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gladu [14]
2 years ago
10

Name: Click or tap here to enter text. Date: Click or tap here to enter text. Graded Assignment Lab Report Create a dichotomous

key that identifies the 10 leaves on the Common Leaves sheet. Look closely at those leaf samples and devise a dichotomous key that helps you identify them. Be sure that your dichotomous key contains only pairs of statements about a single characteristic. For example, a pair of statements might be: A.leaf margin smooth B.leaf margin toothed However, you should avoid pairs of statements that do not address the same characteristic. The following pair, for example, would not be very informative in your key: A.leaf margin smooth B.leaf type needle-like As you develop your key, test it out with the 10 leaves provided on the Common Leaves sheet. When you've developed a key that identifies all 10 leaves, type your statements, “go tos,” and identifications, following the format in the example below. The example is based on this lesson’s dichotomous key for birds. Statement Types/Dichotomous key for birds Identification Name/Number of Leaves Statement 2a The bird has a crest of feathers on the top of its head. go to statement or identify bird blue jay Statement 2b The bird has a smooth head. go to statement or identify bird 3 Once you have completed your dichotomous key, answer the two remaining questions. When you are finished, submit this assignment to your teacher by the due date for full credit. Total score: Click or tap here to enter text. of 26 points (Score for Question 1: Click or tap here to enter text. of 20 points) 1.Complete a dichotomous key for the 10 leaves on the Common Leaves sheet. The chart provided here allows for 11 pairs of statements. Depending on how you build your dichotomous key, you may or may not need all of them, or you may need to add some. Statement Types/Dichotomous key for leaves Identification Name/Number of Leaves Statement 1a Click or tap here to enter text. go to statement or identify leaf Click or tap here to enter text. Statement 1b Click or tap here to enter
Biology
1 answer:
oksian1 [2.3K]2 years ago
4 0

Answer:

How to Make a Dichotomous Key

Explanation:

Step 1: List down the characteristics. ...

Step 2: Organize the characteristics in order. ...

Step 3: Divide the specimens. ...

Step 4: Divide the specimen even further. ...

Step 5: Draw a dichotomous key diagram. ...

Step 6: Test it out.

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small group of 100 people decide to isolate themselves from the world and move to a small and remote deserted island. Out of thi
irina [24]

Answer:

Frequency of p = 0.684

Frequency of p = 0.316

Number of individuals with homozygous dominant (AA) = 47

Number of individuals with heterozygous (Aa)= 43

Number of individuals with homozygous recessive (aa) = 10

Explanation:

Out of 100 people, 10 have albino skin (aa)

So, the frequency of homozygous recessive individuals (q^{2}) is \frac{10}{100} = 0.1

Now, q will be

= \sqrt{q^{2} } = \sqrt{0.1} \\= 0.316

As per Hardy Weinberg's equation -

p + q = 1

Substituting the value of q in above equation, we get -

p + 0.316 = 1p = 1 -0.316\\p = 0.684

Now the frequency of homozygous dominant (AA) will be

p^{2} = 0.684^{2} \\= 0.467

Hence, out of 100 people 0.467 * 100 = 46.7 or 47 people are homozygous dominant (AA)

Like wise out of 100 people 0.1 * 100 = 10 people are homozygous recessive (aa)

As per As per Hardy Weinberg's equation-

p^{2} + q^{2} + 2pq = 1\\

Substituting the values in above equation, we get -

0.467 + 0.316 + 2pq = 1\\2pq = 1 -( 0.467+ 0.1)\\2pq = 0.433

So, out of 100 people 0.433 * 100 = 43.3 or 43 people are heterozygous (Aa)

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