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MatroZZZ [7]
2 years ago
8

If antenna length in Mendaliens is controlled by a single gene and long is dominant to short and two Mendaliens with Long antenn

a have some offspring with short antenna, what is the genotype of the Mendalien parents
Biology
1 answer:
uranmaximum [27]2 years ago
5 0

there are multiple possibilities but one of them is that both parents are heterozygous dominant (one capital and one lowercase letter)

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Round seeds (R) are dominant to wrinkled seeds (r), and yellow seeds (Y) are dominant to green seeds (y). A plant of unknown gen
hammer [34]

Answer:

<h2>RrYy and rryy </h2>

Explanation:

1. As given; Round seeds (R)  are dominant on wrinkled seeds (r),

Yellow seeds (Y) are dominant on green seeds (g).

In a testcross, between an unknown genotype and a homozygous recessive with wrinkled and green seeds (rryy).

offspring are:  

Round and yellow are  53; genotype (R_Y_)

round and green are  49;  genotype (R_yy)  

wrinkle and yellow are 44;  genotype (rrY_)

wrinkled and green are  51 ; genotype  (rryy)

Here, the genotype of parents of these offspring would be RrYy and rryy.

6 0
3 years ago
Higher Ultra-Processed Food Consumption Is Associated with Increased Risk of Incident Coronary Artery Disease in the Atheroscler
Eddi Din [679]

Yes, it is true that Higher Ultra-Processed Food Consumption Is Associated with Increased Risk of Incident Coronary Artery Disease in the Atherosclerosis Risk in Communities Study.

Objectives:

To test the hypothesis that higher intake of ultra-processed foods is associated with higher risk of coronary artery disease.

Methods:

A total of 13,548 adults aged 45-65 y from the Atherosclerosis Risk in Communities study were included in the analytic sample. Dietary intake data were collected through a 66-item FFQ.

Ultra-processed foods were defined using the NOVA classification, and the level of intake (servings/d) was calculated for each participant and divided into quartiles.

We used Cox proportional hazards models and restricted cubic splines to assess the association between quartiles of ultra-processed food intake and incident coronary artery disease.

Results:

There were 2006 incident coronary artery disease cases documented over a median follow-up of 27 y.

Incidence rates were higher in the highest quartile of ultra-processed food intake (70.8 per 10,000 person-y; 95% CI: 65.1, 77.1) compared with the lowest quartile (59.3 per 10,000 person-y; 95% CI: 54.1, 65.0).

Participants in the highest compared with lowest quartile of ultra-processed food intake had a 19% higher risk of coronary artery disease (HR: 1.19; 95% CI: 1.05, 1.35) after adjusting for sociodemographic factors and health behaviours. An approximately linear relation was observed between ultra-processed food intake and risk of coronary artery disease.

Conclusions:

Higher ultra-processed food intake was associated with a higher risk of coronary artery disease among middle-aged US adults. Further prospective studies are needed to confirm these findings and to investigate the mechanisms by which ultra-processed foods may affect health.

Learn more about ultra-processed foods is here : brainly.com/question/26511562

#SPJ4

3 0
2 years ago
Which of these questions is scientific?
erastova [34]

Answer:

Hi elmo!

Explanation:

ANd i think the answer is C!

hope this helps c:

8 0
3 years ago
Are the excretory system and urinary system the same
balandron [24]
The excretory and urinary are not the same because the excretory system is collective for all system performing the function of eliminating waste from the body like excess water,urea, carbon dioxide and lactic acid but the urinary system focuses on removal of excess water and urea from the body
6 0
3 years ago
In a certain triangle ABC , A equals 30°, B equals 50°, and the side AB equals 3 in. Use a protractor and a straightedge to cons
natali 33 [55]
If you were to compute this, Angle C should be equal to 100°.

In a triangle, all angles sum up to 180°.

In triangle ABC, where the m∠A =30° and m∠50°:

m
∠A + m∠B + m∠C = 180°
30° + 50 + m∠C = 180°
80° + m∠C = 180°
m∠C = 180° - 80°
m∠C = 100°

The triangle should look like the picture attached. 

8 0
4 years ago
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