Question

A tall pea plant with round, yellow seeds, heterozygous for all three traits, is allowed to self-fertilize. How many of the offspring would be expected to be tall with round, yellow seeds?

          A tall pea plant with round, yellow seeds, heterozygous for all three traits, is allowed to self-fertilize. How many of the offspring would be expected to be tall with round, yellow seeds?
        

Added by James H.

Biology for AP Courses
Biology for AP Courses
Julianne Zedalis, John Eggebrecht
AceChat toggle button
Close icon
Ace pointing down

Please give Ace some feedback

Your feedback will help us improve your experience

Thumb up icon Thumb down icon
Thanks for your feedback!
Profile picture
A tall pea plant with round, yellow seeds, heterozygous for all three traits, is allowed to self-fertilize. How many of the offspring would be expected to be tall with round, yellow seeds?
Close icon
Play audio
Feedback
Powered by NumerAI
Danielle Fairburn Kathleen Carty
Ivan Kochetkov verified

Keemin Lee and 81 other subject Biology educators are ready to help you.

Ask a new question

*

Labs

-

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

*

Key Concepts

-
Key Concept
Premium Feature
Explore the core concept behind this problem.
Play button
Key Concept
Premium Feature
Explore the core concept behind this problem.
Your browser does not support the video tag.

*

Recommended Videos

-
a-heterozygous-pea-plant-that-is-tall-with-yellow-seeds-ttgg-is-allowed-to-self-fertilize-what-is-the-probability-that-an-offspring-will-be-either-tall-with-yellow-seeds-tall-with-green-seed-67749

A heterozygous pea plant that is tall with yellow seeds, TtGg, is allowed to self-fertilize. What is the probability that an offspring will be either tall with yellow seeds, tall with green seeds, or dwarf with yellow seeds? Use the Sum Rule to calculate the probability asked in the problem.

Keemin L.

in-a-cross-between-homozygous-pure-dominant-yellow-seed-peas-and-green-seed-peas-what-proportion-of-the-offspring-would-be-expected-to-sow-the-green-phenotype-45504

In a cross between homozygous (pure) dominant yellow seed peas and green seed peas what proportion of the offspring would be expected to sow the green phenotype?

Bryan V.

in-peas-yellow-seed-color-is-dominant-a-over-green-seed-color-a-if-a-heterozygous-yellow-seed-pea-plant-is-crossed-with-a-green-seed-pea-plant-what-is-the-probability-that-their-offspring-wi-84076

In peas, yellow seed color is dominant (A) over green seed color (a). If a heterozygous yellow seed pea plant is crossed with a green seed pea plant: What is the probability that their offspring will be heterozygous?

Adi S.


*

Recommended Textbooks

-
Biology for AP Courses

Biology for AP Courses

Julianne Zedalis, John Eggebrecht
achievement 1,248 solutions
Objective Biology for NEET

Objective Biology for NEET

Rajiv Vijay 1st Edition
achievement 1,102 solutions
Introduction to General, Organic and Biochemistry

Introduction to General, Organic and Biochemistry

Frederick A. Bettelheim, William H. Brown, Mary K. Campbell 12th Edition
achievement 1,997 solutions

*

Transcript

-
00:01 In this question, we're told that the heterozygous pea plant that's tall with yellow seeds, the big t, little t, big, g, little g, is allowed to self -fertilized, so we cross it with the exact same.
00:12 And we want to know the probability that an offspring will be either tall with yellow or tall with green or dwarf with yellow.
00:21 So tall with yellow would be big t underscore and then little g, little g, little g.
00:27 Tall with green would be the same thing except big t underscore big g underscore and then finally dwarf with yellow seeds would be little t little t and then big or sorry and then little t little g or little g and it can be any of these and thus using the sum rule we're going to add up the individual probabilities for each of these and the way that we sort of find the probability for each of these is going to be by looking at the gametes one by one or the alleles one by one.
01:00 So for big t underscore, we just need a big t and knowing that it's a one to two to one ratio in a double heterozygous case, we would know that the big t, big t, big t, little t is two, and then little t is two, and then little t is one, which gives us a three and four probability of getting the big t allele.
01:16 Looking at the little g allele, it would just be the one in four...
Need help? Use Ace
Ace is your personal tutor. It breaks down any question with clear steps so you can learn.
Start Using Ace
Ace is your personal tutor for learning
Step-by-step explanations
Instant summaries
Summarize YouTube videos
Understand textbook images or PDFs
Study tools like quizzes and flashcards
Listen to your notes as a podcast
Continue solving this problem
Create a free account to:
  • View full step-by-step solution
  • Ask follow-up questions with Ace AI
  • Save progress and study later
Continue Free
Join the community

18,000,000+

Students on Numerade


Trusted by students at 8,000+ universities

Numerade

Get step-by-step video solution
from top educators

Continue with Clever
or



By creating an account, you agree to the Terms of Service and Privacy Policy
Already have an account? Log In

A free answer
just for you

Watch the video solution with this free unlock.

Numerade

Log in to watch this video
...and 100,000,000 more!


EMAIL

PASSWORD

OR
Continue with Clever