Page 20 - Genetics_From_Genes_to_Genomes_6th_FULL_Part1

Guided Tour xix
114 Chapter 4 The Chromosome Theory of Inheritance

TABLE 4.4 How the Chromosome Theory of Inheritance Explains Mendel’s Laws
(a) The Law of Segregation (b) The Law of Independent Assortment
F 1 Homologous pair Homologous pair
392
for seed shape
F 1 for seed color Chapter 11 Analyzing Genomic Variation
Rr
(Y) Yellow Round (R)
(y) Green Wrinkled (r)
WHAT’S NEXT
r
R r
R
R
R r
r
In this chapter and in Chapters 9 and 10, we have focused different types of proteins that help package and manage the
Meiosis I R R r r on the nucleotide content of genomes, particularly the information carried by DNA. These
Anaphase Meiosis I Comparative Figures proteins have many
Anaphase 6 billion nucleotides organized into 46 chromosomes in roles. Certain proteins help compact the chromosomes to fit
Comparison illustrations lay out the basic
OR
each normal human diploid cell. In the next several chap- in the nucleus. Some proteins ensure that the chromosomal
ters, we examine features of the chromosomes that allow DNA is properly duplicated during each cell cycle, while
differences of often confusing principles.
y
y
y
Y y
Y
Y
Y
these DNA sequences to function properly and to be trans- others govern the distribution of chromosomes to daughter
mitted from one generation to the next. cells. Yet other proteins are responsible for regulating the
We begin by considering how in spite of the enormous availability of genes to the transcriptional machinery so that
Meiosis II
complexity of DNA sequences, the DNA actually constitutes the genes can be expressed into proteins. In Chapter 12, we
Meiosis II
only about one-third of the total mass of a chromosome. The examine how proteins interact with DNA to generate the
Possible remainder of the chromosome is made of thousands of functional complexity of a chromosome.
Possible gametes Yellow Green Green Yellow
gametes round wrinkled round wrinkled
(Y R) (y r) (y R) (Y r)
Round (R) Wrinkled (r) SOLVED PROBLEMS
I. Genomic DNA from a woman’s blood cells is PCR maternally derived chromosome and one on the paternally
amplified by a single pair of primers representing a derived chromosome), as long as the primer can hybridize to
unique locus in the genome. The PCR products are both homologs as is usually the case. The DNA sequence
F 2
F 2 then sequenced by the Sanger method, using one of the trace has two nucleotides at several positions. This fact indi-
R r Y R Y r y R y r
PCR primers as a sequencing primer. The following cates that the woman must be a heterozygote and that the
figure shows a trace of just part of the sequence read. PCR is amplifying both alleles of the locus.
R RR Rr Y R YY RR YY Rr Yy RR Yy Rr a. Notice that both alleles contain multiple repeats of
G T A C
the dinucleotide CA. The most likely explanation
Y r YY Rr YY rr Yy Rr Yy rr for the polymorphism is therefore that the locus
r Rr rr contains an SSR polymorphism whose alleles have
different numbers of CA repeats. One allele has
In an F 1 hybrid plant, the allele for round peas (R) is found on y R Yy RR Yy Rr yy RR yy Rr six repeats; the second allele must have more
one chromosome, and the allele for wrinkled peas (r) is on the CA units.
homologous chromosome. The pairing between the two
homologous chromosomes during prophase through y r Smaller Yy rr yy rR yy rr Larger b. Writing out the first 14 nucleotides of both alleles is
Yy Rr
metaphase of meiosis I makes sure that the homologs will straightforward. If the assumption in part (a) is cor-
a. What kind of polymorphism is most likely represented?
separate to opposite spindle poles during anaphase I. At the One pair of homologous chromosomes carries the gene for rect, then one allele should have more than six CA
end of meiosis II, two types of gametes have been produced: seed shape (alleles R and r). A second pair of homologous
b. With your answer to part (a) in mind, determine the
half have R, and half have r, but no gametes have both alleles. chromosomes carries the gene for seed color (alleles Y and y). repeats. The trace shows evidence for two additional
woman’s genotype at this locus. Indicate all nucleo-
Thus, the separation of homologous chromosomes at meiosis I Each homologous pair aligns at random at the metaphase plate CA repeats in one allele at positions 15–18, for a to-
corresponds to the segregation of alleles. As the Punnett tides that can be read from both alleles and their tal of eight CA repeats.
5′-to-3′ orientation.
square shows, fertilization of 50% R and 50% r eggs with the during meiosis I, independently of the other homologous pair. You can then determine the nucleotides beyond
Thus, two equally likely configurations are possible for the
same proportion of R and r sperm leads to Mendel’s 3:1 ratio in migration of any two chromosome pairs toward the poles
the F 2 generation. c. What kind of molecular event was likely to have gen- the repeats in the shorter allele by subtracting CACA
during anaphase I. As a result, a dihybrid individual will from positions 15–18. The remaining peaks at these
erated this polymorphism?
generate four equally likely types of gametes with regard to
d. How would you know exactly where in the genome
the two traits in question. The Punnett square affirms that positions correspond to ATGT. Note that ATGT can
this locus is found?
independent assortment of traits carried by nonhomologous also be found in the longer allele, but now at nucleo-
chromosomes produces Mendel’s 9:3:3:1 ratio. tides 19–22, just past the two additional CACA re-
e. What is another way in which you could analyze the
PCR products to genotype this locus? peats. You can determine the last four nucleotides in
f. Suppose you wanted to genotype this locus based on the shorter allele by subtracting ATGT from positions
single-molecule DNA sequencing of whole genomes 19–22, revealing TAGG. The sequences of the two
as shown in Fig. 9.24. Would a single read suffice for alleles of this SSR locus (indicating only one strand
genotyping the locus by this alternative method? of DNA each) are thus:
Allele 1: 5′...GGCACACACACACAATGTTAGG...3′
Answer Allele 2: 5′...GGCACACACACACACACAATGT...3′
To solve this problem, you need to understand that PCR will c. The mechanism thought to be responsible for most
simultaneously amplify both copies of a locus (one on the SSR polymorphisms is stuttering of DNA polymerase
har00909_ch04_089-132.indd 114 DNA: © Design Pics/Bilderbuch RF 12/05/17 6:17 PM during DNA replication.
har00909_ch11_365-405.indd 392 6/29/17 7:05 PM
Solving Genetics Problems
The best way for students to assess and increase their un- Review Problems
derstanding of genetics is to practice through problems. Problems are organized by chapter section and in order
of increasing difficulty to help students develop strong
Found at the end of each chapter, problem sets assist stu- problem-solving skills. The answers to select problems
dents in evaluating their grasp of key concepts and allow can be found in the back of this text.
them to apply what they have learned to real-life issues.
Solved Problems
Solved problems offer step-by-step guidance needed to
understand the problem-solving process.