produce both recessive and
heterozygotes as are present here
7.
(a)
(b)
8.
Males are shown by squares and females by circles.
(1)
Both father and mother are tongue rollers (solid symbols represent the expressed character).
(2)
Of the 3 children born, two can roll (solid symbols) and one cannot (hollow symbol)
(3)
The recessive trait (rr) of non rolling in one of the children could have come from nowhere
else but the parents
(4)
The non-roller child in the family chart with genotype rr must have received one ‘r’ gene
from one parent and the other ‘r’ gene from the other parent.
Conclusion is that each of the two parents is heterozygous (Rr).
9.
When normal mother is there and father has polydactyl condition then 50% of children has
polydactyl condition and 50% of children has normal fingers.
10.
(a)
(b)
(c)
11.
(2)
12.
Pedigree analysis in the question is showing dominant X linked genes.
Dominant X linked genes can be detected in human pedigrees through the following clues:
(i)
It is more frequently found in female than in the male of the species.
(ii)
The affected males pass the condition on to all of their daughters but to none of their
sons.
(iii) Females usually pass the condition (defective phonotype) on the one half to their sons
and daughters.
(iv)
A X- linked dominant gene fails to be transmitted to any son from a mother which did
not exhibit the trait itself.
Father has colour blindness.
3 daughters and 2 sons have been born in the family.
1 child has colour blindness out of 5.