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Genetics 2026 worksheet 1 key, Exams of Genetics

The University of Texas at AustinGenetics
Prof. Martin Poenie

Genetics 2026 worksheet 1 key Martin Poeine

Typology: Exams

2025/2026

Uploaded on 05/02/2026

kymo-matcha
kymo-matcha 🇺🇸

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Worksheet 1
1. For an organism that is AaBb, A. show (draw) how their chromosomes will look at Meiosis 1 B. Show
what gametes they will make and in what proportions.
B. ¼ AB, ¼ Ab, aB ¼ , ¼ ab
2. Regarding Humans
A. Humans have how many chromosomes? __46____
B. Humans have how many autosomes? _44______
C. How many autosomes are present in a sperm cell? _22___
D. How many chromosomes are present in a sperm cell? __23__
E. What is the diploid number of chromosomes? __46____
F. What is the haploid number of chromosomes? __23___
3. Consider three independently assorting gene pairs Aa, Bb, and Cc.' What is the probability (give a
fraction) of obtaining each of the following genotypes from parents that are AaBbCC and AaBBCc?'
AABbCC ___1/4 x ½ x 1/2___
aaBBCc __1/4 x ½ x 1/2____
aabbCC ___0___
1
A
A
a
a
B
B
b
b
O
R
A
A
a
a
B
B
b
b
pf3
pf4
pf5
pf8
pf9
pfa

Partial preview of the text

Download Genetics 2026 worksheet 1 key and more Exams Genetics in PDF only on Docsity!

  1. For an organism that is AaBb, A. show (draw) how their chromosomes will look at Meiosis 1 B. Show what gametes they will make and in what proportions. B. ¼ AB, ¼ Ab, aB ¼ , ¼ ab
  2. Regarding Humans A. Humans have how many chromosomes? _46____ B. Humans have how many autosomes? 44______ C. How many autosomes are present in a sperm cell? 22 D. How many chromosomes are present in a sperm cell? 23 E. What is the diploid number of chromosomes? _46____ F. What is the haploid number of chromosomes? 23
  3. Consider three independently assorting gene pairs Aa, Bb, and Cc.' What is the probability (give a fraction) of obtaining each of the following genotypes from parents that are AaBbCC and AaBBCc?' AABbCC 1/4 x ½ x 1/2 aaBBCc __1/4 x ½ x 1/2____ aabbCC 0 A A a a B B b

b O

R

A A a a B B b b

  1. You have grown up 20 plants from 20 yellow seeds. Assume that Y_ is yellow and yy is green. You do not know if those yellow seeds were homozygous or heterozygous. You want to identify the homozygous plants. What can you do to figure this out? You do a test cross where you cross the plants grown from yellow seeds with true-breeding plants that have green seeds What will be the result if the original plant was homozygous? Yellow seeds What will be the result if the original plant was heterozygous? Yellow and Green seeds
  2. If you cross heterozygotes (Yy x Yy) where Y_ are yellow seeds and yy are green, what fraction of the offspring will be will have yellow seeds, what fraction will have green seeds? If you cross heterozygotes (Rr x Rr) where R are round seeds and rr are wrinkled , what fraction of the offspring will have round seeds, what fraction will have wrinkled seeds. If you now set up a Punnet square with a cross between these two using the numbers you obtained, what will be the results? Yellow 3/4 Green 1/ Round 3/4 9/16 yellow round 3/16 Round green Wrinkled 1/4 3/16 yellow wrinkled 1/16 green wrinkled
  3. In some genetically engineered corn plants the dominant gene (BT) produces a protein that is lethal to certain flying insect pests that eat the corn plants. It was also found that the pollen could cause death in some flying insects. If the corn plant is heterozygous for BT, what proportion of the pollen would carry the dominant gene? A. all pollen B. 1/ C. 1/ D. 1/ E. 1/
  4. In a monohybrid cross AA  aa, what proportion of homozygotes is expected among the F 2 offspring? A. 1/ B. 1/ C. 3/ D. All are homozygotes. E. None are homozygotes.
  1. Albinism is caused by an autosomal recessive allele that interferes with skin pigment in mammals. Two normally pigmented parents have an albino girl. What is the probability that three of their next five children will be normally pigmented? One can think about this in terms of having two parts. Part one is how many ways out of 5 children can you get 3 of one and 2 of the other. Part two is what is the probability of getting 3 normal and two albino. The answer will be the number of ways you can get 3 normal and two albino times the probability of getting 3 normal and two albino. The first part of the problem involves determining the number of ways a certain outcome can come about. The total number of children is 5, the question is how many ways in 5 children can you get 3 normally pigmented. We will solve this problem using factorials. The formula for this is n! ------------ ( pRR^ x pSS) in the formula n!/R! x S! calculates the number of ways you can R! x S! 3 normal and 2 albino out of 5 children (pRR^ x pSS^ )calculates the probability of getting 3 normal and 2 albino regardless of order. Here, n = the total number of children (5) so n! = 5! R = the number of normal children (3) so R! = 3!. S = the number of remaining children (ie n– R) so S = 2 and S! = 2! pRR^ Is the probability of the R event raised to the R So, let’s say for albinism A_ is normal aa is albino) what is the probability of a normal child from a cross of heterozygotes Hopefully you came up with ¾ so pR is ¾ and since R is 3 pRR^ = ¾ x ¾ x ¾ or (3/4)^3. Next S refers to the rest of the cases so n-R = S = 2. The probability of an albino is? Hopefully you came up with ¼ so pSS^ = (1/4)^2 5! --------- ((3/4)^3 x (1/2)^2 ) 3! X 2! 5x4x3! 5x4 5x ---------- = ----- = ----- =10 x ((3/4)^3 x (1/4)^2 ) = 10 x (27/64 x 1/16 ) = 10 x (27 /1024) 3! X 2 2 1 = 10 x .0265 =.

