Chapter 5
Question #3:
A pure-breeding round,green (RRyy) pea is crossed with a pure-breeding
wrinkled, yellow (rrYY) pea. All of the F1 are round, yellow peas (RrYy).
These F1 were then “selfed” to observe the F2 generation.
Answer the following:
a.) What are the phenotypic and genotypic ratios of the progeny in the F 2? How
many genotypes are there??
Phenotypic:
RrYy X RrYy
3 Yellow
9 Round, Yellow
1 green
3 Round, green
3 Round
3 Yellow
3 wrinkled, Yellow
1 green
1 wrinkled, green
1 wrinkled
2 genes 4 phenotypes
Genotypic:
¼ RR
RrYy X RrYy
¼ YY
2/4 Yy
¼ yy
2/4 Rr
¼ rr
1/16 RRYY
2/16 RRYy
1/16 RRyy
¼ YY
2/16 RrYY
2/4 Yy
4/16 RrYy
¼ yy
2/16 Rryy
¼ YY
1/16 rrYY
2/4 Yy
2/16 rrYy
¼ yy
1/16 rryy
2 genes9 genotypes
b.) What is the genotypic ratio underlying the 9:3:3:1 phenotypic ratio??
Phenotypes:
Genotypes:
9 Round, Yellow
R_Y_the “underscore” means the
individual can be either homozygous or
heterozygous; either way, the
individual’s phenotype would not
change.
3 Round, green
R_yy
3 wrinkled, Yellow
rrYY
1 wrinkled, green
rryy
c.) Can you devise a simple formula for the calculation of the number of progeny
GENOTYPES in dihybrid, trihybrid, etc… crosses? Repeat for PHENOTYPES!
2 genes How many genotypes do we see?
Hint! Think about the number of genotypes for EACH GENE separately
Think about it… 1 gene3 possible genotypes
2 genes 9 possible genotypes (you just did this in Part A)
and if I told you that 3 genes 27 possible genotypes
See the pattern? 3n, where n= # of genes
2 genes How many phenotypes do we see?
Think about it… 1 gene2 possible phenotypes
2 genes 4 possible phenotypes (you just did this in Part A)
and if I told you that 3 genes 8 possible phenotypes
See the pattern? 2n, where n= # of genes
Question #7:
A female animal with genotype A/a▪B/b is crossed with a double-recessive male
a/a▪b/b. Their progeny include:
442 A/a▪B/b
458 a/a▪b/b
46 A/a▪b/b
54 a/a▪B/b
Explain these proportions and draw a map of the dihybrid parent showing
position of the genes and alleles.
1.) What genotypic ratio would you EXPECT to see if these genes were NOT
linked (assorting independently)?
If NOT linked, then AaBb X aabb (an example of a TESTCROSS) would
result in a ¼ AaBb: ¼ Aabb : ¼ aaBb : ¼ aabb genotypic ratio (1:1:1:1)
Look at the data given…Does it fit a 1:1:1:1 ratio? NOPE!
So, the deviation from the EXPECTED ratio could be due to the genes being
linked. The genotypes seen could be a result of the following:
A
a
B X a
b
a
b
b
GAMETES PRODUCED
A
B
a
b
A
b
a
B
a
A
a
a
a
A
a
b
B
b
b
b
b
b
a
a
B
b
Parental types, so should see most of the
progeny with these genotypes
Recombinant, so should see the LEAST number
of progeny with these genotypes
2.) Draw a map of the dihybrid parent:
Recombination Frequency (RF) = # of recombinants/ total
So, 46+54/1000 = 100/1000 = 10% 10 map units or 10cM
A
a
B
b
10 map units
Question #9:
A fruit fly of genotype B R/b r is testcrossed to b r/b r. In 84% of the meioses,
there are no chiasmata between the linked genes. In 16% of the meioses, there
is one chiasma between the genes. Is the proportion of the progeny that will be
B r/b r
First, what type of progeny is Br/br? A parental or a recombinant?? And how is
that genotype produced from the parental genotypes given??
B
R
b
r
B
r
b
R
b
B
b
b
b
B
b
r
b
b
R
r
R
r
r
r
r
r
PARENTAL
RECOMBINANT
So, Br/br is a recombinant!
Second, 84% of the meioses DO NOT HAVE CHIASMATAmeaning, no
recombination occurring.
16% of the meioses have ONE CHIASMA between the genesrecombination is
occurring in 16% of the meioses. Specifically, ONE chiasma is found. How
many chromatids are involved in a single chiasma?? Only TWO of the FOUR
chromatids are involved… So, 16% recombination—only 2 chromatids involved,
so ACTUAL frequency of recombination is 8%4% are Br/br and 4% are bR/br
a.) 50%?NOPE! Total recombinants should represent way less than 50%, so
one recombinant cannot represent 50%
b.) 4%?YEP!! 
c.) 84%?NOPE! This represents the parental types since 84% of the meioses
do not have chiasmata
d.) 25%?NOPE! 25% is reminiscent of a 1:1:1:1 ratio—This is a ratio that
represents genes assorting independently. These genes are LINKED!
e.) 16%?NOPE! 16% represents the total proportion of recombinants, and
Br/br is only 1 type of recombinant