Biology Questions and Answers
- What is the condition for
Mendel’s second law to be
valid?
Mendel’s second law is ogenes located in different chromnly valid forosomes. (^)
For genes situated in chromosome, i.e., linkthe sameed genes (genes (^)
in linkage) the law is not valid since the
segregation oindependent.f these genes is not
- According to Mendel’s
second law, in the crossing
between homozygous
individuals concerning two
pairs of nonlinked alleles,
AABB x aaBB, what are the
genotypical and phenotypical
proportions in F1 and F2?
Parental genotypes: AABB,Gametes from the parental generation: aaBB. (^)
Ab and aB. ThAaBb gametes (aus F1 will prnd the phenotypicalesent 100%
correspondent form).
As F1 arfrom their crossing can e AaBb individuals thbe: AB, Ab, e gametesaB
and ab. The casual combination ogametes forms the following genotypicalf these (^)
forms: one AABB, two AABb, two AaBb,
four AaBB, one Aabb, one Aabb, oneaaBB, two aaBb and two aabb. The
phenotypical proportion then would be:
nine AB (double dominant); Abb (dominant for the first pair,three
recessive fo(recessive for the second); thrr the first pair, dominantee aaB
for the second); one aabb (double
recessive).
- Considering independent
segregation of all factors, how
many types of gametes does a
VvXXWwYyzz individual
produce? What is the formula
to determinate such number?
The mentioned individual will produce
eight different typ(attention, gametes anes of gametesd not zygotes). (^)
To determine the number of different
gametes pgenotype the number of heterezygousroduced by a given multiple (^)
pairs is counted (in the mentioned case,three) and the result is placed as an
exponent of two (in the example, 2^3 =
8).
- How is it possible to
obtain the probability of
emergence of a given
genotype formed of more than
one pair of different alleles
with independent segregation
from the knowledge of the
parental genotypes?
Taking as eAaBbCc with aaBBCc,xample the crossing o for each f
considered pair of allele verify which genotypes it can fit is possible orm (asto
in an independent analysis) and in
which prBb x BB: BB, Bb (oportion. AA x aa: 1:1). Cc x Aa, aa (1:1).Cc: CC, Cc,
cc (1:2:1). The genotype to which the
probabilityexample aaBbcc. is to b For each pair of thise determined is for (^)
genotype the formation determined: to aa, 0.5; to Bb, probability is0.5; to (^)
cc, 0.25. The final result is obtained by
multiplicationprobabilities, 0.5 x 0.5 of these partialx, 0.5, r (^) esulting
0.0625.
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