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Genetics quiz notes 01, Study notes of Biology

North Carolina State University (NCSU)Biology
Prof. Parks

Genetics quiz notes to help out with punnet squares and prepare for exam #3

Typology: Study notes

2025/2026

Uploaded on 04/08/2026

lain-shaker
lain-shaker 🇺🇸

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BIO 183 Genetics Problem Assignment Name(s): Laian Shaker
Spring2026 25 Points
Show your work.
1. The lubber grasshopper is a very large grasshopper and is black with red
and yellow stripes. Assume that red stripes are expressed from the
homozygous RR genotype, yellow stripes from the
homozygous rr genotype, and both from the heterozygous genotype. (A type
of incomplete dominance) I am going to place the gametes from one
parent along the top and the gametes from the other parent along
the side & going to combine the alleles from the corresponding row
and column to determine the genotypes of each offspring
A. If two grasshoppers were crossed 1 with red stripes and 1 with both
color stripes, what would be the genotypic ratio of the F1 generation
(RR:Rr:rr)? Fill in the Punnett square. After counting the occurrence
of each genotype in the Punnett square. There are two RR
genotypes and two Rr genotypes. The ratio of RR:Rr:rr is 2:2:0.
This simplifies to 1:1:0. The genotypic ratio of the F1 generation
is 1 RR: 1Rr: 0:rr.
R R
R
r
B. What genotypes would be produced by crossing a grasshopper with
both color stripes and one with yellow stripes? Fill in the Punnett
square. The grasshopper with both colors has the genotypes Rr. It can produce
gametes with the R allele and gametes with the r allele. The grasshopper with
yellow stripes has the genotype rr. It can only produce gametes with the r allele. I
am going to place the gametes from one parent along the top and the gametes
from the other parent along the side, then I’m going to combine the occurrence
of each genotype in the Punnet square. There are two Rr genotypes and two rr
genotypes. After counting the occurrences of each genotype in the Punnett
square. There are two Rr genotypes and two rr genotypes. The genotypes
produced are Rr and rr. The genotypes produced are Rr and rr.
R r
RR RR
Rr Rr
pf3
pf4
pf5

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BIO 183 Genetics Problem Assignment Name(s): Laian Shaker Spring2026 25 Points Show your work.

1. The lubber grasshopper is a very large grasshopper and is black with red and yellow stripes. Assume that red stripes are expressed from the homozygous RR genotype, yellow stripes from the homozygous rr genotype, and both from the heterozygous genotype. (A type of incomplete dominance) I am going to place the gametes from one parent along the top and the gametes from the other parent along the side & going to combine the alleles from the corresponding row and column to determine the genotypes of each offspring A. If two grasshoppers were crossed 1 with red stripes and 1 with both color stripes, what would be the genotypic ratio of the F1 generation (RR:Rr:rr)? Fill in the Punnett square. After counting the occurrence of each genotype in the Punnett square. There are two RR genotypes and two Rr genotypes. The ratio of RR:Rr:rr is 2:2:0. This simplifies to 1:1:0. The genotypic ratio of the F1 generation is 1 RR: 1Rr: 0:rr. R R R r B. What genotypes would be produced by crossing a grasshopper with both color stripes and one with yellow stripes? Fill in the Punnett square. The grasshopper with both colors has the genotypes Rr. It can produce gametes with the R allele and gametes with the r allele. The grasshopper with yellow stripes has the genotype rr. It can only produce gametes with the r allele. I am going to place the gametes from one parent along the top and the gametes from the other parent along the side, then I’m going to combine the occurrence of each genotype in the Punnet square. There are two Rr genotypes and two rr genotypes. After counting the occurrences of each genotype in the Punnett square. There are two Rr genotypes and two rr genotypes. The genotypes produced are Rr and rr. The genotypes produced are Rr and rr. R r

