Essential Cell Biology 5th edition

Answers A:53
chromosome during a meiosis, each chromosome
inherited by a grandchild will have been subjected
to about five crossovers since it left the grandfather,
dividing it into six segments. An identical pattern
of SNPs should surround whatever gene causes the
deafness in each of the four affected grandchildren;
moreover, this SNP pattern should be clearly different
from that surrounding the same gene in each of the
seven grandchildren who are normal. These SNPs
would form an unusually long haplotype block—one
that extends for about one-sixth of the length of
Chromosome 12. (One-quarter of the DNA of each
grandchild will have been inherited from the grandfather,
in roughly 70 segments of this length scattered among
the grandchild’s 46 chromosomes.)
ANSWER 19–13 Individual 1 might be either heterozygous
(+/–) or homozygous for the normal allele (+/+). Individual 2
must be homozygous for the recessive deafness allele (–/–).
(Both his parents must have been heterozygous because
they produced a deaf son.) Individual 3 is almost certainly
heterozygous (+/–) and responsible for transmitting the
mutant allele to his children and grandchildren. Given
that the mutant allele is rare, individual 4 is most probably
homozygous for the normal allele (+/+).
ANSWER 19–14 Your friend is wrong.
A. Mendel’s laws, and the clear understanding that we now
have concerning the mechanisms that produce them,
rule out many false ideas concerning human heredity.
One of them is that a first-born child has a different
chance of inheriting particular traits from its parents than
its siblings.
B. The probability of this type of pedigree arising by
chance is one-fourth for each generation, or one in 64
for the three generations shown.
C. Data from an enlarged sampling of family members, or
from more generations, would quickly reveal that the
regular pattern observed in this particular pedigree
arose by chance.
WHITE-EYED FLY
White
gene
Ruby
gene
D. If statistical tests showed that the pattern was not
due to chance, it would suggest that some process of
selection was involved: for example, parents who had
had a first child that was affected might regularly opt
for screening of subsequent pregnancies and selectively
terminate those pregnancies in which the fetus was
found to be affected. Fewer second children would then
be born with the abnormality.
ANSWER 19–15 Each carrier is a heterozygote, and 50%
of his sperm or her eggs will carry the lethal allele. When
two carriers marry, there is therefore a 25% chance that any
baby will inherit the lethal allele from both parents and so
will show the fatal phenotype. Because one person in 100
is a carrier, one partnership in 10,000 (100 × 100) will be a
partnership of carriers (assuming that people choose their
partners at random). Other things being equal, one baby in
40,000 will then be born with the defect, or 25 babies per
year out of a total of a million babies born.
ANSWER 19–16 A dominant-negative mutation gives rise
to a mutant gene product that interferes with the function
of the normal gene product, causing a loss-of-function
phenotype even in the presence of a normal copy of the
gene. For example, if a protein forms a hexamer, and
the mutant protein can interact with the normal subunits
and inhibit the function of the hexamer, the mutation will
be dominant. This ability of a single defective allele to
determine the phenotype is the reason why such an allele
is dominant. A gain-of-function mutation increases the
activity of the gene or makes it active in inappropriate
circumstances. The change in activity often has a phenotypic
consequence, which is why such mutations are usually
dominant.
ANSWER 19–17
A. As outlined in Figure A19–17, if flies that are defective
in different genes mate, their progeny will have one
normal gene. In the case of a mating between a ruby-
White
gene
RUBY-EYED FLY
Ruby
gene
defective
functional
functional
defective
defective
functional
functional
defective
Ruby
product
White
product
MATE
ALL PROGENY ARE RED-EYED
from white parent
from ruby parent
White
gene
defective
functional
Ruby
gene
functional
defective
Figure A19–17
White
product
Ruby
product