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Tatiana [17]
2 years ago
8

In a certain kind of plant, yellow leaves (Y) are dominant and red leaves (y) are recessive. What percentage of the offspring fr

om a cross between a Yy
plant and a yy plant are predicted to have red leaves?
(Hint: Draw a Punnett
square to help you determine the answer.)

A. 50%
B. 0%
C. 25%
D. 75%


Think is A but not sure prove me wrong

Biology
1 answer:
babunello [35]2 years ago
6 0

Answer:

Its A 50%

Explanation:

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sprouted seeds un sprouted seed taken in a jar with lime water in both in which jar lime water changes it colour fast​
nlexa [21]

Answer:

Correct option is White

Explanation:

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4 0
3 years ago
How does recombination contribute to offspring diversity?A. It modifies chromosomes to generate new alleles of genes that code f
omeli [17]

Answer: D. It yields chromosomes that contain both maternal and paternal DNA.

Explanation:

The recombination or crossing over is a genetic process. It takes place during the Prophase I of meiosis. The homologous pair align and held together by synapsis. This results in the mixture of genes combination formed from both parents that can be represented as a single chromosome.

If the recombination do not occur then each chromosome will be either 100% maternal or paternal. Thus recombination assures the recombinant chromosomes which consists of genetic material from both parents. Thus the genetic diversity of offspring increases.

6 0
2 years ago
Bacteria grow in a culture at a rate proportional to the amount present. Initially, 300 strands of the bacteria are in the cultu
OLEGan [10]

Answer:

(a) number of strands (n) = time (t) ÷ proportionality constant (k)

(b) The time needed for the bacterial to double its initial size is 3.36 hours.

Explanation:

(a) Let the rate (time) be represented by t and the amount (number) of strands of bacteria be represented by n

t is proportional to n, therefore, t = kn (k is the proportionality constant)

Since t = kn, then, n = t/k

(b) Initial amount of strands = 300

Amount of strands after 2 hours = 300 + (300 × 20/100) = 300 + 60 = 360

k = t/n = 2/360 = 0.0056 hour/strand

Double of the initial size is 600 (300×2 = 600)

Time (t) needed for the bacterial to double its initial size = kn = 0.0056×600 = 3.36 hours

8 0
3 years ago
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