Uploaded by fronterase8

G3-LAB-ACTIVITY PSYCH3A

advertisement
G3 | PSYCH 3A | 31/1/23
Members:
Codera
Colorado
Cosare
Dela Torre
Harrid
Fronteras
Fernandez
Mendelian Inheritance: Punnett Square & Introduction to Probability
1. Use two 5 peso coins. One coin is equivalent to a pair of genes from a single parent. (ex.
Maternal). The other coin represents the same set of genes from the other parent. ( ex.
paternal)
a. P=head= dominant gene (purple)
b. Recessive gene = p=tail (white)
2. There is heterozygosity in both parents (Pp)
a. Pp is the genetic combination of a single parent
b. Pp is another parent’s genes.
3. For the child of Pp X Pp, fill out the Punnett Square below. Keep a record of the
outcomes in the Expected probability column of table 2.
P
p
P
PP
Pp
p
Pp
pp
4. In our coin-toss experiment, the P allele will be represented by heads, and the p allele by
tails. 50 times total, toss the coins together; note the toss results in the coin toss tally column.
5. Determine the percentage for each genotype (ex.# of PP tosses/50) and record under actual
probability on table 1.
PP = 6x2 = 12%
Pp = 31x2 = 62%
pp= 13x2 = 26%
= 100%
Exercise 2
Mendelian Inheritance Punnett Square and Introduction to Probability
WORKSHEET
Table 1. Genotypic ratios for eye shape
Trial
Number
EE
1
/
2
Ee
ee
/
3
/
4
/
5
/
6
/
7
/
8
/
9
/
10
/
11
12
/
/
13
/
14
/
15
/
16
/
17
/
18
/
19
/
20
/
total
5
9
6
% of each
genotype
25%
45%
30%
Table 2: Data for coin toss
Coin toss tally
Expected
probabilit
y
Actual
Probability
PP (heads, heads)
25%
12%
Pp (heads, tails)
50%
31%
pp (tails, tails)
25%
26%
total
100%
100%
Practice Problems on Mendelian Genetics
1. The gene for normal skin pigmentation is dominant to the allele for albinism in
humans. What is the probability that a child born to two heterozygotes will have a
skin pigment that is normal? What are the odds of having an albino child?
a. What are the odds that the child is a carrier if they are healthy?
A
a
A
AA
Aa
a
Aa
aa
Normal pigment: 75% chance of AA & Aa
Albino: 25% chance of aa
The odds of having an albino child: 67%
2. In an unspecified animal species, one-horn is dominant and no horns are
recessive. Show the cross of an individual heterozygous for horns with another
individual that does not have horns? Summarize the genotypes & phenotypes of
the possible offspring?
H
h
h
Hh
hh
h
Hh
hh
50% chance of Hh (horned)
50% chance of hh (no horn)
3. In humans, the brown-eye (B) allele is dominant to the blue-eye allele (b). If two
heterozygotes mate, what will be the likely genotype and phenotype ratios of the
offspring?
B
b
B
BB
Bb
b
Bb
bb
75% chance of brown-eyed
25% chance of blue-eyed
25% chance of BB
50% chance of Bb
25% chance of bb
4. In seals, the gene for the length of the whiskers has two alleles. The dominant
allele (W) codes long whiskers & the recessive allele (w) codes for short whiskers.
What percentage of offspring would be expected to have short whiskers from the
cross of two long-whiskered seals, one that is homozygous dominant and one that
is heterozygous?
W
w
W
WW
Ww
W
WW
Ww
0% chance of short whiskers
POST LAB QUESTIONS
1. Based on the nips and coin toss activity, why is it crucial to carry out a lot of trials
in an experiment?
● Based on the nips and toss activity, it is essential to run numerous trials in an experiment
to increase the reliability of the results. Your average will get more closely to the true
value as you perform more trials. As a result, the more trials we run, the more accurate
our findings estimate will be. We can ensure that the results are consistent and
unaffected by chance when we do the same experiment more than once. The
experimental probability approaches the theoretical probability more closely as the
number of trials increases.
2. Take a look at your punnett square cross genotypic ratio. These are the expected
results. Examine these expected results against the actual percentages from your
data table. Are they the same? Why or why not?
● Due to the numerous other allele combinations that can be made. The expected results
against the actual percentages from the data table are not the same.
● They are not the same because there are many more combinations of alleles are
possible. As when you flip a coin, there is a chance that the results get close to the
expected values.
Documentation while doing the lab activity:
Download