Download chapter14sganswersfall2008

Study Guide Chapter 14-Answers
Mendel and the Gene Idea
14.1 Mendel’s 2 laws of inheritance___________________________________________
1. Why did Mendel use pea plant as a model organism for his study of genetic
inheritance?
Peas are available in many varieties, and he could strictly control which plants mated
with which. Many of the pea plant traits (flower color, height, etc.) varied in an “eitheror” manner.
2. Mendel tracked only those characters of the pea plants that varied in an “either-or”
manner. What does this mean? For example, the pea plants he studied had purple OR
white flowers; there was no phenotype intermediate (light purple) between these two
varieties.
3A. Alleles: Alternative versions of genes are called alleles. For each character, an
organism inherits __2__(number) alleles for each trait, __1___(number) from each
parent.
3B. What is a dominant allele vs. a recessive allele? Give an example.
P vs p purple flower color is dominate to white flower color allele
3C. If an organism has one of each allele type (one dominant and one recessive) then the
__dominant__(dominant/recessive) allele determines organism’s appearance
4. Genotypes: Homozygous dominant, Homozygous recessive and Heterozygous
Provide at least one example of each genotype.
Homozygous dominant (Two dominant alleles) = AA
Heterozygous recessive (Two recessive alleles) = aa
Heterozygous (One dominant, One recessive allele)= Aa
5. Phenotype vs. genotype
Long fur (L) is dominant to short fur (l). Provide an example of the three possible
genotypes and the phenotype of each genotype.
Homozygous dominant Genotype = LL, Phenotype = Long fur
Heterozygous recessive Genotype = ll, Phenotype = short fur
Heterozygous Genotype –Ll, Phenotype – Long fur
6. Describe the Typical Mendel experiment. Use the following terms:
True-breeding
Parental (P) generation
Hybridization
Hybrids (Heterozygotes)
F1 generation
F2 generation
3:1 phenotype ratio
-Mendel always started with two homozygous (one homozygous dominant one
homozygous recessive) true-breeding organisms in the P generation.
-He crossed P generation true-breeders (hybridization) to get F1 generation, Hybrid
offspring.
-Mendel crossed plants from F1 to produce F2 generation offspring = 3:1 phenotype
offspring ratio observed
7. Construct punnett squares for the following crosses and determine the probability of
offspring genotypes and phenotypes.
A. Rr X Rr
Genotypes: ¼ RR, ½ Rr, ¼ rr
Phenotypes: RR= dominant, Rr = dominant, rr = recessive
¾ dominant, ¼ recessive
B. AA X Aa
Genotypes: ½ AA, ½ Aa
Phenotypes: AA dominant, Aa dominant
ALL dominant
C. Dd X dd
Genotypes: ½ Dd, ½ dd
Phenotypes: Dd dominant, dd recessive
½ dominant, ½ recessive
8. What is a testcross and when is this method utilized?
The breeding of a homozygous recessive organism (aa) with an organism of dominate
phenotype but unknown genotype (AA? or Aa?)
EX: Used when you don’t know the genotype of a dominant phenotype organism (AA or
Aa? = cross it with an organism of KNOWN genotype = homozygous recessive
organism, aa)
CROSS 1 Aa X aa
Offspring = ½ aa and ½ Aa
CROSS 2 AA X aa
Offspring = ALL Aa
So…
if you do the cross and get ½ dominant phenotype offspring and ½ recessive phenotype
offspring, then you know the dominant parent MUST have a Heterozygous genotype.
if you do the cross and get ALL dominant phenotype offspring, then you know the
dominant parent MUST BE Homozygous dominant.
9. Explain the difference between a monohybrid and a dihybrid.
Monohybrid = follow one character P F1 will be hybrids for one trait (have one
dominant allele, one recessive allele) (EX: Pp)
Dihybrid = follow two characters P and R F1 will have one dominant allele, one
recessive allele for each character (EX: PpRr)
14.2 Laws of probability and Mendelian inheritance______________________________
10. Multi-hybrid Cross Calculation
3 characters = trihybrid cross
Parent 1: Purple flowers (PP), Yellow (Yy), Wrinkled (rr)
Parent 2: Purple flowers (Pp), Green (yy), Round (Rr)
Parents: PPYyrr X PpyyRr
Question: What percentage of the offspring from this cross would be predicted to have
the following genotypes: PpyyRr and PPyyRr (phenotype: purple flowers and green
and round seeds)?
Step 1. Consider each character separately (make a punnett square for each character)
Parents: PPYyrr X PpyyRr:
PP X Pp = ½ PP, ½ Pp
Yy X yy = ½ Yy, ½ yy
rr X Rr = ½ Rr, ½ rr
Step 2. Calculate Genotype: Calculate probability for the genotype using the Rule of
Multiplication
State the Rule of multiplication:
Probability that two or more independent events will occur together.
PpyyRr = ½ x ½ x ½ =1/8 (rule of multiplication) chance of offspring having this
genotype
PPyyRr = ½ x ½ x ½ = 1/8 (rule of multiplication) chance of offspring having this
genotype
Additional Multiplication Rule Example
Rr Female X Rr male for a character:
Eggs = ½ will have the R allele and the other ½ will have the r allele
Sperm = ½ will have the R allele and the other ½ will have the r allele
-Chance of RR offspring from this cross = ½ chance of having R egg X ½ change
of R sperm fusing = multiplication rule = ¼ chance!