The Essence So this problem boils down to the probability of 1 outcome times the number of outcomes of that type. The number of outcomes is given by the factorial part of the problem. The probability of single individual being normal ¾ (AA 2Aa) so probability of 3 being normal is ¾ x ¾ x ¾ = 27/64. The probability of 2 being albino is ¼ x ¼ = 1/16 = 4/64 so the probability of getting this outcome (3 normal, 2 albino) is 27/64 x 1/ Convert to decimal 27/64 x 1/16 = .42 x .0625 = .02625 There are 10 of ways of getting that outcome so .02625 x 10 = .2625 = 26%

13 Part B This is the second part of the problem. Now we will consider what happens if 1/6000 people carry the d allele. If 1/6000 people have the allele, then 5999/6000 don’t have the allele (they are DD). Therefor 5999/6000 are DD and therefore the chance that II-2 is heterozygous is ½. But the 1/6000 is Dd which would make the odds of II-2 being heterozygous 2/3. So with this, we can calculate the new probability that II-2 is heterozygous. It is 5999/6000 x ½ + 1/6000 x 2/3 = 18,001 / 36000 Note if you understand and follow this so far, there is a simpler way. The effect of having the recessive allele in the population is to move from II-2 having a probability of carrying the recessive allele from ½ towards a probability of 2/3. The difference between ½ and 2/3 is 1/6. So one can multiply 1/6000 x 1/6 and add that to the ½. So 1/6 x 1/6000 = 1/36000 + 18000/ 36000 = 18,001 / 36000 In either case, now one can finish answering the question of what is the chance that III-1 has the disease. If we new that I-2 was DD the answer would have been ½ x 2/3 x 1/4. Now, given the presence of d in the population it would be 18001/36000 x 2/3 x ¼

  1. The common grandfather of two first cousins has hereditary hemochromatosis, a recessive condition causing an abnormal buildup of iron in the body. Neither of the cousins has the disease, nor do any of their relatives. A. If the first cousins had a child, what is the chance that the child would have hemochromatosis? (Assume that the unrelated, unaffected parents of the cousins are not carriers.) B. How would your calculation change if you knew that one out of every 10 unaffected people in the population (including the unrelated parents of these cousins) was a carrier for hemochromatosis? For the calculation with 1 out of 10 being a carrier. As previously explained, one way to do this problem is calculate the probability as follows: (9/10 x ½ ) + (1/10 x 2/3) = 9/20 + 2/30 = 27/60 + 4/60 = 31/ The other way of doing the calculation is based on understanding that having the disease allele in the population will increase the probability of II-1 and II-4 being heterozygous which would in turn increase the probability of III-1 and III-2 being a carrier from ½ towards 2/3. So the calculation will be ½ + some additional amount based on the number of carriers in the population. The difference between ½ and 2/3 is 1/6. The amount of increase in the probability that III-1 and III-2 will be carriers is 1/10 x 1/6. So the new probability is ½ + (1/10 x 1/6) = 30/60 + 1/60 = 31/ Either way you calculate it, the probability that IV will have the disease is 31/60 x 31/60 x ¼ 1/ ½ or 31/ ½ or 31/

There is a shortcut as long as you understand what is going on. You can look and say the chance of having the D allele is 1/3 of ½ compared to ½ This is 1/6 to 3/6 which is a 1:3 ratio which again = 1/