RR RR

Rr Rr

r r

2. In celery, the leaves can be curly or flat. Curly is dominant and flat is recessive. The other trait is spots on the leaves. Dark green spots are dominant over white spots. Cross a celery plant that is heterozygous for both traits with a plant that has flat leaves and dark green spots (one parent of this plant had white spots). A. Parent genotypes? Let C represent the allele for curly leaves and c represent the allele for flat leaves. Let S represent the allele for dark green spots and s represent the allele for white spots. The first parent is heterozygous for both traits, so its genotype is CcSs. The second parent has flat leaves, which means its genotype for leaf shape is cc. This parent also has dark green spots. We are told that one parent of this plant had white spots. This means the parent with dark green spots must be heterozygous for the spot trait to have produced offspring with white spots (recessive trait). Therefore the genotype for the second parent is ccSc. The parent genotypes are CcSs and ccSs B. Fill in Punnett square CS Cs cS cs cS cs C. What is the ratio of genotypes? From the Punnett square, we can count the occurrences of each genotype: CcSS: 1 Rr rr Rr rr CcSS CcSs ccSS ccSs CcSs CCss ccSs ccss

● Punnet Square: A Punnett square shows the following probabilities for offspring genotypes: ○ DD (healthy): ¼ ○ Dd (healthy): ¼ ○ dd(affected): ¼ ● Phenotypes: ○ Healthy: DD and Dd (total probability of ¼ + ¼ =½ ) ○ Affected: dd(probability of ¼) ● Probability of one healthy and one affected offspring ○ The probability of the first offspring being healthy and the second being affected is (½) * (¼) =(⅛) ● The probability of the first offspring being affected and the second being healthy is (¼) * (½)=⅛ ● Since the order does not matter, we add these probabilities: ⅛ +⅛ = 2/8=¼ ● Answer=¼

4. A normal sighted XX parent (whose XY parent was color-blind) is crossed with an XY parent who is colorblind. (Color-blindness is located on the X chromosome) X^c X (mother) X X^c Y (father) A. What is the probability that they will have an offspring that is color-blind? (x^c X^c or X^c Y) which is 2 out of 4 possible outcomes, which is ½. B. If they have three offspring, what is the probability that they will have 2 color-blind offspring and 1 that is able to see colors (in any order)? Probabilities for a single child: ● Probability of being color-blind (P(C))=½ ● Probability of having normal vision (P(N))=½ ● Binomial Probability: we need to find the probability of having exactly 2 colors-blind offspring and 1 normal offspring in any order. This can be calculated using the binomial probability formula: P (x successes in n trials)= (n/x) p^x (1-p)^n-x ● Where ○ N+number of trials (offspring)= ○ x=number of successes (color-blind offspring)= ○ p=probability of success (being color-blind)=½ ○ 1-p= probability of failure (having normal vision)=½

● Calculations: ● Number of ways to have 2 color-blind and 1 normal off spring ○ (3/2)= 3!/2!(3-2)! + 3!/2!1!= ○ The possible orders are C C N, C N C, N C C ○ Probability of each specific order (e.g., C C N): ○ (½)^2 x (½)^1 =¼ x ½ = ⅛ ○ Total probability+number of orders x probability of one order = 3

x ⅛= ⅜

5. A wild-type fruit fly (heterozygous for gray body color and normal wings) is mated with a black fly with vestigial wings. The offspring have the following phenotypic distribution: wild type = 700, black vestigial = 650, black normal = 320, and gray vestigial = 330. What is the recombination frequency between these genes for body color and wing type? ● Parental Genotypes: ○ Wild-type fruit fly is heterozygous for gray body color ( denoting gray as G and black as g) and normal wings (let’s denote normal as N and vestigial as n) So, the genotypes is GgNn ○ The other parent is black with vestigial wings, so its genotypes is ggnn ● Offspring Phenotypes and Counts ○ Wild type (gray, normal wings): 700 ○ Black vestigial: ○ Black normal: ○ Gray vestigial: 330 ● Recombination Frequency: ○ Recombination frequency is calculated as the number of recombinant offspring divided by the total number of offspring, multiplied by 100%. Recombinant offspring are those that have a different combination of traits than the parent ● Identifying Parental and Recombinant offspringL ○ The parents’ phenotypes are gray normal (from the heterozygous parent) and black vestigial (from the