-Chance of rr offspring from this cross = ½ chance of having r egg X ½ change of
r sperm fusing = multiplication rule = ¼ chance!
-Chance of Rr offspring from this cross = ½ chance of having r egg X ½ change
of R sperm fusing = multiplication rule = ¼ chance!
-Chance of Rr offspring from this cross = ½ chance of having R egg X ½ change
of r sperm fusing = multiplication rule = ¼ chance
Step 3. Calculate Phenotype: Use the Rule of Addition to determine the probability of
offspring that have the genotype ppyyrr (phenotype: white flowers and green and
wrinkled seeds)?
State the rule of addition:
Probability that any one of two or more exclusive events will occur.
PpyyRr = ½ x ½ x ½ =1/8
PPyyRr = ½ x ½ x ½ = 1/8
Add them together to get the chance of having offspring with this PHENOTYPE
=2/8 =reduce to 1/4
Additional Addition Rule Example:
There are two ways to get a Rr offspring from the cross described in the answer for the
previous step. So we will use the addition rule to determine the fraction of Heterozygotes
expected from this cross. Chance Rr offspring from that cross = ¼ + ¼ =1/2 (addition
rule)
11. Pea plants heterozygous for pea shape and color (YyRr) are allowed to self-pollinate.
What fraction of the offspring would be predicted to have the genotype yyrr? green and
wrinkled phenotype.
(to calculate use the method for calculating offspring probability outlined in the previous
question)
Parents= YyRr X YyRr
Step 1. monohybrid cross for each character (construct a punnett square for each)
Yy X Yy = 1/4YY, 1/2 Yy,1/4 yy
Rr X Rr = 1/4Rr, 1/2 Rr, ¼ rr
Step 2.Multiplication rule:
yyrr= 1/4 X 1/4 = 1/16
1/16 of the offspring will be both green and wrinkled
Step 3: Does not apply to this question because there is only one possible offspring
genotype.
14.3 Inheritance patterns are more complex than predicted by Mendel________________
12. Describe how complete dominance differs from both co-dominance and incomplete
dominance. Provide and example of each.
Complete = Menelian characteristics where one allele is totally dominant over another
allele. Phenotypes of homozygous dominant and heterozygous individual are
indistinguishable (Ex: PP vs. Pp)
Codominance = when both alleles for a character are dominant
(Ex: M and N molecules on RBC. MM genotype = M molecules only, NN genotype = N
molecules only, NM genotype = BOTH N and M molecules on RBC)
Incomplete Dominance = one allele is not completely dominant over the other allele.
Resulting in a combined or mixed phenotype in the heterozygotes.
For example, if you cross-pollinate homozygous red and homozygous white carnation
flower plants, the dominant allele that produces the red color is not completely dominant
over the recessive allele that produces the white color. The resulting offspring are pink.
Cross true-breeding parents in P generation (RR red and WW white)
F1 = hybrid offspring (RW pink) because in these flowers the red allele only contributes
½ the amount of red pigment needed for a red flower. So to have a red flower, 2 Red
alleles are required.
F2 = take two F1 organisms (RW x RW) and cross = ¼ RR red, ¼ WW white, and ½ RW
pink offspring.
13. How many alleles determine human blood type? How many different blood type
phenotypes are possible?
3 alleles: A, B and O
4 phenotypes B, A, AB, O
6 genotypes BB, BO, AA, AO AB, OO
14. Can a person with type A blood receive blood from a person with type B blood?
WHY/WHY NOT?
No, B type blood has B carbohydrates molecules on its surface. A person with type A
blood will not recognize these B carbohydrates and their immune system will think they
are foreign invaders = blood clumps
15. A woman with type O blood has a child with type B blood. If the “father” has Type O
blood, could he be this child’s father? NO! If the mother has type O blood, her genotype
must be OO and she can only give her child an O allele. The same is true for a father with
type O blood, he can only give the child an O allele. The child would have to have O type
blood for this man to possibly be the father….so “He is NOT the father!”
If the “father” had type AB blood, could he have fathered this child?
Again, Mom give the child an O allele. This man can give the child an A allele OR a B
allele. If he gave the child the B allele, the child would have the BO genotype, giving
him type B blood. So it is possible that he could be the father of this child.
14.4 Human traits and Mendelian inheritance___________________________________
16. Construct a pedigree following the trait for free earlobe F (dominant) vs. attached
earlobe ff (recessive) for the following family:
Grandparents: Ed and Lucy
Children: Peggy, Sue, Dave, Martha
Peggy marries Alec
Sue marries Dan
Dave marries Lisa
Martha is single
Grandchildren:
Peggy and Alec have two boys: Jim and Bob
Sue and Dan have no children
Dave and Lisa have one girl: Betty
Ed, Peggy and Bob have attached earlobes. Everyone else has free earlobes. Determine
each person’s genotype for the earlobe trait.
Ed (ff) and Lucy (Ff, we know she MUST be heterozygous because she has a child
(Peggy) with attached earlobes!)
Peggy (ff) and Alec (Ff): Jim (Ff), Bob (ff)
We know Alec MUST be heterozygous because he has a child (Bob) with attached
earlobes!)
Sue (Ff) and Dan (FF or Ff)
We do not know if Dan is FF or Ff because he and Sue have no children.
Dave (Ff) and Lisa (FF or Ff): Betty (FF or Ff)
We do not know if Lisa is FF or Ff because their child also has the dominant phenotype
so her genotype is unknown.
Martha (Ff)